Solutions to Exercises from

Kenneth Brown’s Cohomology of Groups

Christopher A. Gerig, Cornell University (College of Engineering)

August 2008 - May 2009

I appreciate emails concerning any errors/corrections: cgerig@berkeley.edu. Any

errors would be due to solely myself, or at least the undergraduate-version of myself
when I last looked over this. Remark made on 1/28/14.

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 major: Applied & Engineering Physics
year: Sophomore Undergraduate

Hey Ken, thanks for taking me under your wing.

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Contents
0 Errata to Cohomology of Groups 4

1 Chapter I: Some Homological Algebra 5

2 Chapter II: The Homology of a Group 12

3 Chapter III: Homology and Cohomology

with Coefficients 21

4 Chapter IV: Low-Dimensional Cohomology

and Group Extensions 31

5 Chapter V: Products 41

6 Chapter VI: Cohomology Theory

of Finite Groups 47

7 Chapter VII: Equivariant Homology and

Spectral Sequences 55

8 Chapter VIII: Finiteness Conditions 58

9 Chapter IX: Euler Characteristics 59

10 Additional Exercises 61

11 References 76

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0 Errata to Cohomology of Groups
pg62, line 11 missing a paranthesis ) at the end.
pg67, line 15 from bottom missing word, should say “as an abelian group”.
pg71, last line of Exercise 4 hint should be on a new lineP(for whole exercise).
pg85, line 9 from bottom incorrect function, should be g∈C/H g −1 ⊗ gm.
pg114, first line of Exercise 4 misspelled endomorphism with an extra r.
pg115, line 5 from bottom missing word, should say “4.4 is a chain map”.
pg141, line 4 from bottom missing a hat b on the last H ∗ .
pg149, line 13 the ideal I should be italicized.
pg158, line 20 there should be a space between SL2 (Fp ) and (p odd).
pg160, line 5 from bottom b ∗ (H, M ).
missing the coefficient in cohomology, H
pg165, line 26 the comma in the homology should be lowered, Hq (C∗,p ).

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1 Chapter I: Some Homological Algebra
P P
2.1(a): The
P P set A = P{g − 1 | g ∈ G, g 6= 1} is linearly independent
P because αg (g − 1) = 0 ⇒ αg g =
αg = αg 1 + 0g, and since ZG is a freePZ-module, αg has P
a unique expression, yielding
α
Pg = 0 ∀gP∈ A. To showP that I =
P span(A),P let α g g ∈ I and hence αg = 0. Thus we can write
αg g = αg g − 0 = αg g − αg = αg (g − 1). Since A is a linearly independent set which
generates I, it is a basis for I.

2.1(b): Consider the P A = ({s −
1 | sP

Pleft ideal ∈ S})Pover ZG.
A ⊆ I since ε(A) = P s ε ( r
Pg g g)(s − 1) = Ps ε( P g rg g)ε(s − 1) = y · 0 = 0.
If x ∈ I then x =
P α g
i iP− β g
j jP such thatP iα = Pβj . P
x
P = ( α g
i i − α
P i ) − ( β g
j j − β j ) + ( αi − βj )
Thus x = αi (gi − 1) − βj (gj − 1) and it suffices to show g − 1 ∈ A for any g ∈ G so that x ∈ A and
I ⊆ A. Since G = hSi, we have a representation x = s±1 ±1
1 · · · sn . By using ab − 1 = a(b − 1) + (a − 1)
−1 −1
and c − 1 = −c (c − 1), the result follows immediately.

2.1(c): Suppose S ⊂ G | I = ({s − 1 | s ∈ S}). Then every element of I is a sum of elements of

Pn−1
the form g − g 0 | g = g 0 s±1 . For g ∈ G, g − 1 ∈ I, so we have a finite sum g − 1 = i=1 (gi − gi0 ).
Since this is a sum of elements in G where G is written multiplicatively, ∃ i0 | gi0 = g, say i0 = 1.
Pn−1 Pn−1
Thus g − 1 = g − g10 + i=2 (gi − gi0 ) ⇒ g10 − 1 = i=2 (gi − gi0 ). Another iteration yields g2 = g10 and
g = g1 = g10 s±1 ±1 0 ±1 ±1 0 0 0
1 = g2 s1 = g2 s2 s1 . At the last iteration, gn−2 − 1 = gn−1 − gn−1 ⇒ gn−1 = 1, gn−2 =
0

gn−1 = gn−1 sn−1 = 1sn−1 . Through this method we obtain a sequence g1 , g2 , ..., gn−1 , gn | gi = gi+1 s±1
0 ±1 ±1
i
and g1 = g, gn = 1. ∴ g has a representation in terms of elements of S and so G = hSi since g was
arbitrary.

2.1(d): If G is finitely generated, then by part(b) above, I is finitely generated. For the converse,
Noting from part(a) that I = hg − 1ig∈G as a Z-module,
suppose I = (a1 , ..., an ) is a left ideal over ZG. P
each ai can be represented as a finite sum ai = j zj (gj − 1). Since each ai is generated by finitely many
elements, and there are finitely many ai , I is finitely generated as a left ideal by elements s − 1 where
s ∈ G. Therefore, we apply part(c) to have G = hs1 , . . . , sk i and thus G is a finitely generated group.

2.2: Let G = hti with |G| = n and let t be the image of T in R = Z[T ]/(T n − 1) ∼ = ZG. The el-
ement T − 1 is prime in Z[T ] because Z[T ]/(T − 1) ∼ = Z which is an integral domain (An ideal P of
a commutative ring R is prime iff the quotient ring R/P is an integral domain; Proposition 7.4.13[2]).
By Proposition 8.3.10[2] (In an integral domain a prime element is always irreducible), T − 1 is irre-
ducible in Z[T ]. We also could have obtained this result by applying Eisenstein’s Criterion with the
substitution T = x + 3 and using the prime 2. Since Z[T ] is a Unique Factorization Domain, the specific
factorization T n − 1 = (T − 1)(T n−1 + T n−2 + · · · + T + 1) with the irreducible T − 1 factor is unique,
Pn−1
considering the latter factor i=0 T i as the expansion of its irreducible factors [note: if n is prime then
Pn−1 i
i=0 T = Φn (T ), a cyclotomic polynomial which is irreducible
Pn−1 i
in Z[T ] by Theorem 13.6.41[2]]. Thus
every f ∈ R is annihilated by t − 1 iff it is divisible by N = i=0 t (and vice versa), and so the desired
free resolution of M = Z = ZG/(t − 1) is:
t−1 N t−1
· · · → R −→ R −→ R −→ R → M → 0 .
F
3.1: The right cosets Hgi are H-orbits of G with the H-action as group multiplication. Since G = Hgi
Z[Hgi ] ∼ Z[H/Hgi ]. G is a free H-set because hg = g ⇒ hgg −1 =
L L
where gi ranges over E, ZG = =
−1
gg ⇒ h = 1, i.e. the isotropy groups Hg are trivial. Therefore, ZG is a free ZH-module with basis E.
P
3.2: Let hSi = H ⊆ G and consider Z[G/H]. Now x ∈ Z[G/H] has the expression x = zi (gi H),
and there exists an element fixed by H, namely, x0 = g0 H = H where g0 ∈ H. H is annihilated by
I = Kerε = ({s − 1}) since (s − 1)H = sH − H = H − H = 0 ∀s ∈ S, and so Ix0 = 0. We have g − 1 ∈ I
since ε(g − 1) = ε(g) − ε(1) = 1 − 1 = 0. Hence (g − 1)x0 = 0 ⇒ gx0 = x0 ∀g ∈ G ⇒ Gx0 = x0 . Finally,
GH = H ⇒ G ⊆ H. ∴ G = H = hSi.

4.1: Orienting each n-cell en gives a basis for Cn (X). If X is an arbitrary G-complex, then with

5
 ηi eni ∈ Cn (X), g ∈ G can reverse the orientation of en by inversion (fixing the cell). Thus G need not
P
permute the basis, and hence Cn (X) is not necessarily a permutation module.

4.2: Since X is a free G-complex, it is necessarily a Hausdorff space with no fixed points under the
G-action. First, assume G is finite and take the set of elements in Gx0 , which are distinct points
gi x0 with 1x0 = x0 for an arbitrary point x0 ∈ X. Applying the Hausdorff condition, we have open
sets Ugi T containing gi x0 where each such set is disjoint from U1 containing x0 . Form the intersection
W = ( i gi−1 Ugi ) ∩ U1 which contains x0 . Since gk W ⊆ Ugk , we have gk W ∩ W = ∅ for all nonidentity
gk ∈ G, and so W is the desired open neighborhood of x0 [This result does not follow for arbitrary G
since an infinite intersection of open sets need not be open].
Assume the result has been proved for G infinite.
Let ϕ : X →`X/G be the quotient map, which sends the disjoint collection of gi W ’s to ϕ(W ). Since
ϕ−1 ϕ(W ) = i gi W , gi W → ϕ(W ) is a bijective map (restriction of ϕ) and thus it is a homeomorphism
(ϕ continuous and open). This covering space is regular because G acts transitively on ϕ−1 (Gx) by def-
inition. Elements of G are obviously deck transformations since Ggx = Gx, hence G ⊆ Aut(X). Given
Γ ∈ Aut(X) with Γ(a) = b, those two points are mapped to the same orbit in X/G (since ϕ ◦ Γ = ϕ),
and so ∃ g ∈ G sending a to b. By the Lifting Lemma (uniqueness) we have Γ = g, hence Aut(X) ⊆ G ⇒
G is the group of covering transformations.
If X is contractible then G ∼ = π1 (X/G)/π1 (X) ∼ = π1 (X/G)/0 = π1 (X/G) and ϕ is the universal cover of
X/G, so X/G is a K(G, 1) with universal cover X.
It suffices to show that the G-action is a “properly discontinuous” action on X when G is infinite:
Every CW-complex with given characteristic maps fj,n : (Bjn , Sjn−1 ) → (σ̄jn , ∂ σ̄jn ) admits a canonical
open cover {Uσ } indexed by the cells, where Ua and Ub are disjoint open sets for distinct cells of equal
dimension (for instance, if X is a simplicial complex we can take Uσ = St(σ̂) which is the open star
of the barycenter of σ in the barycentric subdivision of X). More precisely, for one cell σkn ⊂ X n in
each G-orbit of cells define its “barycenter” as σ̃kn ≡ fk,n (bk ), where bk = 0 ∈ B n is the origin of the
n-disk; for the rest of the cells {σin = gi σkn } in each G-orbit define their “barycenters” as σ̃in ≡ fi,n (bi ),
where bi ∈ Bin is chosen so that σ̃in = gi σ̃kn . Considering the 0-skeleton X 0 , its cells σj0 are by defini-
tion open and so the canonical open cover of X 0 is the collection {Uσj0 = σj0 }. Proceeding inductively
(with Uσ open in X n−1 ), consider the n-skeleton X n = X n−1 j σjn and note that the preimage under
S
−1
fj,n of the open cover of X n−1 is an open cover of the unit circle Sjn−1 . Take an open set fj,n (Uσ )
n−1 n
in Sj and form the “open sector” Wj,σ ⊂ Bj which is the union of all line segments emanating
−1 −1
from bj and ending in fj,n (Uσ ), minus bj and fj,n (Uσ ); each Uσ determines such a Wj,σ . As fj,n is a
n n
homeomorphism of Int(Bj ) with σj , we have such open sectors fj,n (Wj,σ ) in the n-cell. Noting the
weak topology on X, the set Uσ0 = Uσ j fj,n (Wj,σ ) is open in X n iff its complement in X n is closed
S

iff (X n − Uσ0 ) ∩ σ̄ki ≡ σ̄ki − (Uσ0 ∩ σ̄ki ) is closed in σ̄ki for all cells in X n . For i < n, Uσ0 ∩ σ̄ i = Uσ ∩ σ̄ i
because fj,n (Wj,σ ) ⊂ σjn which is disjoint from the closure of all other cells, and the complement of this
intersection in σ̄ i ⊂ X n−1 is closed because Uσ is open by inductive hypothesis. Therefore, it suffices to
show that σ̄kn − (Uσ0 ∩ σ̄kn ) is closed in σ̄kn ∀ k, which is equivalent 0 n
Sσ ∩ σ̄k to be
under topology of cells for U
n 0 n n n n
S
open in σ̄k . For arbitrary k we have Uσ ∩ σ̄k = (Uσ ∩ σ̄k ) j (fj,n (Wj,σ ) ∩ σ̄k ) = (Uσ ∩ σ̄k ) fk,n (Wk,σ )
−1 −1
and taking the preimage we have Y = fk,n (Uσ0 ∩ σ̄kn ) = fk,n (Uσ ) ∪ Wk,σ . The n-disk is compact, the
n
CW-complex X is Hausdorff, a closed subset of a compact space is compact (Theorem 26.2[6]), the
image of a compact set under a continuous map is compact (Theorem 26.5[6]), and every compact subset
of a Hausdorff space is closed (Theorem 26.3[6]); thus fk,n is a closed map and hence a quotient map for
σ̄kn (by Theorem 22.1[6]). It suffices to check that Y ⊂ Bkn is open, for then fk,n (Y ) = Uσ0 ∩ σ̄kn is open
−1
in σ̄kn by definition of a quotient map. By construction, Y = {x ∈ Bkn − bk | r(x) ∈ fk,n (Uσ )} where
n n n−1 x−bk −1
r : Bk − bk → ∂Bk ≡ Sk is the radial projection r(x) = ||x−bk || . As r is continuous and fk,n (Uσ ) is
−1
open in ∂Bkn , we have Y = r−1 (fk,n (Uσ )) open in Bkn − bk and hence in Bkn (by Lemma 16.2[6]). Our
new collection for X n is the open sets Uσ0 (where σ ⊂ X n−1 ) plus the open n-cells Uσ0 jn = σjn ; this is the
canonical open cover of X n and hence completes the induction.
Any point x ∈ X will lie in an i-cell σ which lies in the open set Uσ0 , and we take this as our desired
neighborhood of x: since any open set of our constructed cover is bounded by barycenters, and g ∈ G
maps barycenters to barycenters by construction, we have gUσ0 = Ugσ 0
which is disjoint from Uσ0 by
construction for all g 6= 1.

6
4.3: Given G = Z ⊕ Z we have the torus T with π1 T ∼ = G and its universal cover ρ : R2 → T .
2 2
After drawing the lattice Z ⊂ R , pick a square and label that surface L, with the bottom left cor-
ner as the basepoint x0 and the bottom side as the edge es and the left side as the edge et (so the
corners are x0 , sx0 , stx0 , tx0 going counterclockwise around L, and the top and right sides of L are re-
spectively tes and set ). Following Brown’s notation [in this section], x0 generates C0 (R2 ) and es , et
generate C1 (R2 ) with ∂1 (es ) = (s − 1)x0 and ∂1 (et ) = (t − 1)x0 . Lastly, L generates C2 (R2 ) with
∂2 (L) = es + set − tes − et = (1 − t)es − (1 − s)et . Thus the desired free resolution of Z over ZG is:
2 ∂ 1 ∂ ε
0 → ZG −→ ZG ⊕ ZG −→ ZG −→ Z → 0 .

4.4:

5.1: The homotopy operator h in terms of the Z-basis g[g1 | · · · |gn ] for Fn is h(g[g1 | · · · |gn ]) = [g|g1 | · · · |gn ].

5.2: Using G = Z2 = {1, s}, the elements of the normalized bar resolution F¯∗ = F∗ /D∗ are [s|s| · · · |s],
and each element forms a basis for the corresponding dimension, giving the identification F¯n = ZG.
Denoting si = s ∀ i,

s[s| · · · |s] ,
 i=0
di [s1 |s2 | · · · |sn ] = [s1 | · · · |si−1 |si si+1 | · · · |sn ] = 0 , 0<i<n

[s| · · · |s] , i=n


The middle equation resulted from si si+1 = s2 = 1 (so the element lies in D∗ ). The boundary operator
∂n then becomes s − 1 for n odd and s + 1 for n even.
∴ the normalized bar resolution is:
s−1 s+1 s−1 ε
· · · → ZG −→ ZG −→ ZG −→ ZG → Z → 0 .

5.3(a): [[Geometric Realization of a Semi-Simplicial Complex ]]

For each (n+1)-tuple σ = (g0 , . . . , gn ), let ∆σ be a copy of the standard n-simplex with vertices v0 , . . . , vn .
Let di σ = (g0 , . . . , ĝi , . . . , gn ) and let δi : ∆di σ → ∆σ be`the linear embedding which sends v0 , . . . , vn−1
to v0 , . . . , v̂i , . . . , vn . Consider the disjoint union X0 = σ ∆σ (topologize it as a topological sum) and
def
define the quotient space X = X0 /∼ using the equivalence relation generated by (σ, δi x) ∼ (di σ, x),
where we rewrite ∆σ as σ × ∆σ for clarity of the relation properties.
We assert that the geometric realization X is a CW-complex with n-skeleton X n = ( dim∆σ ≤n ∆σ )/∼.
`

X 0 is the collection of vertices and hence a 0-skeleton, and so we proceed by induction on n [sketch]:
(n)
In X n the equivalence relation ∼ identifies a point on a boundary ∂∆σ with a point in X n−1 , and
(n) (n)
it doesn’t touch the interior points of ∆σ . This means that the n-cells are {∆σ } with the attaching
maps induced by di ∀ i. Refer to Theorem 38.2[4] for the analogous construction with adjunction spaces,
providing
S Hausdorffness of X n and weak topology w.r.t. {X i }i<n . Thus, X is a CW-complex as the
i
union X with the weak topology.
We define the G-action on the simplices by left multiplication on their associated tuples; it is free since
tuples are unique and so the only element which fixes a tuple (and hence a simplex) is the identity
element of G. This makes X a G-complex. We deduce that X is contractible because for each simplex,
X contains its cone which is contractible, so taking hσ = (1, g0 , . . . , gn ) we can use the straight-line
homotopy between δ0 : ∆σ → ∆hσ and the constant map ∆σ → ∆hσ at v0 (for any point in the domain
simplex of this homotopy H which lies on a subsimplex, the homotopy associated to that subsimplex is
just the restriction of H, and hence the straight-line homotopy is well defined).
To form the desired isomorphism between the cellular chain complex C(X) and the standard resolution
F∗ , it suffices to determine the boundary operator on C(X) and see that it provides commutativity of
the diagram F∗ → C(X), P noting that Ci (X) ∼ = Fi by the correspondence ∆σ ↔ σ. Now C(X0 ) has the
boundary ∂[v0 , . . . , vn ] = i (−1)i [v0 , . . . , v̂i , . . . , vnP
] and each resulting (n-1)-simplex is the image under
i
a δi . Thus C(X) has boundary maps ∂(∆σ ) = i (−1) ∆di σ , and these parallel those of F∗ , giving
commutativity of the diagram.

Note that this provides a solution to Exercise I.4.4: For any group G we can form the contractible

7
G-complex X as above, and since X → X/G is a regular covering map by the result of exercise I.4.2
above, the orbit space X/G is a K(G, 1), called the “classifying space.”

5.3(b): For the normalized standard resolution F̄∗ , we simply follow part(a) while making these further
identifications in X0 to collapse degenerate simplices. For each σ = (g0 , . . . , gn ) let si σ = (g0 , . . . , gi , gi , . . . , gn ),
and when forming the quotient X0 → X we also collapse ∆si σ to ∆σ via the linear map Li which
sends v0 , . . . , vn+1 to v0 , . . . , vi , vi , . . . , vn (so the only simplices of X are those whose associated tu-
ples have pairwise distinct coordinates). Thus the equivalence relation in part(a) is also generated by
(σ, Li x) ∼ (si σ, x). During the inductive process for X n , if ∆σ is a degenerate simplex then X n−1 will
already contain it and so those simplices need not be considered as n-cells. No problems arise when using
the homotopy because for tuples of the form τ = (1, g0 , . . . , gn ) | g0 6= 1, the cone ∆τ ∗ v0 = ∆hτ = ∆τ
has the identity map δ0 still being nulhomotopic [note: we actually have a deformation retraction since
∆(1) remains fixed instead of looping around ∆(1,1) as in part(a)].

6.1: Given S a finite CW-complex X with a map f : X → X such that every open cell satisfies
f (σ) ⊆ τ 6=σ τ where dimτ ≤ dimσ, we have the condition f (σ) ∩ σ = ∅ and there are no fixed
points. Viewing the open n-cell σ on the chain level in Hn (X (n) , X (n−1) ) = Cn (X), f] (σ) does not
consist of σ and so the respective matrix has the value 0 at the row-column intersection for σ. Therefore,
P = 0 ∀ n since the diagonal ofP
tr(f] , Cn (X)) the matrix for the basis elements is zero. By the Hopf Trace
Theorem, (−1)i tr(fi , Hi (X)/torsion) = (−1)i tr(f] , Ci (X)) and so the Lefschetz number Λ(f ) = 0.
If X is a homology (2n − 1)-sphere, then Hi (X) is nontrivial only in dimensions 0 and 2n − 1 (in
which case it is isomorphic to Z). Thus, by the Lefschetz Fixed Point Theorem, (−1)0 tr(f0 , Z) +
(−1)2n−1 tr(f2n−1 , Z) = 1 − d = 0 ⇒ d = 1 and f∗ : H2n−1 (X) → H2n−1 (X) is the identity.

6.2: The group action G → Homeo(S 2n ) yields a degree map

φ : G → Aut(H2n (S 2n )) ∼
= Aut(Z) = {±1} = Z/2Z
which sends g ∈ G to the degree d = deg(g) of its associated homeomorphism g : S 2n → S 2n [note:
deg(g) · deg(g −1 ) = deg(g · g −1 ) = deg(id) = 1 ⇒ |d| = 1].
Consider nontrivial G 6= Z/2Z and assert that this φ is not injective:
Kerφ = 0 ⇒ 3 ≤ |G| = |Kerφ| · |Imφ| = 1 · |Imφ|. Since Imφ ⊆ Z/2Z, |Imφ| = 1 or 2 . In either case
we arrive at a contradiction (since 1, 2 < 3). ∴ ∃ g ∈ Kerφ | g 6= id ⇒ degg = 1. Now assume this
action is free, and use the notation fi : Hi (S 2n ) → Hi (S 2n ). By the Lefschetz Fixed Point Theorem,
(−1)0 tr(f0 ) + (−1)2n tr(f2n ) = 1 + d = 0 ⇒ degf = d = −1 ∀ f 6= id. Our contradiction has now been
reached (taking f = g from above).

7.1: Given the finite cyclic group G = hti, the free resolution F of Z over ZG with period two (chain
complex with rotations of S 1 ), and the bar resolution F 0 , we obtain a commutative diagram where
f : F → F 0 is the desired augmentation-preserving chain map:
··· / ZG t−1 / ZG N / ZG t−1 / ZG ε / Z /0

f3 f2 f1 f0 idZ
    
··· / F30 / F20 / F10 / F00 /Z /0
We define f inductively as fn+1 = kn fn ∂, where k is a contracting homotopy for the augmented complex
associated to F 0 , and each map is determined by where it sends the basis element:
f0 (1) = k−1 idZ ε(1) = k−1 idZ (1) = k−1 (1) = (1) ≡ [ ]
f1 (1) = k0 f0 ∂(1) = k0 f0 (t − 1) = k0 {tf0 (1) − f0 (1)} = k0 {t[ ] − [ ]} = [t] − [1]
Pn−1 Pn−1
f2 (1) = k1 f1 ∂(1) = k1 f1 (N ) = k1 { i ti f1 (1)} = i ([ti , t] − [ti , 1])
Pn−1
f3 (1) = k2 f2 ∂(1) = k2 f2 (t − 1) = i ([t, ti , t] + [1, ti , 1] − [t, ti , 1] − [1, ti , t])

7.2: Here is the axiomatized version, Lemma 7.4:

Under the additive category A, let (C, ∂) and (C 0 , ∂ 0 ) be chain complexes, let r be an integer, and let
(fi : Ci → Ci0 )i≤r be a class of morphisms such that ∂i0 fi = fi−1 ∂i for i ≤ r. If Ci is projective relative
0
to the class E of exact sequences for i > r, and Ci+1 → Ci0 → Ci−10
is in E for i ≥ r, then (fi )i≤r extends
0
to a chain map f : C → C and f is unique up to homotopy. (Theorem 7.5 follows immediately)

8
7.3(a): Given an arbitrary category C, let A ∈ Ob(C) be an object and let hA = HomC (A, −) : C →
(Sets) be the covariant functor represented by A, with uA ∈ hA (A) as the identity map A → A. Let
T : C → (Sets) be an arbitrary covariant functor. Any natural transformation ϕ : hA → T yields the
commutative diagram (with f : A → B in hA (B) arbitrary):
hA (f )
hA (A) −−−−→ hA (B)
 
ϕ ϕ
y y
T (f )
T (A) −−−−→ T (B)
For any v ∈ T (A) suppose we have the natural transformation with ϕ(uA ) = v. By commutativity,
T (f )(v) = (ϕ ◦ hA (f ))(uA ) = ϕ(f ◦ uA ) = ϕ(f ), and hence the transformation is unique [determined by
where it sends the identity]. For existence of the natural transformation ϕ(f ) = T (f )(v), we assert that
it satisfies the commutative diagram, using arbitrary g : B → C (and f as above):
T (g)[ϕ(f )] = T (g)[T (f )(v)] = T (g ◦ f )(v) = ϕ(g ◦ f )
ϕ[hA (g)(f )] = ϕ(g ◦ f ) = T (g)[ϕ(f )]
Thus, HomF (hA , T ) ∼
= T (A) where F is the category of functors C → (Sets), and we have finished proving
Yoneda’s Lemma.
L
7.3(b): An M-free functor F : C → Ab is isomorphic to α ZhAα , where Aα ∈ M [M is a sub-
class of Ob(C)] and ZhA : C → Ab is the composite of hA and the functor (Sets) → Ab which associates
to a set the free abelian group it generates. Given the additive category A whose objects are covariant
functors C → Ab and whose maps are natural transformations of functors, let E be the class of M-exact
sequences in A. Consider the mapping problem (for all rows in E):
F
ψ 0
ϕ
~ 
T0 /T / T 00
i j
By Yoneda’s Lemma (part(a) above), each component ZhAα of F with any natural transformation in the
above diagram is completely determined by the identity uAα , and so these identities “form a basis” for
F . In particular, for the identity uA we obtain the exact sequence T 0 (A) → T (A) → T 00 (A) of abelian
groups from the above mapping problem since the associated row lies in E with A ∈ M. Thus, for each
identity we have ϕ(uAα ) ∈ Kerj = Imi, which implies ∃ xα ∈ T 0 (Aα ) | i(xα ) = ϕ(uAα ), and so we
form ψ by ψ(uAα ) = xα . This means that F (an M-free functor) is projective relative to the class E of
M-exact sequences.

7.3(c): There is a natural chain map in A from M-free complexes to M-acyclic complexes, and it
is unique up to homotopy [using the categorial definitions from parts (a) and (b)]. This statement is a
result of the combination of part(b) and Exercise 7.2 above, and is precisely the Acyclic Model Theorem
in a rephrased form (M is the set of models).

7.4: Under the category of R-modules, let (C, δ) and (C̄, δ̄) be cochain complexes, let r be an inte-
ger, and let (fi◦ : C̄ i → C i )i≤r be a class of morphisms such that fi−1◦
δ̄i−1 = δi−1 fi◦ for i ≤ r. If C i is
◦
injective relative to the class E of exact sequences for i > r [yielding a cochain complex of injectives],
and C̄ i−1 → C̄ i → C̄ i+1 is in E ◦ for i ≥ r [an acyclic cochain complex], then (fi◦ )i≤r extends to a cochain
map f ◦ : C̄ → C and f ◦ is unique up to homotopy.
(The analogous “Theorem 7.5” follows immediately)

8.1: Obviously the trivial group is one, since Z[{1}] = Z and Z is a projective Z-module; so assume G is
nontrivial. Give Z the trivial module structure (so that for r ∈ ZG, r · a = ε(r)a ∀ a ∈ Z). Considering
ε
the short exact sequence of modules 0 → I → ZG → Z → 0, we must find a splitting µ : Z → ZG for Z
to possibly be ZG-projective.
P Any such map is determined by where 1 ∈ ZPis sent; say µ(1) P = x.PThen,
for nontrivial
P α = r g
i i , µ(αP· 1) = α · µ(1) =
P α · x = αx.
P But µ(α
P · 1) = µ( ri ) = µ(1) ri = ( ri )x,
so αx = ( ri )x. Thus (α − ri )x = 0 ⇒ ri gi = ri ⇒ ri (gi − 1) = 0 [this sum can be viewed

9
as having no gi = 1, and hence it lies in I]. Restricting our choice of nontrivial α to one which is not an
integer, there is some nontrivial ri0 associated to gi0 6= 1 and hence we must have gi = 1 ∀ i (by freeness
of I). But then G is the trivial group, and we are done.

8.2: Assume P is a projective ZG-module and consider the subgroup H ⊆ G. Then F = P ⊕ K

where F is a free ZG-module, by Proposition I.8.2[1]. By restriction of scalars from ZG to its subring
ZH [r · n = f (r)n with homomorphism f : ZG → ZH preserving identities], F has an inherent ZH-
module structure. Since P is a direct summand of such a free module, it is a projective
L ZH-module, by
Proposition I.8.2[1]. Alternatively, we can note from Exercise 3.1 above that ZG = ZH and so F is a
direct sum of H-modules, hence ZH-free.

8.3(a): Given F as a non-negative acyclic chain complex of projective modules Pn (over an arbitrary
∂¯n
ring R), it will be contractible if each short exact sequence 0 → Zn → Pn → Zn−1 → 0 splits (where ∂¯ is
induced by the boundary ∂), by Proposition I.0.3[1]. The case n = 0 is trivial using the zero map, and
so arguing inductively we assume that ∂¯n−1 splits. Since Ker∂¯n−1 = Zn−1 is a direct summand of the
projective module Pn−1 and a projective module is a direct summand of a free module by Proposition
I.8.2[1], Zn−1 is necessarily a direct summand of a free module, and hence is projective by Lemma I.7.2[1].
Therefore, ∂¯n must have a splitting in the aforementioned short exact sequence by Proposition I.8.2[1].

8.3(b): If R is a principal ideal domain, then submodules of a free R-module are free by Theorem
II.7.1[5]. Since a projective module is a direct summand of a free module by Proposition I.8.2[1], it is
in particular a submodule and hence is free (over R as a PID). Therefore, submodules of a projective
module over a PID are free and hence projective (free modules are projective by Lemma I.7.2[1]). The
non-negativity hypothesis of Corollary I.7.7 can be dropped if we then restrict ourselves to PIDs, because
we can follow part(a) above but not use induction since Zn−1 is already projective, being a submodule
of the projective chain module Pn−1 (i.e. we don’t need any “starting point” in the resolution to obtain
the desired splitting).

8.4: Every permutation module admits the decomposition QX ∼

L
= Q[G/Gx ] and a direct sum of
projective modules is projective iff each summand is projective (by Lemma XVI.3.6[5]). Thus it suffices
to show that Q[G/Gx ] is a projective QG-module, where G is an arbitrary group and Gx is finite. Note
that HomQG (Q[G/Gx ], −) is a left-exact functor (Corollary 10.5.32[2]); it is given by M → M Gx because
any homomorphism ϕ is determined by ϕ(Gx ), and ϕ(Gx ) = ϕ(g · Gx ) = g · ϕ(Gx ) for any g ∈ Gx .
Thus, we must show that HomQG (Q[G/Gx ], −) takes surjective homomorphisms M → M̄ to surjective
homomorphisms M Gx → M̄ Gx [because then the functor P is exact and then Q[G/Gx ] is projective by
definition]. If m̄ ∈ M̄ Gx , lift m̄ to m ∈ M . Then |G1x | g∈Gx gm is also a lifting of m̄ which lies in M Gx
(because |G1x | gm 7→ |G1x | g m̄ = |G1x | · |Gx |m̄ = m̄), and so M Gx → M̄ Gx is surjective. Thus, QX is
P P

a projective QG-module, where X is a G-set and Gx is finite for all x ∈ X.

8.5: If G is finite and k is a field of characteristic zero, consider any short exact sequence of kG-
ϕ
modules of the form 0 → M 0 → M → P → 0. Since k-vector spaces are free modules over k, and free
modules are projective (by Lemma I.7.2[1]), P is k-projective and hence the sequence [as k-modules]
splits by Proposition I.8.2[1]. Choosing a splitting f : P →P M for the underlying sequence of k-vector
1 −1
spaces, we form the homomorphism φ : P → M by x 7→ |G| g∈G gf (g x).
Since f is a k-module homomorphism, it suffices to show that φ is equivariant (compatible with G-action)
for it to thus be a kG-module homomorphism:
1
gf (g −1 g0 x) = |G|
1
g0 hf (h−1 x)
P P
φ(g0 x) = |G|
−1
[where h = g0 g] P
1
gf (g −1 g0 x)) = g0 ( |G|
1
hf (h−1 x)) = φ(g0 x)
P
g0 φ(x) = g0 ( |G|
P P −1
[because g∈G h = g∈G g, as g0 permutes the elements of G]
By Proposition 10.5.25[2] it suffices to show that ϕφ = idP for φ to thus be a splitting (noting that
ϕf = idP ):
1 −1 1 −1 1 −1
P P P
ϕ(φ(x)) = ϕ( |G| g∈G gf (g x)) = |G| g∈G ϕ(gf (g x)) = |G| g∈G gϕ(f (g x)) =

10
 1
gg −1 x = 1 1
P P
|G| g∈G |G| g∈G x= |G| |G|x =x

Since φ is a kG-splitting, P is kG-projective by Proposition I.8.2[1].

8.6: Suppose P is an R-module such that ϕ : P ∗ ⊗R P → HomR (P, P ) is surjective, P where this
map is given byPϕ(u ⊗ m)(x) = u(x) P · m. Then in particular we have idP = ϕ( f i ⊗ e i ), so that
x = idP (x) = ϕ(fi ⊗ ei )(x) = fi (x)ei . Thus, P is projective by Proposition I.8.2[1], and it is
finitely generated by the ei elements (noting that the summation is finite for tensor products).

8.7: Assume P is a finitely generated projective R-module and M is any (left) R-module, and take
the canonical isomorphism ϕ : P ∗ ⊗R M → HomR (P, M ) from Proposition I.8.3[1] which is given by
ϕ(u ⊗ m)(x) = u(x) · m. For any z ∈ P ∗ ⊗R P we define the map ψz : HomR (P, M ) → P ∗ ⊗R M as
ψz (f ) = (P ∗ ⊗ f )(z), which is a homomorphism since tensors are distributive over sums.
Method 1 : View ϕ−1 as a natural transformation HomR (P, −) → P ∗ ⊗R −, where HomR (P, −)
and P ∗ ⊗R − are exact covariant functors from the category R-mod to the category Ab by Corollary
10.5.41[2] and Corollary 10.5.32[2], noting that P ∗ is projective by Proposition I.8.3[1] and hence is a
flat module by Corollary 10.5.42[2]. By Yoneda’s Lemma, ϕ−1 is uniquely determined by ϕ−1 (idP ) = z,
and the proof of the lemma (Exercise I.7.3(a) above) states that ϕ−1 (f ) = (P ∗ ⊗ f )(z) = ψz (f ). Thus,
the inverse homomorphism ϕ−1 is a map of the form ψz (independent of M ).
∗
Method 2 :PBy Proposition I.8.2[1] we can choose elements ei ∈ P and P fi ∈ P such that for every
x ∈ P, x = fi (x)ei and fi (x) = 0 for cofinitely P many x. Set z P = fi ⊗ ei . We have ψz ϕ = id
∗
because ψzP (ϕ(u ⊗ m)) = P (P ⊗ u · m)(z) = f i ⊗ u(ei ) · m = f i u(ei ) ⊗ m = u ⊗ m [note:
·
u(x) = u( fi (x)ei ) = fi (x) · u(ei ) as u is an R-module homomorphism]; we also haveP ϕψz = id
∗
P P
because ϕ(ψ z (f )) = ϕ[(P ⊗ f )(z)] = ϕ( f i ⊗ f (e i )) = f i · f (e i ) = f [note: f (x) = f ( fi (x)ei ) =
fi (x) · f (ei ) as f is also an R-module homomorphism]. Thus, the inverse ϕ−1 is a map of the form ψz
P
(independent of M ).

8.8: Since P is finitely presented, we can form the obvious exact sequence F1 → F0 → P → 0 with F0 and
F1 free of finite rank (the generators and relators, respectively). By Theorem 10.5.33[2], HomR (−, D)
is a left exact contravariant functor, and so we obtain an exact sequence 0 → P ∗ → F0∗ → F1∗ of right
R-modules [notation: M ∗ = HomR (M, R)]. Since P is a flat module, we can tensor this exact sequence
with P to obtain the exact sequence 0 → P ∗ ⊗R P → F0∗ ⊗R P → F1∗ ⊗R P . Since Fi [i = 0, 1] is
free, it is necessarily projective (by Lemma I.7.2[1]) and hence Fi∗ ⊗R P ∼= HomR (Fi , P ) by Proposition
I.8.3[1]. Since F1 is contained in the quotient of F0 which gives P , HomR (F1 , P ) = 0 and thus we have
the isomorphism P ∗ ⊗R P ∼ = HomR (F0 , P ).
Now HomR (−, P ) as a functor on the original presentation sequence gives rise to the exact sequence 0 →
HomR (P, P ) → HomR (F0 , P ) → HomR (F1 , P ) = 0, and thus we have the isomorphism HomR (P, P ) ∼ =
HomR (F0 , P ).
∼
=
Since the discovered isomorphism P ∗ ⊗R P → HomR (P, P ) is surjective, P is a projective R-module by
Exercise I.8.6 above.

11
2 Chapter II: The Homology of a Group
2.1: For S an arbitrary G-set, ZS ∼
L
= Z[G/Gs ]. By the Orbit-Stabilizer Theorem we have a bijection
between Gs and G/Gs . When passing to the quotient for the group of co-invariants, the subset Gs ⊆ S
L element Gs∼⊆L
is sent to the S/G since s is identified with gs in S/G (giving trivial G-action). Therefore
(ZS)G ∼ ∼ ∼
L
= ( Z[G/G ])
s G = (Z[G/G ])
s G = Z[Gs] = Z[S/G].

2.2: ∼
L A weaker hypothesis “Let X be an arbitrary G-complex without inversions” suffices. Then C∗ (X) =
Z[Xi ] is a direct sum of permutation modules where Xi is the basis set of i-cells, so by the previous
exercise, C∗ (X)G ∼ = (Z[Xi ])G ∼ Z[Xi /G] ∼
L L
= = C∗ (X/G) which has a Z-basis with one basis element
for each G-orbit of cells of X.

2.3(a): With the G-module M and normal subgroup H / G, MH = M/IM where I is the augmentation
ideal of ZH. The induced G/H-action is given by gH(m + IM ) = gm + IM . The properties of an action
are obviously satisfied, and it is well defined because if g2 is another coset representative of g1 H, then
g2 = g1 h and g2 m+IM = g1 hm+IM = g1 hm−g1 m+g1 m+IM = (h−1)g1 m+g1 m+IM = g1 m+IM .

2.3(b): We can form the group homomorphism ϕ : MG → (MH )G/H using part(a) by m 7→ m + IM ,
which is well-defined because m = gm 7→ gm + IM = gH(m + IM ) = m + IM ; it is a homomor-
phism since ϕ(m1 m2 ) = ϕ(m1 m2 ) = m1 m2 + IM = (m1 + IM )(m2 + IM ) = (m1 + IM )(m2 + IM ) =
ϕ(m1 )ϕ(m2 ). For the inverse φ we use m + IM 7→ m which is well-defined since given the equivalent
elements m + IM and gm + (h − 1)m0 + IM in (MH )G/H ,
φ(gm + (h − 1)m0 + IM ) = gm+hm0 −1m0 = m+m0 −m0 = m = φ(m + IM ); it is a homomorphism be-
cause φ(m1 + IM m2 + IM ) = φ(m1 m2 + IM ) = m1 m2 = m1 m2 = φ(m1 + IM )φ(m2 + IM ). Thus,
we have the isomorphism MG ∼ = (MH )G/H .

2.3(c): Let Z[G/H] ⊗ZG M be a G/H-module, where Z[G/H] is the obvious (G/H, G)-bimodule which
forms the tensor product and gives it the desired module structure. The map Z[G/H] × M → MH given
by (a, m) 7→ am̄ is clearly G-balanced, and so by the universal property of tensor products (Theorem
10.4.10[2]) there exists the group homomorpism ϕ : Z[G/H] ⊗ZG M → MH given by a ⊗ m 7→ am̄, and
it is clearly a G/H-module homomorphism. There is a well-defined map φ : MH → Z[G/H] ⊗ZG M
defined as m̄ 7→ 1 ⊗ m because of the identity 1H ⊗ hm = 1Hh ⊗ m = 1H ⊗ m, and it is a G/H-module
homomorphism because φ(gH · m̄) = φ(g m̄) = 1H ⊗ gm = Hg ⊗ m = gH ⊗ m = gH1H ⊗ m = gH · φ(m̄)
[noting that gH = Hg since H is normal, and G/H acts on MH by part(a) above]. Since ϕ and φ are
inverses of each other, they are isomorphisms and we obtain MH ∼ = Z[G/H] ⊗ZG M .

3.1: Let g1 , . . . gn ∈ G be pairwise-commutative elements and consider z = (−1)sgnσ [gσ(1) | · · · |gσ(n) ] ∈

P
Cn (G), where σ ranges over all permutations of {1, . . . , n}. The sign of a permutation is defined here to
be the number of swaps betweenP adjacent integers to bring the permuted set back to the identity. Looking
at the boundary ∂z where ∂ = (−1)i di , a particular dj with j 6= 0, n will provide elements in Cn−1 (G)
of the form [· · · |gk gk0 | · · · ]. Each of these appears twice because gk gk0 = gk0 gk , but the paired elements
will have opposite signs and hence will cancel each other. For j = 0, n we have elements of the form
[gi | · · · ] with one gk missing from each, and each of these elements also appears twice because d0 will
take off gk from the beginning of some element while dn will take off gk from the end of some other
element. The paired elements differ by the sign (−1)n due to the boundary map, and they also differ
by the sign (−1)n−1 due to the permutation which takes the first slot and sends it to the last slot; since
(−1)n (−1)n−1 = (−1)2n−1 = −1, these paired elements will also cancel each other. Thus, ∂z = 0 and z
is a cycle in Cn (G).

3.2: Suppose Z admits a projective resolution of finite length over ZG where G = Zn . Then ∃ i0 | Hi G =
0 ∀ i > i0 , and we make note that Hi G is independent of the choice of resolution (see Section II.1[1]).
Yet by an earlier calculation II.3.1[1] (using an infinite resolution), Hi G ∼
= Zn for all positive odd integers
i. Thus we have arrived at a contradiction.

3.3: If G has torsion, say Zn ⊆ G, and Z admits a projective resolution of finite length over ZG,

12
then since a projective ZG-module is projective as a ZH-module for any subgroup H ⊆ G (by Exercise
I.8.2), we would obtain a corresponding finite projective resolution over Z[Zn ]. But this cannot occur
due to the previous exercise, and hence we arrive at a contradiction.

4.1: If Y is a path-connected space and has a contractible regular covering space X with covering
group G, then X is its universal cover with free G-action as translation, and π1 Y = G (so Y ∼ = X/G is
a K(G, 1)-space). The singular chain module Cnsing (X) is a free Z-module with basis the set of singular
simplices σ n : ∆n → X (continuous maps of the standard simplex into the space). A G-action on this
basis is given by the composition gσ n : ∆n → X → X, and thus Cnsing (X) is a free ZG-module with
one basis element for every G-orbit of singular simplices. Denoting the ith face of σ n as σ n Fi where
Fi = [v0 , . . . , v̂i , . . . , vn ] : ∆n−1 → ∆n is the inclusion map, the induced G-action on the simplices maps
faces of σ n to faces of gσ n . Thus, the boundary operator is equivariant and the singular [augmented]
chain complex C∗sing (X) is a free G-module chain complex. As X is contractible, the complex is exact
and so it is a free resolution of Z over ZG.
Consider the projection ϕ : C∗sing (X) → C∗sing (Y ) given by σ 7→ qσ, where q : X → X/G ∼ = Y is
the regular covering map. Noting that ϕ(gσ) = qgσ = qσ = ϕ(σ), the projection induces the map
ϕ̄ : C∗sing (X)G → C∗sing (Y ) which sends basis elements to basis elements. Since C∗sing (Y ) has a Z-basis
with one basis element for each G-orbit of singular cells of X as does C∗sing (X)G [by a property (II.2.3)
of the coinvariants functor], ϕ̄ is an isomorphism. Therefore, H∗ G ∼ = H∗ Y as the homologies of a group
are independent of the choice of resolution up to canonical isomorphism.

4.2:

4.3:

5.1: Let Y be an n-dimensional connected CW-complex such that πi Y = 0 for i < n (n ≥ 2), let π = π1 Y ,
and let X be the universal cover of Y (so that π1 X = 0). Since πi X ∼ = πi Y for i > 1, πi X is trivial for
i < n and so by the Hurewicz Theorem Hi X = 0 for 0 < i < n and the Hurewicz map h : πn X → Hn X
is an isomorphism. In addition, we have a partial free resolution Cn (X) → · · · → C0 (X) → Z → 0 whose
nth homology group is Zn X = Hn X (noting that X is n-dimensional). Lemma II.5.1[1] now gives us an
exact sequence 0 → Hn+1 π → (Hn X)π → Zn Y = Hn Y → Hn π → 0, where Y ∼ = X/π because every
universal cover is regular with covering transformation group π1 Y (by Corollary 81.4[6]). Finally, noting
that the coinvariants functor takes isomorphisms to isomorphisms (by right-exactness), the commutative
∼
diagram (πn X)π = / (Hn X)π

∼
=
 
(πn Y )π / Hn Y
yields the desired exact sequence 0 → Hn+1 π → (πn Y )π → Hn Y → Hn π → 0.

5.2(a): Let G = hS ; r1 , r2 , . . .i = F (S)/R where R is the normal closure in F (S) of the words ri .
Consider the abelianization map ri rj 7→ [ri ] + [rj ] from R to the relation
Qn module Rab with the denotation
[ri ] = ri mod[R, R]. Any element x ∈ R has a representation x = i=1 (fi ri±1 fi−1 ) where fi ∈ F = F (S).
Now F acts by conjugation on R and so induces an F -action on Rab , and R acts trivially on Rab
due to the definition of abelianization;P it is immediate that we obtain the (G = F/R)-action on Rab :
n
g · [ri ] = [f ri f −1 ]. Subsequently, [x] = i=1 (gi · [ri ]) and hence Rab is generated as a G-module by the
images of the presentation words.

5.2(b): Let Y = ( s S 1 ) ∪r1 e2 ∪r2 e2 ∪ · · · be the 2-complex associated to the given presentation
W

of G in part(a), and let Ỹ be its universal cover. Consider the augmented cellular chain complex
∂ ∂
(∗) 0 → C2 (Ỹ ) →2 C1 (Ỹ ) →1 C0 (Ỹ ) → Z → 0
L L
Note that C0 (Ỹ ) = ZG and C1 (Ỹ ) = |S| ZG and C2 (Ỹ ) = |R| ZG. By Proposition II.5.4[1] we have
the exact sequence
0 / Rab / C1 (Ỹ ) ∂1 / C0 (Ỹ ) /Z /0

13
and hence Rab = Ker∂1 is always in the chain complex for Ỹ . If Y is a K(G, 1) then Ỹ is acyclic and
(*) is exact by Proposition I.4.2[1], so C2 (Ỹ ) = Rab and hence Rab is a free ZG-module.
Now suppose Ỹ is not acyclic, so that Y is not a K(G, 1). Then H2 Ỹ is nontrivial (since Hi Ỹ = 0 for
i > 2 by (*) and H1 Ỹ = 0 by the simply-connected property of Ỹ ) which implies that the boundary map
∂2 is not injective, and by exactness we can refer to this non-injective map as C2 (Ỹ ) → Ker∂1 = Rab .
Therefore, there exists a nontrivial ZG-relation amongst the words ri in Ker∂2 , so {[ri ]} is not ZG-
independent in Rab and hence does not generate Rab freely.

5.2(c): Let G = hS; ri be an arbitrary one-relator group and write r = un ∈ F = F (S), where
n ≥ 1 is maximal. By a result of Lyndon-Schupp, the image t of u in G has order exactly n, and we let
C = hti = Zn . If n > 1 then the relation module Rab is not freely generated by r mod[R, R] since this
generator is fixed by C, but a result of Lyndon shows that no other relations hold (i.e. the projection
Z[G/C] → Rab is an isomorphism).
Let Y be a bouquet of circles indexed by S, and let Ỹ be the connected regular covering space of Y
corresponding to the normal subgroup R of F = π1 Y . Choosing a basepoint ṽ ∈ Ỹ lying over the
vertex of Y , we identify G with the group of covering transformations of Ỹ ; as explained in [1] on
pg15, Ỹ is a (1-dimensional) free G-complex. Since ũ ends at tṽ, the lifting r̃ is the composite path
tṽ o
ũ
ṽO
tũ tn−1 ũ

t2 ṽ / tn−1 ṽ
Thus the map S 1 → Ỹ corresponding to r̃ is compatible with the action of C, where C acts on S 1 as a
group of rotations (i.e. tk is multiplication by e2πik/n ). Consider the 2-complex X obtained by attaching
2-cells to Ỹ along the loops gr̃, where g ranges over a set of representatives for the cosets G/C. Given
a 2-cell σ, each g ∈ G sends ∂σ homeomorphically to ∂σ 0 for some 2-cell σ 0 , and so we just pick our
favorite extension g : σ → σ 0 . This G-action makes X a G-complex since the permutations of 2-cells are
determined by the permutations of their boundary loops. Thus, if σ is the 2-cell attached along r̃ then
only C ⊂ G will fix σ, since ti simply rotates the loop ∂σ (i.e. Gσ = C). Let Γ = ( s S 1 ) ∪r e2 be the
W
S
standard 2-complex associated to the presentation of G; its universal cover is Γ̃ = Ỹ g∈G σg where σg
is attached along gr̃, and C2 (Γ̃) = ZG. Thus X is the quotient of Γ̃ by identifying σg with σgti for all i
(for each g), and C2 (X) = Z[G/C] ∼ = Rab . If n = 1 then C = {1} and X = Γ̃. Lyndon’s theorem about
one-relator groups says that Rab is freely generated by the image of r, provided r is not a power, which
is equivalent to Γ being a K(G, 1) by part(b) above, and hence equivalent to X being contractible (X is
the Cayley complex associated to the presentation of G).

5.3(a): With G = F/R and following Kenneth Brown’s proof of Theorem II.5.3, let F = F (S), let
Y be a bouquet of circles indexed by S, and let Ỹ be the connected regular covering space of Y cor-
responding to the normal subgroup R of F = π1 Y . Choosing a basepoint ṽ ∈ Ỹ lying over the vertex
of Y , we identify G with the group of covering transformations of Ỹ . For any f ∈ F we regard f as
a combinatorial path in the CW-complex Y and we denote by f˜ the lifting of f to Ỹ starting at ṽ.
This path f˜ ends at the vertex f¯ṽ, where f¯ is the image of f in G. Define the function d : F → C1 Ỹ
by letting df be the sum of the oriented 1-cells which occur in f˜. Since the lifting of f1 f2 is the path
f˜1 f¯1 f˜2
ṽ −→ f¯1 ṽ −→ f¯1 f¯2 ṽ, we have d(f1 f2 ) = df1 + f¯1 df2 for all f1 , f2 ∈ F . Thus, if we regard the G-module
C1 Ỹ as an F -module via the canonical homomorphism q : F → G, then d is a derivation [since the
F -action is given by restriction of scalars: f1 · df2 = f¯1 df2 = q(f1 )df2 ].

5.3(b): For any free group F we can apply part(a) above with R = {1} to get the desired deriva-
tion d : F → Ω where G = F/1 = F and Ω = C1 Ỹ = ZF (S) which is the free module with basis (ds)s∈S
[note: ds−1 = −s−1 ds].
The above note is a result of d(1) = d(1 · 1) = d(1) + 1d(1) =P2d(1) ⇒ d(1) = 0.
We write the total free derivative df of f as the sum df = s∈S (∂f /∂s)ds, where ∂f /∂s ∈ ZF is the
partial derivative of f with respect to s [the coefficient of ds when df is expressed in terms of the basis
(ds)].
It is immediate that ∂/∂s : F → ZF is a derivation because d is a derivation: f f 0 7→ d(f f 0 ) =

14
 0 0 0
P P
P + f df = [(∂f /∂s) + f (∂f /∂s)]ds 7→ (∂f /∂s) + f (∂f /∂s). For t ∈ S we have dt = (∂t/∂s)ds =
df
s6=t 0ds + 1dt, and so ∂t/∂s = δs,t .

Example: S = {s, t} ⇒ ∂(ts−1 ts2 )/∂s = ∂ts−1 /∂s + ts−1 (∂ts2 /∂s) =
[∂t/∂s + t(∂s−1 /∂s)] + ts−1 [∂t/∂s + t(∂ss/∂s)] =
0 + t[−s (∂s/∂s)] + ts 0 + ts−1 t[∂s/∂s + s(∂s/∂s)] = −ts−1 1 + ts−1 t[1 + s1] = ts−1 ts + ts−1 t − ts−1
−1 −1

5.3(c): Consider any free group F = F (S) and derivation d : F → M where M is an F -module. By the
representation
P f = s±1 ±1
1 · · · sn and the definition of a derivation d(gh) = dg + gdh, we have the equation
df = s∈S ws ds through trivial induction on n, where ws = ∂f /∂s by definition of the partial derivative.
θ ∂ ε
5.3(d): Consider θ in the exact sequence 0 → Rab → ZG(S) → ZG → Z → 0 of G-modules, where
ZG(S) is free with
P basis (es )s∈S and ∂es = s̄ − 1 [bar denotes image in G], and consider ϕ : R → ZG
(S)

given by r 7→ s∈S (∂r/∂s)es where (∂r/∂s) is the image of ∂r/∂s under the canonical map ZF → ZG.
In order to show that θ is induced by ϕ we must verify exactness of the above sequence, and so we start
by calculating the partial derivatives of the representation r = sb11 · · · sbnn ∈ Rab (where si 6= sj6=i ). Since
Pb−1 Pb
∂sb /∂s = j=0 sj for b > 0 and ∂sb /∂s = − j=1 s−j for b < 0 and ∂(Asb )/∂s = A(∂sb /∂s) with s - A,
we obtain
( bi−1 Pbi −1 j
sb11 · · · si−1 j=0 si bi > 0
∂r/∂si = b1 bi−1 Pbi −j
−s1 · · · si−1 j=1 si bi < 0

Injectivity of ϕ|Rab follows immediately from the freeness of ZG(S) and the fact that any nontrivial r
P
has some nontrivial bi (hence ∂r/∂si 6= 0). It suffices to show that ∂ϕ(r) = (∂r/∂s)(s − 1) = 0 for
±bi−1 ±j ±b
±b1
r ∈ Rab . For a particular i, (∂r/∂si )(si −1) = s1 · · · si−1 ( si − si ) = s±b
1±j
· · · si−1i−1 (si±bi −1),
P P 1
1
b1 bi−1 b1 bi bi+1
and (∂r/∂si )(si − 1) + (∂r/∂si+1 )(si+1 − 1) = −s1 · · · si−1 + s1 · · · si si+1 [suppressing the ±]. Thus,
(∂r/∂s)(s − 1) = −1 + 0 + · · · + 0 + s̄b11 · · · s̄bnn = −1 + r̄ = −1 + 1 = 0, and exactness is satisfied.
P
If R is the normal closure of a subset T ⊆ F , then the projection ZG(T ) → Rab given by g · et 7→ g · [t]
and the above exact sequence provides us with a partial free resolution:
ZG(T )
∂2
/ ZG(S) ∂1 / ZG ε / Z /0
;

##- θ
Rab
where the matrix of ∂2 is the “Jacobian matrix” (∂t/∂s)t∈T,s∈S .

5.4: Sketch (via Ken Brown): Let G = F/R and use the same notation as in Exercise 5.3(a) above.
Consider the following chain map in dimensions ≤ 2 using the bar resolution B∗
B2
∂2
/ B1 ∂1
/ B0 ε /Z /0

Γ2 Γ1 Γ0
  
ZG(R) / C1 (Ỹ ) / C0 (Ỹ ) /Z /0
We have the identification C1 (Ỹ ) = ZG(S) = IR where I is the augmentation ideal of ZF (so IR is a
free G-module on the images s − 1), and ZG(R) is the free G-module with basis (er )r∈R which maps
onto Rab = H1 Ỹ ⊂ C1 (Ỹ ). The specific chain map is given by Γ2 : [g1 |g2 ] 7→ −g1 g2 · r(g1 , g2 ) and
Γ1 : [g] 7→ f (g) − 1 and Γ0 : [ ] 7→ 1, where r(g1 , g2 ) ∈ R and f (g) ∈ F such that f (g) = g ∈ G and
f (g)f (h) = f (gh)r(g, h). Applying the coinvariants functor, the group homomorphism C2 (G) → Rab
given by [g|h] 7→ r(g, h) mod[R, R] induces the isomorphism ϕ : H2 G → R ∩ [F, F ]/[F, R] by passage to
subquotients.
A specific chain map γ in the other direction is given by γ2 : r 7→ hγ1 (r − 1) and γ1 : s − 1 7→ [s̄] and
γ0 : 1 7→ [ ], where h : B1 → B2 is the contracting homotopy h(g · [h]) = [g|h]. Regarding C2 (G) as an
γ1 h
F -module via f · [g|h] = [f¯g|h], the map D : F ,→ P ZF → I → IR → B1 → B2 → (B2 )G ≡ C2 (G) is a
derivation such that Ds = [1|s̄]. Moreover, Df = s∈S [∂f /∂s|s̄], where the symbol [·|·] is Z-bilinear.
Then D|R : R → C2 (G) is a homomorphism (since R acts trivially on C2 (G)) which induces ϕ−1 by
passage to subquotients.

15
 Qn
Pn a1 , ¯. . . , an , b1 , .¯. . , bn ∈ F ¯such that
Let r = i=1 [ai , bi ] ∈ R. The formula ϕ−1 (r mod[F, R]) =
−1 ¯
i=1 {[Ii−1 |āi ] + [Ii−1 āi |b̄i ] − [Ii−1 āi b̄i āi |āi ] − [Ii−1 |b̄i ]}, where Ii = [a1 , b1 ] · · · [ai , bi ], is proven in the
universal example where F is the free group on a1 , . . . , an , b1 , . . . , bn and R is the normal closure of r.
Using the constructed formula for ϕ−1 and the product rule for derivations, the desired formula arises
from Dr = i Ii−1 · D[ai , bi ] and D[a, b] = [1|a] + [a|b] − [aba−1 |a] − [aba−1 b−1 |b].
P

5.5(a): With the presentation G = hs1 , · · · , sn | r1 , · · · , rm i we associate the 2-complex Y = ( s S 1 ) ∪r1

W
e2 · · · ∪rm e2 so that π1 Y ∼
= G. By computing the Euler characteristic χ(Y ) two different ways (by The-
orem 22.2[4]) we obtain the equation (−1)i rkZ (Hi Y ) = (−1)i ci , where rkZ is the rank and ci is the
P P
number of i-cells. Then 1 − rkZ (Gab ) + rkZ (H2 Y ) = 1 − n + m, and so rkZ (H2 Y ) = m − n + r where
r = rkZ (Gab ) = dimQ (Q ⊗ Gab ). Now H2 Y = Ker∂2 is a free abelian group (subgroup of cellular 2-chain
group), and by applying Theorem II.5.2[1] we get a surjection H2 Y → H2 G (from the exact sequence in
the theorem). Thus H2 G can be generated by m − n + r elements.

5.5(b): Since Gab = 0 and the number of generators equals the number of relations (in the finite
presentation), m − n + r = m − m + 0 = 0. Thus by part(a), H2 G can be generated by at most 0
elements, and so H2 G = 0.

5.5(c): Given Gab = 0, H2 G ∼ = Z2 ⊕ Z2 , and n as the number of generators, let m be the number
of relations in the presentation of G. Then H2 G is generated by the two elements (0, 1) and (1, 0), and
r = 0, so by part(a), 2 ≤ m − n + 0 ⇒ m ≥ n + 2. Therefore, any n-generator presentation must involve
at least n + 2 relations.

5.6(a): From the group extension 1 → N ,→ G  Q → 1 we have G/N ∼ = Q, and from Hopf’s formula
we have H2 G ∼= R ∩ [F, F ]/[F, R] and H2 Q ∼ = S ∩ [F, F ]/[F, S], where G = F/R and Q = F/S with
R ⊆ S ⊆ F (so N ∼ = S/R). (H1 N )Q ∼ = (Nab )Q = Nab /h{(q − 1) · n[N, N ]}i = Nab /h{gng −1 n−1 [N, N ]}i =
∼
(N/[N, N ])/([G, N ]/[N, N ]) = N/[G, N ], where the Q-action on Nab is induced by the conjugation
action of G on N , and the latter isomorphism follows from the Third Isomorphism Theorem. Now
[G/N, G/N ] = {g1 N g2 N g1−1 N g2−1 N } = {g1 g2 g1−1 g2−1 N } = [G, G]N/N , so we have H1 Q ∼ = Qab =
(G/N )/[G/N, G/N ] ∼ = G/(N [G, G]), where the latter isomorphism follows from the Third Isomorphism
Theorem.
Thus, the desired 5-term exact sequence is obtained by showing the exactness of the sequence
α β γ δ
R ∩ [F, F ]/[F, R] → S ∩ [F, F ]/[F, S] → N/[G, N ] → G/[G, G] → G/(N [G, G]) → 0
where γ and δ are induced by the injection and surjection of the group extension. From δ : g[G, G] 7→
g[G, G]N = gN [G, G] we have Kerδ = N/[G, G] and Imδ = G/(N [G, G]). From γ : n[G, N ] 7→ n[G, G]
we have Kerγ = N ∩ [G, G]/[G, N ] and Imγ = N/[G, G] = Kerδ. As deduced above, N/[G, N ] =
(S/R)/[F/R, S/R] ∼ = S/(R[F, S]) and so from β : s[F, S] 7→ s[F, S]R = sR[F, S] we have Kerβ =
R ∩ [F, F ]/[F, S] and Imβ = S ∩ [F, F ]/(R[F, S]) ∼ = N ∩ [G, G]/[G, N ] = Kerγ. Finally, from α :
r[F, R] 7→ r[F, S] we have Imβ = R ∩ [F, F ]/[F, S] = Kerβ.

5.6(b): Applying part(a) to the group extension 1 → R ,→ F  G → 1 we obtain the exact se-
quence
H2 F
α / H2 G β
/ (H1 R)G γ
/ H1 F δ / H1 G /0

∼
= ∼
= ∼
= ∼
=
   
0 R/[F, R] F/[F, F ] G/[G, G]
where the first vertical isomorphism follows from Example II.4.1[1] and the second vertical isomorphism
arises in the solution to part(a). By the First Isomorphism Theorem and exactness of the sequence,
H2 G ∼= H2 G/Kerβ ∼ = Imβ ∼ = Kerγ = R ∩ [F, F ]/[F, R].

5.7(a): S 3 is a closed orientable 3-manifold, and it has a group structure under quaternion multpli-
cation (S 3 < H as the elements of norm 1). The finite subgroup G of S 3 provides a multiplication action,
and it is free because the only solution in H to the equation gx = x for nontrivial x is g = 1. From
a result in the solution to Exercise I.4.2, the G-action is a “properly discontinuous” action and so the

16
quotient map ρ : S 3 → S 3 /G is a regular covering space.
Following a proof by William Thurston, S 3 /G is Hausdorff: considering two points α, β of S 3 in distinct
orbits, form respective neighborhoods Uα and Uβ such that they are disjoint and neither neighborhood
3
contains any translates of α or β (this can S be done by the Hausdorff property of S ). Taking the union
K of these two neighborhoods, Int(K − g6=1 gK) yields a neighborhood of α and a neighborhood of β
which project to disjoint neighborhoods of Gα and Gβ in S 3 /G [note: we needed to refine K because if
Uα intersects gUβ , then after projecting to the orbit space, ρ(gUβ ) = ρ(Uβ ) intersects ρ(Uα )].
S 3 /G is a closed connected 3-manifold since it has S 3 as a covering space (Theorem 26.5[6] provides the
compactness, Theorem 23.5[6] provides the connectedness, and the property of evenly-covered neighbor-
hoods provides the nonboundary and manifold structure).
Since the actions g : S 3 → S 3 are fixed-point free, we have deg(g) = (−1)3+1 = 1 by Theorem 21.4[4]
and hence G acts by orientation-preserving homeomorphisms [note: orientation is in terms of local ori-
entations µx ∈ H3 (S 3 , S 3 − x) ∼
= H3 (S 3 ) ∼
= Z, as defined in [3] on pg234]. Local orientations µGx = µρ(x)
3 3
of S /G = ρ(S ) are given by the images Γx (µx ) = Γgx (µgx ) under the isomorphisms of local homology
groups Γx : H3 (S 3 , S 3 − x) → H3 (ρ(S 3 ), ρ(S 3 ) − ρ(x)) which arise from the Excision Theorem and the
local-homeomorphism property of covering spaces; the ‘local consistency condition’ follows in the same
respect (where a ball B containing Gx and Gy has as preimage under ρ a union of balls, each of which is
homeomorphic to B, and such a homeomorphic ball containing g1 x and g2 y provides the local consistency
condition for S 3 ). Therefore, S 3 /G is orientable.
Since S 3 is simply-connected, π1 S 3 = 0 and so G ∼ = π1 (S 3 /G)/π1 (S 3 ) ∼
= π1 (S 3 /G). Applying Poincaré
Duality and the Universal Coefficient Theorem we obtain H2 (S /G) = H 1 (S 3 /G) ∼
3 ∼ = Hom(H1 (S 3 /G), Z) ∼
=
Hom(Gab , Z) = 0 [noting that G is finite]. A theorem of Hopf (Theorem II.5.2[1]) gives us an exact se-
quence which includes the surjection H2 (S 3 /G) → H2 G → 0, hence H2 G = 0.

5.7(b): The binary icosahedral group G (of order 120) is the preimage in S 3 which maps onto the
alternating group A5 under S 3 → SO(3) [up to isomorphism with the group of icosahedral-rotational
symmetries] as explained in [3] on pg75. By part(a), H2 G = 0. Consider the group extension 1 → K →
G → A5 → 1 where K is the central kernel of order 2 (corresponding to G mapping onto A5 ). The
associated 5-term exact sequence becomes 0 → H2 (A5 ) → (H1 K)A5 → 0 because H1 G ∼ = Gab = 0,
and thus we obtain the isomorphism H2 (A5 ) ∼ = (H1 K)A5 . The A5 -action on H1 K ∼ = Kab = K ∼= Z2
is induced by the conjugation G-action on K ∼ = Z2 which is the trivial action (since K ≤ Z(G)), and
therefore H2 (A5 ) ∼
= H1 K ∼
= Z2 .

5.7(c): Consider the abstract group G = hx, y, z ; x2 = y 3 = z 5 = xyzi which is a finite presenta-
tion with the same number of generators as relations. We show that G is perfect (G = [G, G]) so that
H1 G ∼= Gab = 0 and H2 G = 0 by Exercise 5.5(b) above. Now G/[G, G] is an abelian group with the
relations 2x̄ = 3ȳ = 5z̄ = x̄ + ȳ + z̄. From this we see that x̄ = ȳ + z̄, so we need not look at x̄.
Subsequently, 2ȳ + 2z̄ = 3ȳ ⇒ ȳ = 2z̄ and so we need not look at ȳ. Finally, 5z̄ = 3(2z̄) = 6z̄ ⇒ z̄ = 0
and so all generators vanish, i.e. G/[G, G] = 0.
Alternatively, the commutator subgroup is [G, G] ⊆ G, and to prove the opposite inclusion it suffices to
show that x, y, z all lie in [G, G] ≡ G0 .
z }| {
x2 = xyz ⇒ x = yz ⇒ x2 = yzyz = y 3 ⇒ z −1 yzy −1 = z −2 y ⇒ z 2 [z −1 , y] = y
z }| {
The two overbraced equations allow us to finish by showing that z ∈ G0 .
yzyz = y 3 ⇒ y 2 = zyz ⇒ [y, z] = z −1 yz −1 ⇒ y = z[y, z]z
We then have z = xyz ⇒ z 3 = xyz −1 = yz · z[y, z]z · z −1 = z[y, z]z · z 2 [y, z] ⇒ z 2 = [y, z]z 3 [y, z] =
5
z }| {
z 3 z −3 · [y, z]z 3 [y, z] ⇒ z −1 = g[y, z] ⇒ z = [y, z]−1 g −1 ∈ G0 with g = z −3 [y, z]z 3 ∈ G0 by normality of
the commutator subgroup.

With the cyclic subgroup C = hxyzi we have G/C = A5 and hence the group extension 1 → C → G →
A5 → 1 [A5 = hx, y, z ; x2 = y 3 = z 5 = xyz = 1i of order 60]. The associated 5-term exact sequence
now yields the isomorphism H2 (A5 ) ∼
= H1 C = C, noting the trivial A5 -action (since C is generated by a
central element) and noting the cyclicity Cab = C. Thus, by part(b) we deduce that |C| = 2 and hence
|G| = 2 · 60 = 120. In fact, G = SL2 (F5 ) is the binary icosahedral group!

17
6.1: Given N / G, let F be a projective resolution of Z over ZG and consider the complex FN . Since a
projective ZG-module is also projective as a ZH-module for any subgroup H ⊆ G (by Exercise I.8.2), F
is a projective resolution of Z over ZN and so H∗ (FN ) = H∗ N . By Exercise II.2.3(a), FN is a complex
of G/N -modules and so H∗ (FN ) inherits a G/N -action, with FN → FN given by x 7→ (gN )x = gx. For
Corollary II.6.3 we have the conjugation action α : N → N given by n 7→ gng −1 (for g ∈ G), and we have
the augmentation-preserving N -chain map τ : F → F given by x 7→ gx [it commutes with the boundary
operator ∂ of F since ∂ is equivariant, and it satisfies the condition τ (nx) = gnx = gng −1 gx = α(n)τ (x)].
By Proposition II.6.2[1], if α is conjugation by g ∈ N then H∗ (α) is the identity (hence trivial action),
and so τ 0 : FN → FN is given by τ 0 (x) = (gN )x = gx which agrees with the above map.

6.2: For any finite set A let Σ(A) be the group of permutations of A. For |A| ≤ |B|, choose an injection
i : A ,→ B and consider the injection Σ(A) ,→ Σ(B) obtained by extending a permutation on A to be the
identity on B − iA. In order to show that the induced map H∗ Σ(A) → H∗ Σ(B) is independent of the
choice of i, it suffices to show that any two injections i1 and i2 give conjugate maps ī1 , ī2 : Σ(A) ,→ Σ(B),
because the conjugation map Σ(B) → Σ(B) induces the identity map on homology H∗ Σ(B) → H∗ Σ(B)
by Proposition II.6.2[1]. Let τ be the permutation which takes i1 (a) to i2 (a) for all a ∈ A and is an
arbitrary permutation (B − i1 A) → (B − i2 A). Then for a permutation ī1 (σ̃) = σ ∈ Σ(B), the per-
mutation τ στ −1 is equal to ī2 (σ̃). Thus ī1 and ī2 are conjugates of each other by τ , and the result follows.

6.3(a): Given the homomorphism α : G → G0 , the n-tuples in Cn (G) are sent to the n-tuples in Cn (G0 )
coordinate-wise via α, where [gh] 7→ [α(gh)] = [α(g)α(h)]. Thus H1 (α) maps ḡ to α(g), where ḡ de-
notes the homology class of the cycle [g]. We also have the explicit isomorphism H1 G → Gab given by ḡ 7→
∼
g mod[G, G]. It is immediate that we have the commutative diagram H1 G
= / Gab

H1 (α) α∗
 ∼ 
H1 G0
= / G0
ab
∗ 0 0
with α : g mod[G, G] 7→ α(g) mod[G , G ], which is precisely the map obtained from α by passage to the
quotient. Thus, the isomorphism H1 ( ) ∼
= ( )ab is natural.

6.3(b): Suppose G = F/R and G0 = F 0 /R0 with F = F (S) and F 0 = F 0 (S 0 ) free, and suppose
α : G → G0 lifts to α̃ : F → F 0 . Let Y and Ỹ be associated to the presentation of G as in Exercise
II.5.3(a), and similarly for Y 0 and Ỹ 0 with G0 . Now α̃ yields a map Y → Y 0 that sends the combinatorial
path s = Ss1 to the combinatorial path α̃(s). ***Incomplete***

7.1: Consider G = G1 ∗A G2 with αk : A → Gk not necessarily injective, and let Ḡ1 = β1 (G1 ),
Ḡ2 = β2 (G2 ), Ā = β1 α1 (A) = β2 α2 (A) be the images of G1 , G2 , A in G [where βk : Gk → G arises from
the amalgamation diagram for G]. Form the amalgam H = Ḡ1 ∗Ā Ḡ2 and the commutative diagram:
Ā
i2
/ Ḡ2

i1 j2
  γ2
Ḡ1 /H
j1
ϕ

γ1 
,G
where γk is the natural inclusion and ik is the obvious injection [subsequently, we have γ1 i1 = γ2 i2 and
hence the unique map ϕ from the universal mapping property].

18
We also have the commutative diagram: A
α2
/ G2
r2
α1 β2
 
G1 /G Ḡ2
β1
φ
j2
r1

Ḡ1 /H
j1
where rk is βk with the codomain restricted to form the inclusion [subsequently, j1 r1 α1 (a) = j1 β1 α1 (a) =
j1 i1 β1 α1 (a) = j2 i2 β1 α1 (a) = j2 i2 β2 α2 (a) = j2 r2 α2 (a) and hence we have the unique map φ from the
universal mapping property].
From the diagrams (and dropping subscripts), φβ = jr and ϕj = γ.
ϕφ(β(g)) = γ(r(g)) = γ(β(g)) = β(g)
φϕ(j[β(g)]) = φγ[β(g)] = φ(β(g)) = j(r(g)) = j(β(g))
Thus ϕφ = idG and φϕ = idH , and so ϕ = φ−1 is an isomorphism (H ∼ = G).
Consequently, any amalgamated free product is isomorphic to one in which the maps A → Gk are injec-
tive.

7.2:

7.3: Consider the Special Linear Group SL2 (Z) ∼ = Z4 ∗Z2 Z6 which is the subgroup of all 2x2 inte-
gral matrices with determinant 1. Applying the Mayer-Vietoris sequence for groups and using the fact
that H∗ (Zk ) is trivial in positive even dimensions and is isomorphic to Zk in positive odd dimensions,
α β γ
we get the exact sequence 0 → H2n (SL2 (Z)) → Z2 → Z4 ⊕ Z6 → H2n−1 (SL2 (Z)) → 0.
Noting that the only nontrivial map ϕ : Z2 = hti → Z4 = hsi is the canonical embedding defined by
t 7→ s2 , we assert that the induced map under H2n−1 is the same embedding: Considering the two
periodic free resolutions of Z, there exists an augmentation-preserving chain map f between them by
Theorem I.7.5[1] and we have a commutative diagram
Z[Z2 ]
1+t
/ Z[Z2 ] t−1 / Z[Z2 ]

fn+1 fn
 1+s+s2 +s3
 
Z[Z4 ] / Z[Z4 ] s−1 / Z[Z4 ]
The left-side square yields (1+s+s2 +s3 )fn+1 (1) = fn (1+t) = fn (1)+ϕ(t)fn (1) = (1+s2 )fn (1), and by
exactness of the bottom row we have 0 = (s − 1)(1 + s2 )fn (1) = (s3 − s2 + s − 1)fn (1) ⇒ fn (1) = 1 + s,
hence (1 + s + s2 + s3 )fn+1 (1) = (1 + s2 )(1 + s) = 1 + s + s2 + s3 ⇒ fn+1 (1) = 1. Then after moving
to quotients, the cycle elements (for odd-dimensional homology) are mapped via ϕ∗ (1) = 1 + 1 = 2
while the boundary elements are mapped via ϕ∗ (1) = 1, and the result follows [applies to all odd n, and
f0 (1) = 1].
In general, an injection H ,→ G = H × K onto a direct summand will pass to an injection under
j
any covariant functor T because the composition identity H ,→ G  H yields the identity T (H) →
T (H × K) → T (H) which implies j is injective [this can be applied to H = Z2 and G = Z6 = Z2 × Z3 ].
Thus we have β(t) = (s21 , s32 ) injective and so H2n (SL2 (Z)) ∼
= Imα = Kerβ = 0. From the MV-sequence
we see that H2n−1 (SL2 (Z)) is a finite abelian group of order dividing |Z4 ⊕ Z6 | = 24 = 23 · 3, hence
contains only 2-torsion and 3-torsion by the Primary Decomposition Theorem. Considering the 3-torsion
in the MV-sequence, 0 → 0 → 0 ⊕ Z3 → H2n−1 (SL2 (Z))(3) → 0, we have H2n−1 (SL2 (Z))(3) = Z3 by
exactness [this row can be extracted from the MV-sequence because finitely generated abelian groups
are direct sums of their Sylow pi -subgroups (by the Primary Decomposition Theorem) and maps be-
tween the abelian groups will send primary components (the Sylow pi -subgroups) to respective primary
components]. For 2-torsion we consider the 2-torsion subgroup and its MV-sequence, and we obtain a
commutative diagram

19
 0 / Z2 / Z4 ⊕ Z2 / H2n−1 (Z4 ∗Z2 Z2 = Z4 ) ∼
= Z4 /0 /0

∼
= ∼
= ∼
= ψ ∼
= ∼
=
     
0 / Z2 / Z4 ⊕ Z2 / H2n−1 (SL2 (Z))(2) /0 /0
Then by the Five-Lemma, ψ is an isomorphism and so H2n−1 (SL2 (Z))(2) = Z4 .
Thus, H2n−1 (SL2 (Z)) ∼
= Z4 ⊕ Z3 ∼
= Z12 .


Z
 i=0
⇒ ∼
Hi (SL2 (Z)) = Z12 i odd

0 i even


20
3 Chapter III: Homology and Cohomology
with Coefficients
0.1: Let F be a flat ZG-module and M a G-module which is Z-torsion-free (ie. Z-flat), and consider the
tensor product F ⊗ M with diagonal G-action. Since (F ⊗ M ) ⊗G − = (F ⊗ M ⊗ −)G = F ⊗G (M ⊗ −),
it suffices to show that X = F ⊗G (M ⊗ −) is an exact functor so that F ⊗ M is ZG-flat (by Corollary
10.5.41[2]). But by the same corollary M ⊗ − is Z-exact (and a G-module) and F ⊗G − is ZG-exact, so
X is exact and the result follows.

0.2: Let F be a projective ZG-module and M a Z-free G-module, and consider the tensor product
F ⊗ M with diagonal G-action. Since HomG (F ⊗ M, −) ∼ = Hom(F ⊗ M, −)G ∼
= Hom(F, Hom(M, −))G ∼ =
HomG (F, Hom(M, −)) where the second isomorphism is adjoint associativity (Theorem 10.5.43[2]), it
suffices to show that X = HomG (F, Hom(M, −)) is an exact functor so that F ⊗ M is ZG-projective
(by Corollary 10.5.32[2]). But by the same corollary Hom(M, −) is Z-exact (and a G-module) and
HomG (F, −) is ZG-exact, so X is exact and the result follows.

G
P N : MG → M induced from
1.1(a): For a finite group G and a G-module M we have the norm map
the map M → M which is multiplication by the norm element N = g∈G g. Noting that N m = |G|m
for both m ∈ MG and m ∈ M G , we see that |G| · KerN = 0 (as N m = N m/∼ = 0 by definition of
kernel) and |G| · CokerN = 0 (as CokerN = M G /N M and N m modN M = 0).

1.1(b): Suppose M is an induced module (M = ZG ⊗ A) where A is an abelian group and G acts

by g · (r ⊗ a) = gr ⊗ a. Then MG = (ZG)G ⊗ A = Z ⊗ A and M G = (ZG)G ⊗ A = Z · N ⊗ A, where
N is the norm element. The norm map N : MG → M G is now given by z ⊗ a 7→ zN ⊗ a (for z ∈ Z)
which is clearly a bijection. It is an isomorphism because N [z1 ⊗ a1 + z2 ⊗ a2 ] = N z1 ⊗ a1 + N z2 ⊗ a2 =
z1 N ⊗ a1 + z2 N ⊗ a2 = N [z1 ⊗ a1 ] + N [z2 ⊗ a2 ].
L
1.1(c): Let M be a projective ZG-module;
L it is a direct summand of a free module F = L i ZG. By
application of part(b) above with A = i Z we see that N is an isomorphism for F because F = i ZG =
L
(ZG ⊗ Z) = ZG ⊗ (
L
Z). Now (M ⊕ N ) G
∼
= Z ⊗ ZG (M ⊕ N ) ∼
= (Z ⊗ ZG M ) ⊕ (Z ⊗ ZG N ) ∼
= M G ⊕ NG ,
i i
and (M ⊕ N )G = M G ⊕ N G under the coordinate-wise G-action (of F) since g · (m, n) = (g · m, g · n) =
(m, n) ⇒ g · m = m , g · n = n. Thus MG ⊕ NG ∼ = M G ⊕ N G , and since the norm map is bilinear we
∼
have MG = M . G

1.2: Using the standard cochain complex, an element of C 1 (G, M ) is a function f : G → M , and
under the coboundary map it is sent to (δf )(g, h) = g · f (h) − f (gh) + f (g). The kernel of this map
consists of functions which satisfy f (gh) = f (g) + g · f (h), and these are derivations, so Z 1 (G, M ) ∼
=
Der(G, M ). Since an element of C 0 (G, M ) is simply m ∈ M , and under the coboundary map it is sent
to (δm)(g) = g · m − m, which is a principal derivation, we have B 1 (G, M ) ∼ = PDer(G, M ). Thus,
H 1 (G, M ) ∼
= Der(G, M )/PDer(G, M ).
If G acts trivially on M then PDer(G, M ) = 0 and Der(G, M ) = Hom(G, M ), so H 1 (G, M ) ∼ = Hom(G, M ) =
Hom(Gab , M ) = Hom(H1 G, M ), where the second-to-last equality comes from the fact that any group
homomorphism from G to an abelian group factors through the commutator subgroup [G, G] by Propo-
sition 5.4.7[2]. In particular, H 1 (G) = 0 for any finite group G.

1.3: Let A be an abelian group with trivial G-action, and let F → Z be a projective resolution
of Z over ZG. Then F ⊗G A = (F ⊗ A)G ∼ = Z ⊗G (F ⊗ A), and since the diagonal G-action on
F ⊗ A is simply the left G-action on F (since G acts trivially on A), we can apply tensor associa-
tivity (Theorem 10.4.14[2]) to obtain Z ⊗G (F ⊗ A) ∼ = (Z ⊗G F ) ⊗ A ∼ = FG ⊗ A. Thus there is a
universal coefficient sequence 0 → Hn (G) ⊗ A → Hn (G, A) → TorZ1 (Hn−1 (G), A) → 0 by Proposition
I.0.8[1]. Also, HomG (F, A) = Hom(F, A)G ∼ = Hom(FG , A), where the last isomorphism arises because
(gu)(m) = g · u(g −1 m) = u(g −1 m) and so we must have g −1 m = m ∈ F for gu = u. Thus there is
also a universal coefficient sequence 0 → Ext1Z (Hn−1 (G), A) → H n (G, A) → Hom(Hn (G), A) → 0 by
Proposition I.0.8[1].

21
1.4(a): “Let f : C 0 → C be a weak equivalence between arbitrary complexes, and let Q be a non-
negative cochain complex of injectives. Then the map HomR (f, Q) : HomR (C, Q) → HomR (C 0 , Q) is a
weak equivalence.”
To prove this, note that the mapping cone C 00 = C ⊕ ΣC 0 of f is acyclic by Proposition I.0.6[1], where
(ΣC 0 )p = Cp−1
0
is the 1-fold suspension of C. The mapping cone of HomR (f, Q) is HomR (C 00 , Q) because
HomR (C 0 , Q)n ⊕ ΣHomR (C, Q)n = q HomR (Cq0 , Qq+n ) ⊕ q HomR (Cq , Qq+n−1 ) =
Q Q
0 0 −1
C, Q)n = HomR (C 00 , Q)n
Q
q HomR (Cq ⊕ Cq+1 , Qq+n ) = HomR (C ⊕ Σ

noting that C 0 ⊕ Σ−1 C = (Cq0 ⊕ Cq+1 ) = (Cq−1

0
⊕ Cq ) = ΣC 0 ⊕ C = C 00 . Thus it suffices to show that
HomR (C , Q) is acyclic (by Proposition I.0.6[1]), i.e., that Hn (HomR (C 00 , Q)) ≡ [C 00 , Q]n = 0 ∀ n ∈ Z.
00

By the uniqueness part of the result of Exercise I.7.4, [C 00 , Q]n ≡ [Σn C 00 , Q] is indeed 0, since any map
on Q is zero in negative dimensions (so all extensions off of that zero map are homotopy equivalent).

1.4(b): Let ε : F → Z be a projective resolution and let η : M → Q be an injective resolution.

By part(a) above and noting that ε is a weak equivalence (regarded as a chain map with M concentrated
in dimension 0), we have a weak equivalence HomR (F, Q) ← HomR (Z, Q). Similarly, by Theorem I.8.5[1]
we have a weak equivalence HomR (F, M ) → HomR (F, Q).
In particular, H ∗ (G, M ) = H ∗ (QG ) because H ∗ (G, M ) = H ∗ (HomG (F, M )) and H ∗ (HomG (F, M )) ∼
=
H ∗ (HomG (Z, Q)) = H ∗ (Hom(Z, Q)G ), noting that HomZ (Z, Q) ∼ = Q.

2.1: Given projective resolutions F → M and P → N of arbitrary G-modules M and N , there is

∼
=
an isomorphism of graded modules Γ : F ⊗G P → P ⊗G F given by f ⊗ p 7→ (−1)degf ·degp p ⊗ f , where
∼
we consider diagonal G-action on Fi ⊗ Pj = Pj ⊗ Fi [generally, degx = n for x ∈ Cn ]. If we show that Γ
(a degree 0 map) is a chain map, then it is a homotopy equivalence (hence a weak equivalence) and so
TorG ∼ G 0
∗ (M, N ) = H∗ (F ⊗G P ) = H∗ (P ⊗G F ) = Tor∗ (N, M ). Denote by d and d the boundary operators
0
of F and P , respectively, and denote by D and D the boundary operators of F ⊗G P and P ⊗G F ,
respectively. Then

D0 Γ(f ⊗ p) = D0 [(−1)degf ·degp p ⊗ f ] = (−1)degf ·degp (d0 p ⊗ f + (−1)degp p ⊗ df )

= (−1)degf ·degp d0 p ⊗ f + (−1)degp·(degf +1) p ⊗ df
and
ΓD(f ⊗ p) = Γ[df ⊗ p + (−1)degf f ⊗ d0 p]
= (−1)degp·(degf −1) p ⊗ df + (−1)degf +degf ·(degp−1) d0 p ⊗ f
= (−1)degp·(degf +1) p ⊗ df + (−1)degf ·degp d0 p ⊗ f
Thus D0 Γ = ΓD and so Γ is a chain map.
(This simultaneously provides a solution to Exercise I.0.5)

3.1: Let P be a projective R-module, and let C be a short exact sequence of S-modules which can
be regarded as R-modules via restriction of scalars. Since P is projective, HomR (P, C) is a short exact
∼
=
sequence, and so it suffices to show that the isomorphism of functors HomS (S ⊗R P, −) → HomR (P, −)
is natural [because then HomS (S ⊗R P, C) is a short exact sequence which implies that S ⊗R P is a
projective S-module]. Given a module homomorphism ψ : M → N , we must check commutativity of the
diagram
α /
HomS (S ⊗R P, M ) HomS (S ⊗R P, N )
ϕ1 ϕ2
 
HomR (P, M )
β
/ HomR (P, N )
where α and β are given by f 7→ ψ ◦ f , and ϕi (i = 1, 2) is given by f 7→ f ◦ i under the universal
mapping property with i : P → S ⊗R P , i(p) = 1 ⊗ p. Now ϕ2 [α(F )] = ϕ2 [ψ ◦ F ] = (ψ ◦ F ) ◦ i and
β[ϕ1 (F )] = β[F ◦ i] = ψ ◦ (F ◦ i) = (ψ ◦ F ) ◦ i. Therefore, ϕ2 α = βϕ1 and the result follows:
Extension of scalars takes projective R-modules to projective S-modules.

3.2: Let Q be an injective R-module, and let C be a short exact sequence of S-modules which can be re-
garded as R-modules via restriction of scalars. Since Q is injective, HomR (C, Q) is a short exact sequence,

22
 ∼
=
and so it suffices to show that the isomorphism of functors HomS (−, HomR (S, Q)) → HomR (−, Q) is
natural [because then HomS (C, HomR (S, Q)) is a short exact sequence which implies that HomR (S, Q)
is an injective S-module]. Given a module homomorphism ψ : M → N , we must check commutativity
of the diagram
α /
HomS (N, HomR (S, Q)) HomS (M, HomR (S, Q))
ϕ1 ϕ2
 
HomR (N, Q)
β
/ HomR (M, Q)
where α and β are given by f 7→ f ◦ ψ, and ϕi (i = 1, 2) is given by f 7→ π ◦ f under the universal
mapping property with π : HomR (S, Q) → Q , π(f ) = f (1). Now ϕ2 [α(F )] = ϕ2 [F ◦ ψ] = π ◦ (F ◦ ψ)
and β[ϕ1 (F )] = β[π ◦ F ] = (π ◦ F ) ◦ ψ = π ◦ (F ◦ ψ). Therefore, ϕ2 α = βϕ1 and the result follows:
Co-extension of scalars takes injective R-modules to injective S-modules.

3.3: Given S which is flat as a right R-module, let Q be an injective S-module, let C be a short exact
sequence of S-modules, and consider HomR (C, Q) where C and Q are regarded as R-modules via re-
striction of scalars. Since S is R-flat, S ⊗R C is a short exact sequence, and since Q is S-injective,
HomS (S ⊗R C, Q) is a short exact sequence. It suffices to show that the isomorphism of functors
∼
=
HomS (S ⊗R −, Q) → HomR (−, Q) is natural [because then HomR (C, Q) is a short exact sequence which
implies that Q is an injective R-module]. Given a module homomorphism ψ : M → N , we must check
commutativity of the diagram
α /
HomS (S ⊗R N, Q) HomS (S ⊗R M, Q)
ϕ1 ϕ2
 
HomR (N, Q)
β
/ HomR (N, Q)
where α is given by f 7→ f ◦ (S ⊗R ψ), β is given by f 7→ f ◦ ψ, ϕ1 is given by f 7→ f ◦ iN for the natural
map iN : N → S ⊗R N , and ϕ2 is given similarly by f 7→ f ◦ iM . Now ϕ2 [α(F )] = ϕ2 [F ◦ (S ⊗R ψ)] =
(F ◦ (S ⊗R ψ)) ◦ iM = F ◦ ((S ⊗R ψ) ◦ iM ) and β[ϕ1 (F )] = β[F ◦ iN ] = (F ◦ iN ) ◦ ψ = F ◦ (iN ◦ ψ). Also,
iN [ψ(m)] = 1 ⊗ ψ(m) = (S ⊗R ψ)(1 ⊗ m) = (S ⊗R ψ)[iM (m)]. Therefore, ϕ2 α = βϕ1 and the result
follows:
Restriction of scalars takes injective S-modules to injective R-modules if S is a flat right R-module.

3.4: Given S which is projective as a left R-module, let P be a projective S-module, let C be a short
exact sequence of S-modules, and consider HomR (P, C) where C and P are regarded as R-modules via
restriction of scalars. Since S is R-projective, HomR (S, C) is a short exact sequence, and since P is
S-projective, HomS (P, HomR (S, C)) is a short exact sequence. It suffices to show that the isomorphism
∼=
of functors HomS (P, HomR (S, −)) → HomR (P, −) is natural [because then HomR (P, C) is a short exact
sequence which implies that P is a projective R-module]. Given a module homomorphism ψ : M → N ,
we must check commutativity of the diagram
α /
HomS (P, HomR (S, M )) HomS (P, HomR (S, N ))
ϕ1 ϕ2
 
HomR (P, M )
β
/ HomR (P, N )
where α is given by f 7→ φ ◦ f for φ(g) = ψ ◦ g (g : S → M ), β is given by f 7→ ψ ◦ f , ϕ1 is given
by f 7→ πM ◦ f under the universal mapping property with π : HomR (P, M ) → M , πM (f ) = f (1),
and ϕ2 is given similarly by f 7→ πN ◦ f . Now ϕ2 [α(F )] = ϕ2 [φ ◦ F ] = πN ◦ (φ ◦ F ) = (πN ◦ φ) ◦ F
and β[ϕ1 (F )] = β[πM ◦ F ] = ψ ◦ (πM ◦ F ) = (ψ ◦ πM ) ◦ F . Also, πN [φ(f )] = πN [ψ ◦ f ] = (ψ ◦ f )(1) =
ψ[f (1)] = ψ[πM (f )]. Therefore, ϕ2 α = βϕ1 and the result follows:
Restriction of scalars takes projective S-modules to projective R-modules if S is a projective left R-module.

4.1: Let R = Z/nZ, and note that the ideals of R are the ideals Ix = (x) mod(n) for x|n ∈ Z by the 4th
Isomorphism Theorem. It suffices to show that every map ϕ : I → R extends to a map R → R so that R
is self-injective by Baer’s Criterion (Proposition III.4.1[1]). Given Ix and ϕ(x mod(n)) = r mod(n), we
can write n = yx so that ϕ(yx mod(n)) = 0. But ϕ(yx mod(n)) = yϕ(x mod(n)) mod(n) = yr mod(n)

23
and thus yr = mn = myx ⇒ r = xm. Then, since ϕ(x mod(n)) = [x mod(n)][m mod(n)] we can
extend ϕ to the domain R by setting ϕ(1 mod(n)) = m mod(n).
(a): Let A be an abelian group such that nA = 0, and let C ⊆ A be a cyclic subgroup of order
|C| = n. We can regard A as an R-module because xn = x0 ∈ R and R acts on A by xi · a = ia.
As C is a [self-injective] subgroup of A, we have an inclusion C ,→ A of an injective R-module into
an R-module. By definition of “injective module” (pg.782-783 of [5], statement XX.4.I1), every exact
sequence of modules 0 → Q → M → M 0 → 0 splits for injective Q, hence Q is a direct summand of M .
Therefore, C is a direct summand of A.
(b): Given A as above (i.e. an arbitrary abelian group of finite exponent), we regard A as an R-
module. Since n is minimal (to annihilate A) there exists an element of order n and hence a cyclic
subgroup C ∼ = R of order n in A. If we can show that A = A0 ⊕ A00 for A0 of smaller exponent then
by induction on n we have that A0 is a direct sum of cyclic groups; thus it remains to show that A00
is a direct sum of modules (each isomorphic to R) and is a direct summand of A. An application of
Zorn’s Lemma on the set of direct sums with summands in A isomorphic to R (noting that the set is
nonempty because it contains C) provides the maximal element A00 ; we can use this lemma because an
upper bound for any chain would be the direct sum of those elements (the direct sums) in that chain. A
ring is Noetherian iff every ideal is finitely generated (by Theorem 15.1.2[2]); thus Z is Noetherian (being
a Principal Ideal Domain). Now R = Zn is also Noetherian, because a quotient of a Noetherian ring by
an ideal is Noetherian (by Proposition 15.1.1[2]). It is a fact that a ringLis Noetherian iff an arbitrary
direct sum of injective modules (over that ring) is injective. Thus A00 = i R is R-injective, so A00 is a
direct summand of A, and A is a direct sum of cyclic groups.
This result is known as Prufer’s Theorem for abelian groups.

5.1: For any H-module M consider the G-module IndG

L
HM = g∈G/H gM where this equality fol-
lows from Proposition III.5.1[1]. The summand gM is a gHg −1 -module and hence gHg −1 is the isotropy
group of this summand in IndG G ∼ G
H M . By Proposition III.5.3[1], IndH M = IndgHg −1 gM .
In particular, by Proposition III.5.6[1] we have the K-isomorphism
−1 0 0 −1
L
IndK gHg ∼ L 0 IndK 0 gg H(gg ) 0
g∈E K∩gHg −1 ResK∩gHg −1 gM = gg ∈E K∩gg H(gg 0 )−1 ResK∩gg 0 H(gg 0 )−1 gg M

−1
gHg
Thus the K-module IndK
K∩gHg −1 ResK∩gHg −1 gM depends up to isomorphism only on the class of g ∈ E
in K\G/H.

5.2(a): For any H-module M and G-module N consider the tensor product N ⊗ IndG H M which has
the diagonal G-action. By Proposition III.5.1[1] and the fact that tensor products commute with direct
sums, we have N ⊗ IndG M ∼
= N ⊗ (
L
gM ) ∼
=
L
(N ⊗ gM ) which has N ⊗ M as a direct
H g∈G/H g∈G/H
G
summand in the underlying abelian group. Treating this as an H-module ResH N ⊗ M with a diago-
nal action so that H is its isotropy group, we have N ⊗ IndG ∼ G G
H M = IndH (ResH N ⊗ M ) by Proposition
III.5.3[1].
In particular, for M = Z we have N ⊗ Z[G/H] ∼ = IndG G
H ResH N .

5.2(b): For any H-module M and G-module N consider U = Hom(IndG H M, N ) which has the “di-
agonal” G-action given by (gu)(m) = g · u(g −1 m). By Proposition III.5.1[1] and the Q fact that the
Hom-functor “commutes” with direct sums/products, we have U ∼ = Hom( gM, N ) ∼
L
= Hom(gM, N ),
where the indices on the sum/product symbols are implicitly the coset representatives in G/H. Thus
U admits a direct product decomposition (πg : U  Hom(gM, N )). Using the denotation πg (u) = ug ,
we have [(πg g0 )(u)](m) = [πg (g0 u)](m) = g0 · ug (g0−1 m) = g0 · ug−1 g (m) = g0 · [πg−1 g (u)](m), and so
0 0

πg g0 ∼ πg−1 g , with m ∈ IndG

H M . This decomposition has U  Hom(M, N ) as one of the surjections
0

in the underlying abelian group. Treating this surjection as an H-module π1 : U  Hom(M, ResGHN)
so that H is its isotropy group, we have Hom(IndG ∼ G G
H M, N ) = CoindH Hom(M, ResH N ) by Proposition
III.5.8[1].
An analogous proof will provide Hom(N, CoindG ∼ G G
H M ) = CoindH Hom(ResH N, M ).

5.3: Let F be a projective G-module and M a Z-free G-module, and consider the tensor product F ⊗ M
with diagonal G-action. By Corollary III.5.7[1], ZG⊗M is a free G-module since M is free as a Z-module.

24
 L
Since F is projective, it is a direct summand of a free G-module F = i ZG, so that F =LF ⊕ K. As
the tensor product commutes with arbitrary direct sums (Corollary XVI.2.2[5]), F ⊗ M = i (ZG ⊗ M )
and hence is a free G-module. Finally, F ⊗ M = (F ⊕ K) ⊗ M = (F ⊗ M ) ⊕ (K ⊗ M ) and so F ⊗ M is
a direct summand of F ⊗ M , hence G-projective.
Note that this gives a new proof of the result of Exercise III.0.2 above.

5.4(a): If L |G : H| = ∞ then there are infinitely many distinct coset representatives in G/H. Now
IndG
H M = g∈G/H gM by Proposition III.5.1[1], and it has a transitive G-action which permutes the
PN
summands. Consider an arbitrary [nontrivial] element x = i=1 gi mi with all mi 6= 0 (this also refers
to the sum over all coset representatives with cofinitely many mi = 0). We may take the summand
entry g1 m1 of x and a summand g 0 M which doesn’t appear in the representation of x (i.e. g 0 6= gj for
1 ≤ j ≤ N ), and there then exists g 00 ∈ G such that g 00 · g1 m1 = g 0 m because the G-action is transitive
on the summands. Thus x is not fixed by G (since g 00 · x 6= x), and so (IndG G
H M ) = 0.
Note that if the index is finite, then this result does not hold. For instance, take 2Z ⊂ Z = hxi which
has index |Z : 2Z| = |Z2 | = 2. Then IndZ2Z M = M ⊕ xM where x is the coset representative of x(2Z)
which generates Z2 . Since x2i ·(m1 , xm2 ) = (x2i m1 , x(x2i m2 )), we must have (IndZ2Z M )Z ⊆ M 2Z ⊕xM 2Z .
Since x2j+1 · (m̄1 , xm̄2 ) = x · (m̄1 , xm̄2 ) = (m̄2 , xm̄1 ), we must have m̄1 = m̄2 and hence (IndZ2Z M )Z =
{(m̄, xm̄) | m̄ ∈ M 2Z }. For M = Z4 with 2Z-action defined as x2i · m = (−1)i m, the largest submodule
on which 2Z acts trivially [so that −m = m] is Z2 . Thus (IndZ2Z Z4 )Z 6= 0 is our desired example.

5.4(b):

5.4(c): Assume statement (i), so that there is a finitely generated subgroup G0 ⊆ G such that |G0 : G0 ∩
gHg −1 | = ∞ for all g ∈ G (with H ⊆ G). Using the analogue of Proposition III.5.6[1] for coinduction and
∼ L G0 gHg −1
0 ∼
G
passing to G0 -coinvariants, we obtain (ResG G0 CoindH M )G0 = ( g∈E CoindG0 ∩gHg −1 ResG0 ∩gHg −1 gM )G =
L G0 0
g∈E (CoindG0 ∩gHg −1 M )G where this latter isomorphism follows from commutativity of the tensor
0

product with direct sums (and E is the set of representatives for the double cosets KgH). It is a
fact that if G is finitely generated and |G : H| = ∞ then (CoindG HML)G = 0 for any H-module M
G
[part(b) above], so by statement (i) we have (ResG G0 Coind H M )G 0 = 0 = 0 for any H-module M .
Noting that restriction of scalars gives the action g 0 · x = ϕ(g 0 )x where ϕ is a map G0 → G, we have
(CoindG G G G
H M )G ⊆ (ResG0 CoindH M )G0 . Thus (CoindH M )G = 0 and (i) implies (ii).
G
Assume statement (ii), so that (CoindH M )G = 0 for all H-modules M . Then in particular (for M = Z)
there is only one element of (CoindG H Z)G , and that element must be zero, so (ii) implies (iii).
Assume statement (iii), so that the element of (CoindG H Z)G represented by the augmentation map
ε ∈ CoindG H Z is zero. We first note that [in general] if n̄0 = 0 ∈ NG for some n0 ∈ N then it is
also zero in NG0 for some finitely generated subgroup G0 ⊆ G; this is because NG = N/hgn − ni and so
n0 can be written as a finite Z-linear combination of elements of the form gn − n, which implies that
we can take G0 to be the subgroup generated by those specific g’s. In particular, ε̄ = 0 ∈ (CoindG H Z)G0
for some finitely generated subgroup G0 of G. Using the double coset formula (analogue of Proposition
III.5.6[1]) and treating Z and other modules appropriately over specific groups (to ignore restriction),
0
we must have ε̄g = 0 ∈ (CoindG G0 ∩gHg −1 Z)G0 for all g ∈ G, where εg denotes the component of the
augmentation map in the specific summand of coinduction. Now if |G0 : G0 ∩ gHg −1 | < ∞ then
0
∼ G0 ∼
CoindG 0 0 −1
], giving εg = g0 ∈K g 0 ⊗ εg (g 0 ) = ( g 0 ) ⊗ 1
P P
G0 ∩gHg −1 Z = IndG0 ∩gHg −1 Z = Z[G /G ∩ gHg
where K is the set of coset representatives for the quotient G0 /G0 ∩ gHg −1 . Then g 00 εg = εg ∀ g 00 ∈ G0 ,
0
so ε̄G 6= 0 ∈ (CoindG 0
G0 ∩gHg −1 Z)G0 . Thus we must have |G : G ∩ gHg
0 −1
| = ∞, so (iii) implies (i).

5.5: Let G be a finite group and let k be a field, and consider the free module kG. We have kG ∼ =
kG ⊗k k = IndG ∼ G
{1} k = Coind{1} k = Homk (kG, k), where the second-to-last equation follows from the ana-
logue over k of Proposition III.5.9[1] since |G| < ∞. Then HomkG (−, kG) ∼ = HomkG (−, Homk (kG, k)) ∼ =
Homk (−, k) where the last isomorphism follows from the universal property of co-induction. It is a fact
that every k-vector space is an injective k-module [if the vector space W with basis B is a subspace of a
vector space V , then we can extend B to a basis of V and then V = W ⊕ U where U is the vector space
spanned by the additional basis vectors extended from B]; thus k is injective ⇒ Homk (−, k) is exact ⇒
HomkG (−, kG) is exact ⇒ kG is self-injective as a kG-module.

25
A Noetherian ring is a commutative ring which satisfies the Ascending Chain Condition on ideals (i.e.
no infinite increasing chain of ideals), and any field k is Noetherian because the only ideals are {0} and
k (giving {0} ⊆ k) by Proposition 7.4.9[2]. Now kG is Noetherian (for G finite) because it is a finite-
dimensional k-vector space and so any infinite ascending chain of subspaces would require cofinitely many
of those subspaces to have dimension greater than |G|, a contradiction; since ideals of kG are necessarily
k-subspaces, the result follows. It is a fact that a ring R is Noetherian
L iff an arbitrary direct sum of
injective R-modules is injective. Thus the free module F = i kG is injective and so any projective
kG-module is kG-injective (because a projective module is a direct summand of a free module, and a
direct summand of an injective module is injective).
Assuming the claim is true that any kG-module is a submodule of a kG-projective module, then by
definition of “injective” it follows that any injective kG-module is a direct summand of a kG-projective
module, hence kG-projective; it remains to prove this claim. We have a canonical kG-module injective
map M → HomkH (kG, M ) where the kG-module M can be regarded as a kH-module by restriction of
scalars (see pg64 of [1]). But HomkH (kG, M ) = CoindG ∼ G
H M = IndH M = kG ⊗kH M , where this second-
to-last equation follows from the analogue over k of Proposition III.5.9[1] since |G| < ∞. Using L H = {1},
this says that M is a submodule of kG ⊗ M . But M is treated as a k-vector space (M ∼
k = i k), so
kG ⊗k M ∼ = i (kG ⊗k k) ∼
L L
= i kG; i.e. kG ⊗k M is a free [hence projective] kG-module.

6.1(a): We first note that an arbitrary direct sum of projective resolutions is projective, which follows
from the fact that an arbitrary direct sum of projective modules is projective and from the exactness of
the row for each summand.LWe thenLnote that homologyLcommutes L with direct sums, and this follows
from the obvious facts Ker( i di ) = i Ker(di ) and Im( i di ) = i Im(di ). From these two facts and
noting that the tensor product commutes with direct sums, we see that the Tor-functor commutes with
direct sums,
TorR
L L ∼ L ∼ L H∗ (M ⊗R Ni ) = L TorR (M, Ni ).
∗ (M, i Ni ) = H∗ (M ⊗R ( i Ni )) = H∗ ( i (M ⊗R Ni )) = i i ∗

For the amalgamation G = G1 ∗A G2 consider the short exact sequence of permutation modules 0 →
Z[G/A] → Z[G/G1 ] ⊕ Z[G/G2 ] → Z → 0. By Proposition III.6.1[1] we have the long exact sequence
/ Hn (G, Z[G/A]) / Hn (G, Z[G/G1 ] ⊕ Z[G/G2 ]) / Hn (G, Z) /

∼
= ∼
= ∼
=
  
Hn (A) Hn (G1 ) ⊕ Hn (G2 ) Hn (G)
where the vertical isomorphisms follow from the fact H∗ (H) = H∗ (G, Z[G/H]), and the middle isomor-
phim utilizes commutativity of direct sums which follows from above because H∗ (G, −) = TorG∗ (Z, −).
This long exact sequence is the Mayer-Vietoris sequence for the amalgam G.

6.1(b): For the amalgamation G = G1 ∗A G2 consider the short exact sequence of permutation modules
0 → Z[G/A] → Z[G/G1 ]⊕Z[G/G2 ] → Z → 0. Applying −⊗M still yields a short exact sequence because
all of the modules are free Z-modules (we consider the sequence of permutation modules as a free reso-
lution of Z, hence a homotopy equivalence by Corollary I.7.6[1]); this homotopy equivalence for Z gives
a homotopy equivalence for Z ⊗ M = M (since functors preserve identities), hence a weak equivalence
(which can be stated as a resolution). By Proposition III.5.6[1] we have Z[G/A] ⊗ M ∼ = IndG G
A ResA M ,
G G ∼ G
and by Shapiro’s Lemma we apply H∗ (G, −) to obtain H∗ (G, IndA ResA M ) = H∗ (A, ResA M ). Similar
results follow for the other modules, and so by Proposition III.6.1[1] the exact sequence (which resulted
from the sequence of permutation modules after application of − ⊗ M ) yields a long exact sequence [the
Mayer-Vietoris sequence for homology with coefficients]
Hn (A, ResG AM )
/ Hn (G1 , ResG G
G M ) ⊕ Hn (G2 , ResG M )
/ Hn (G, M )
1 2

Now consider the original sequence of permutation modules, but instead apply Hom(−, M ) which still
yields a short exact sequence (same reason as mentioned above). By the analogue over co-induction of
Proposition III.5.6[1] (or the result of Exercise III.5.2(a) above) we have Hom(Z[G/A], M ) ∼
= CoindG G
A ResA M ,
and by Shapiro’s Lemma we apply H (G, −) to obtain H (G, CoindA ResA M ) ∼
∗ ∗ G G ∗ G
= H (A, ResA M ). Similar
results follow for the other modules, and so by Proposition III.6.1[1] the exact sequence (which resulted
from the sequence of permutation modules after application of Hom(−, M ) yields a long exact sequence
[the Mayer-Vietoris sequence for cohomology with coefficients]

26
H n (G, M ) / H n (G1 , ResG n G
G1 M ) ⊕ H (G2 , ResG2 M )
/ H n (A, ResG
AM )
∂ ε
7.1(a): Consider the exact sequence · · · → C1 →1 C0 → M → 0, where each Ci is H∗ -acyclic. We
can apply the dimension-shifting technique using the short exact sequences 0 → Kerε → C0 → M → 0
and 0 → Ker∂i → Ci → Ker∂i−1 → 0 to obtain the isomorphism Hn (G, M ) ∼ = H1 (G, Ker∂n−2 ) ∼
=
Ker{H0 (G, Ker∂n−1 ) → H0 (G, Cn−1 )} = Ker{(Ker∂n−1 )G → (Cn−1 )G }. Now consider the diagram be-
low concerning CG
∂¯n+1 ∂¯n
(Cn+1 )G / (Cn )G / (Cn−1 )G
: 9
γ1 α β γ2

%% %%
(Zn )G (Zn−1 )G
with Zi = Ker∂i , noting that the composition βα is exact [right-exactness of (−)G on 0 → Ker∂n →
Cn → Ker∂n−1 → 0] and Kerγ2 ∼ = Hn (G, M ). Thus Im∂¯n+1 = Imα = Kerβ (since γ1 is surjective) and so
¯
Hn (CG ) = Ker∂n /Kerβ. Noting that (Zn−1 )G = (Cn )G /Kerβ by the First Isomorphism Theorem, take
the kernel of γ2 [denoted K/Kerβ] and take its preimage under β to obtain K in (Cn )G . Since K maps
to zero under γ2 β = ∂¯n we have K ⊆ Ker∂¯n . We also have Ker∂¯n ⊆ K [hence they are equal] because
by commutativity of the maps Ker∂¯n maps to K/Kerβ and hence lies in K. Thus Ker∂¯n /Kerβ = Kerγ2
and so H∗ (G, M ) ∼= H∗ (CG ).
δ δ
7.1(b): Consider the exact sequence 0 → M → C 0 → 0
C1 →
1
· · · , where each C i is H ∗ -acyclic. We
can apply the dimension-shifting technique using the short exact sequences 0 → M → C0 → Kerδ0 → 0
and 0 → Kerδi−1 → Ci → Kerδi → 0 to obtain the isomorphism H n (G, M ) ∼ = H 1 (G, Kerδn−2 ) ∼ =
0 0 G G
Coker{H (G, Cn−1 ) → H (G, Kerδn−1 )} = Coker{(Cn−1 ) → (Kerδn−1 ) }. Using a similar approach
as in part(a) above, we see that H ∗ (G, M ) ∼
= H ∗ (C G ).

7.2: This will reprove Proposition III.2.2[1] on isomorphic functors.

Method 1 : Let M be Z-torsion-free, so that M ⊗ − is an exact functor (M is Z-flat). Then
H∗ (G, M ⊗ −) is a homological functor because given a short exact sequence of modules C, M ⊗ C
is a short exact sequence and F ⊗G (M ⊗ C) is a short exact sequence of chain complexes (F is projec-
tive, hence flat), so the corresponding long exact homology sequence gives us the desired property (by
Lemma 24.1[4] and Theorem 24.2[4]). Similarly, TorG 0
∗ (M, −) = H∗ (F ⊗G −) is a homological functor,
0
where F → M is a projective resolution. Both functors are effaceable [erasible] in positive dimensions,
since the chain complexes F ⊗G (M ⊗ P ) and F 0 ⊗G P are exact for P projective. In dimension 0,
H0 (G, M ⊗ N ) ∼ = Z ⊗G (M ⊗ N ) ∼ = (M ⊗ N )G = M ⊗G N ∼ = TorG 0 (M, N ). Therefore, by Theorem
∼
III.7.3[1] we have an isomorphism of ∂-functors H∗ (G, M ⊗ −) = TorG ∗ (M, −). [The case for cohomology
is similar].
Method 2 : The chain complex associated to the group TorG 0
∗ (M, N ) is given by · · · → F0 ⊗G N →
0 0
M ⊗G N → 0, where F → M is a projective resolution. This can be rewritten as · · · → (F0 ⊗N )G → (M ⊗
N )G → 0 which yields TorG 0
∗ (M, N ) = H∗ (CG ), where C is the chain complex (Fi ⊗ N ). This is indeed an
exact sequence because the universal coefficient theorem yields H∗ (F ⊗N ) = H∗ (F 0 )⊗N = N (nontrivial
0

only in dimension 0). Now Fi0 is projective and hence a summand of a freeL module F = Fi0 ⊕ K = j ZG.
L
Then H∗ (G, F⊗N ) ∼ = H∗ (G, Fi0 ⊗N )⊕H∗ (G, K ⊗N ) and H∗ (G, F⊗N ) ∼ = j H∗ (G, ZG⊗N ) = 0, noting
that induced modules are H∗ -acyclic by Corollary III.6.6[1]. Thus H∗ (G, Fi0 ⊗ N ) = 0 and each Fi0 ⊗ N is
H∗ -acyclic. We can now apply Exercise III.7.1(a) which implies TorG ∼
∗ (M, N ) = H∗ (CG ) = H∗ (G, M ⊗N ).
[The case for cohomology is similar].

7.3: For dimension-shifting in homology, we can choose the induced module M = ZG ⊗Z M which
maps onto M by ϕ(r ⊗ m) = rm; it is an H∗ -acyclic module by Corollary III.6.6[1]. This map is Z-split
because it composes with the natural map i : M → M to give the identity, m 7→ 1 ⊗ m 7→ 1m = m.
For dimension-shifting in cohomology, we can choose the coinduced module M = HomZ (ZG, M ) which
provides the embedding M → M given by m 7→ (r 7→ rm); it is an H ∗ -acyclic module by Corollary
III.6.6[1]. This map is Z-split because it composes with the natural map π : M → M to give the identity,
m 7→ (r 7→ rm) 7→ [r 7→ rm](1) = 1m = m.

27
8.1: Let H be a central subgroup of G and let M be an abelian group with trivial G-action. Then
the isomorphism c(g) : (H, M ) → (gHg −1 , M ) becomes the identity on (H, M ) given by (h 7→ ghg −1 =
h, m 7→ gm = m). By Corollary III.8.2[1], this conjugation action of G on (H, M ) induces an action of
G/H on H∗ (H, M ) given by gH · z = c(g)∗ z. Letting α : H → gHg −1 = H denote the group map of
c(g), and letting F be a projective resolution of Z over ZG, we can choose the chain map τ : F → F
to be the identity, since it satisfies the condition τ (hx) = hx = α(h)x = α(h)τ (x). Thus the chain map
F ⊗H M → F ⊗H M given by x ⊗ m 7→ x ⊗ gm = x ⊗ m is the identity, and so the induced map c(g)∗ is
the identity on H∗ (H, M ) which gives the trivial G/H-action gH ·z = c(g)∗ z = z. Similarly, by Corollary
III.8.4[1], we have an induced action of G/H on H ∗ (H, M ) given by gH · z = (c(g)∗ )−1 z. Using the same
chain map τ , we have the cochain map HomH (F, M ) → HomH (F, M ) given by f 7→ [x 7→ gf (x) = f (x)]
which is the identity. Thus the induced map c(g)∗ is the identity on H ∗ (H, M ) which gives the trivial
G/H-action gH · z = (c(g)∗ )−1 z = z.
Alternatively, the trivial G/H-action follows immediately from the fact that functors preserve identities,
where H∗ and H ∗ are the functors in question and c(g) is the identity map in question.

8.2: Let α : H ,→ G be an inclusion, let M be an H-module, let i : M → IndG H M be the canon-

ical H-map i(m) = 1 ⊗ m, and let π : CoindG H M → M be the canonical H-map π(f ) = f (1). We
can take the chain map τ : F → F to be the identity (F is a free resolution of Z over ZG) since
τ (hx) = hx = α(h)x = α(h)τ (x) and F can be regarded as a free resolution over ZH.
Consider (α, i)∗ on the chain level, induced by the map F ⊗H M → F ⊗G IndG H M given by x ⊗ m 7→
x ⊗ (1 ⊗ m). This map is a homotopy equivalence because we can use the universal property of ten-
sor products to define its inverse x ⊗ (g ⊗ m) = xg ⊗ (1 ⊗ m) 7→ xg ⊗ m, the composite map being
x ⊗ (g ⊗ m) 7→ xg ⊗ m 7→ xg ⊗ (1 ⊗ m) = x ⊗ g · (1 ⊗ m) = x ⊗ (g ⊗ m). In particular we have a
weak equivalence which yields the isomorphism H∗ (H, M ) ∼ = H∗ (G, IndG
H M ) of Shapiro’s Lemma given
by (α, i)∗ .
Now consider (α, π)∗ on the cochain level, induced by the map HomH (F, M ) → HomG (F, CoindG HM)
given by [x 7→ f (x)] 7→ [x 7→ (g 7→ gf (x))] with g ∈ ZG. This map is a homotopy equivalence because
we can use the universal property of co-induction to define its inverse [x 7→ (g 7→ gf (x))] 7→ [x 7→
(1 7→ f (x))] = [x 7→ f (x)]. In particular we have a weak equivalence which yields the isomorphism
H ∗ (H, M ) ∼= H ∗ (G, CoindG ∗
H M ) of Shapiro’s Lemma given by (α, π) .

9.1: Considering homology, let x ⊗H m represent z ∈ H∗ (H, M ). Computing corG H z on the chain
level yields x ⊗H m 7→ x ⊗G m, while computing corG gHg −1 gz on the chain level yields gx ⊗gHg −1 gm 7→
gx ⊗G gm = x ⊗G m. Since the images are equal, corG G
gHg −1 gz = corH z.
Considering homology, let x ⊗G m represent z ∈ H∗ (G, M ). Computing resG gHg −1 z on the chain level
0 G
P
yields x ⊗G m 7→ 0 −1
g ∈gHg \G P g · (x ⊗ gHg −1 m). Computing g · res H z on the chain level yields
x⊗G m 7→ g0 ∈H\G g 0 ·(x⊗H m) 7→ g0 ∈H\G (gg 0 )·(x⊗gHg−1 m) = g00 ∈gHg−1 \G g 00 ·(x⊗gHg−1 m), where
P P

this last equality arises from g(Hg 0 ) = gHg −1 gg 0 = (gHg −1 )g 00 with g 00 = gg 0 being the coset representa-
tive. Since the images are equal (the sums are the same for all coset representatives), g·resG G
H z = resgHg −1 z.
Considering cohomology, let fG represent z ∈ H ∗ (G, M ). Computing g · resG H z on the chain level yields
[x 7→ fG (x)] = [x 7→ g −1 fG (gx)] 7→ [x 7→ g −1 fH (gx)] 7→ [x 7→ gg −1 fgHg−1 (g −1 gx)] = [x 7→ fgHg−1 (x)],
while computing resG gHg −1 z on the chain level yields [x 7→ fG (x)] 7→ [x 7→ fgHg −1 (x)]. Since the images
are equal, g · resH z = resG
G
gHg −1 z.
Considering cohomology, let f represent z ∈ H ∗ (H, M ). Computing corG H z on the chain level yields
0
corG
P
f 7→ g 0 ∈G/H g · f , where G acts diagonally on f ∈ Hom(F, M ). Computing gHg −1 gz on the
0 00
P P
chain level yields g · f 7→ g0 ∈G/gHg−1 (g g) · f = g00 ∈G/H g · f , where this last equality arises from
g 0 (gHg −1 )g = g 0 gH = g 00 H with g 00 = g 0 g being the coset representative. Since the images are equal
(the sums are the same for all coset representatives), corG G
gHg −1 gz = corH z.

9.2: The transfer map H1 (G) → H1 (H) can be regarded as a map of abelian groups Gab → Hab .
If ḡ denotes the representative of Hg then ρg = gḡ −1 , where ρ : G → H is the unique map of left H-sets
which sends ever coset representative to 1. Now the transfer map is induced by the composite chain map
tr
F (G)G → F (G)H → F (H)H , where the latter map concerns the chain map τ : F (G) → P F (H) given
by (g0 , g1 ) 7→ (ρg0 , ρg1 ). Using bar notation, this composite chain map is given by [g] 7→ g0 ∈E g 0 [g] 7→

28
 −1
τ (g 0 , g 0 g) = g0 ∈E (g 0 g 0 , ρ(g 0 g)) = g0 ∈E (1, ρ(g 0 g)) = g0 ∈E [ρ(g 0 g)], where E is a set of rep-
P P P P
g 0 ∈E
resentatives for the right cosets Hg 0 and we note then that g 0 = g 0 . Note that for homology classes,
[g
P1 ] + [g2 ] =0 [g1 g2 ] Q because of the boundary map ∂2 [g1 |g2 ] = [g2 ] − [g1 g2 ] + [g1 ]; thus (as a homology class)
0
g 0 ∈E [ρ(g g)] = [ g 0 ∈E ρ(g g)]. Using the isomorphism H1 (G) → Gab given by [g] 7→ g mod [G, G], the
transfer map Gab → Hab is computed as g mod [G, G] 7→ g0 ∈E g 0 g(g 0 g)−1 mod [H, H].
Q

Example: The transfer map Z → nZ is multiplication by n, since

x 7→ [1x(1x)−1 ][xx(xx)−1 ] · · · [xn−2 x(xn−2 x)−1 ][xn−1 x(xn−1 x)−1 ] = 1 · 1 · · · 1 · xn (1)−1 = xn , where x is
the generator of Z.

10.1: The symmetric group G = S3 on three letters is the group of order 3! = 6 whose elements are the
permutations of the set {1, 2, 3}. The Sylow 3-subgroup is generated by the cycle (1 2 3), and a Sylow
2-subgroup is generated by the cycle (1 2). Noting the semi-direct product representation S3 = Z3 o Z2
where Z2 acts on Z3 by conjugation, we have H ∗ (S3 ) = H ∗ (S3 )(2) ⊕ H ∗ (S3 )(3) ∼ = H ∗ (S3 )(2) ⊕ H ∗ (Z3 )Z2
by Theorem III.10.3[1]. Now S3 is the unique nonabelian group of order 6, so D6 ∼ = S3 and we can use
Exercise AE.9 which implies that the Z2 -action on H2i−1 (Z3 ) ∼ = H 2i (Z3 ) is multiplication by (−1)i (we
can pass this action to cohomology by naturality of the UCT). Thus H n (Z3 )Z2 is isomorphic to Z3 for
n = 2i where i is even, and is trivial for n odd and n = 2i where i is odd. Taking any Sylow 2-subgroup
H ∼= Z2 , Theorem III.10.3[1] states that H ∗ (S3 )(2) is isomorphic to the set of S3 -invariant elements of
∗
H (H). In particular we have the monomorphism H 2i−1 (S3 )(2) ,→ H 2i−1 (H) = 0, so H 2i−1 (S3 )(2) = 0.
gHg −1
An S3 -invariant element z ∈ H 2i (H) ∼ = Z2 must satisfy the equation resH K z = resK gz, where K
denotes H ∩ gHg −1 . If g ∈ H then gHg −1 = H and the above condition is trivially satisfied for all z
(hz = z by Proposition III.8.1[1]). If g ∈ / H then K = {1} because H is not normal in S3 and only
contains two elements, so the intersection must only contain the trivial element. But then the image of
both restriction maps is zero, so the condition is satisfied for all z; thus H 2i (S3 )(2) = Z2 . Alternatively,
a theorem of Richard Swan states that if G is a finite group such that Sylp (G) is abelian and M is a
∗ NG (Sylp (G))
trivial G-module, then Im(resG Sylp (G) ) = H (Sylp (G), M ) . It is a fact that NS3 (Z2 ) = Z2 (refer
to pg51[2]), so taking G = S3 and H = Syl2 (S3 ) ∼ = Z2 and M = Z we have Im(resSH3 ) = (Z2 )Z2 = Z2 in
the even-dimensional case. Since any invariant is in the image of the above restriction map (by Theorem
III.10.3[1]), the result H 2i (S3 )(2) = Z2 follows.


 Z n=0

∼
Z
6 n ≡ 0 mod 4 , n 6= 0
⇒ H n (S3 ) =


 Z 2 n ≡ 2 mod 4
0 otherwise


10.2(a): Let H be a subgroup of G of finite index, let C be the double coset HgH, and let T (C) be the en-
domorphism H ∗ (G, f (C)) of H ∗ (H, M ) where f (C) is the G-endomorphism of IndG H M given by 1 ⊗ m 7→
P −1 H gHg −1
c∈C/H c ⊗ cm. To show that T (C)z = corH∩gHg−1 resH∩gHg−1 gz, it suffices to check this equation
in dimension 0 (by Theorem III.7.5[1]). The right side maps m ∈ M H to
P
h∈H/H∩gHg −1 h(gm) =
0 H 0
P P
hg∈HgH/H (hg)m = g 0 ∈C/H g m ∈ M , where g = hg as a coset representative. Now T (C) in
dimension 0 is given by the composite map
α β f∗ β −1 α−1
H 0 (H, M ) → H 0 (G, CoindG 0 G 0 G 0
H M ) → H (G, IndH M ) → H (G, IndH M ) → H (G, CoindH M ) →
G
0
H (H, M )
P
where α is the Shapiro isomorphism m 7→ (s 7→ m), and β is the canonical isomorphism F 7→ x∈G/H x⊗
F (x−1 ), and f ∗ is induced by f (C), and α−1 is the inverse F 7→ F (1), and β −1 is the inverse x ⊗ m 7→
β −1 [x ⊗ m](s · x) which is sxm if sx ∈ H and is 0 if sx ∈
/ H. This composite is given by
T (C) : m 7→ (s 7→ m) 7→ x∈G/H x ⊗ m 7→ x c∈C/H x(c−1 ⊗ cm) = x c xc−1 ⊗ cm 7→
P P P P P
P P −1 −1
[xc ⊗ cm](s · xc−1 ) 7→ x c β −1 [xc−1 ⊗ cm](xc−1 ) ∈ M H
P P
x cβ

To simplify this last term, note that the image of β −1 [xc−1 ⊗ cm] is nontrivial iff xc−1 ∈ H ⇔ x ∈
C, and for each x there is at most one c such that xc−1 ∈ H. Thus the double sum reduces to

29
 xc−1 · cm =
P P
x∈G/H , x∈C x∈C/H xm, and this is precisely the image of the right-side map.
Note: α was determined by noting that any element f of (CoindG G
H M ) must satisfy f (xg) = f (x) and
hence f is determined by f (1) = m. So f is given by g 7→ m, but it must also commute with the H-action
which means that hg 7→ hm and hence hm = m, i.e. m ∈ M H . Thus α(m) = (s 7→ m), s ∈ G.

10.2(b): If z ∈ H ∗ (H, M ) is G-invariant where H ⊆ G is of finite index as above, then T (C)z =

gHg −1 −1
corH H H
H∩gHg −1 resH∩gHg −1 gz = corH∩gHg −1 resH∩gHg −1 z = |H : H ∩ gHg | = a(C)z, where the second-to-
last equality follows from Proposition III.9.5[1].

10.2(c): Let X = {z ∈ H ∗ (H, M ) | T (C)z = a(C)z ∀ C} where C is any H-H double coset
and a(C) = |C/H| = |H : H ∩ gHg −1 |. Since the image of the restriction map resG H lies in the
set of G-invariant elements of H ∗ (H, M ), and such elements lie in X by part(b) above, we have
Im(resGH ) ⊆ X. In the situation of Theorem III.10.3[1] and Proposition III.10.4[1], consider the ele-
ment w = corG n n
H z ∈ H (G, M ) where z is an arbitrary element of X. Then either H (H, M ) is anni-
n
hilated by |H| [H = Sylp (G)] in which case w ∈ H (G, M )(p) , or |G : H| is invertible in M [hence
gHg −1
in H n (G, M )]. Using Proposition III.9.5[1] we obtain resG H
P
Hw = g∈H\G/H corH∩gHg −1 resH∩gHg −1 gz =
−1
P P P
g∈H\G/H T (C)z = g∈H\G/H a(C)z = ( g∈H\G/H |H : H ∩ gHg |)z = |G : H|z. Since either
|G : H| is prime to p or is invertible in M , it follows that z = resH w where w0 = w/|G : H|. Thus
G 0

X ⊆ Im(resG H ) ⇒ X = Im(resH ).
G

30
4 Chapter IV: Low-Dimensional Cohomology
and Group Extensions
2.1: If d is a derivation [crossed homomorphism], then d(1) = d(1 · 1) = d(1) + 1 · d(1) = 2 · d(1) and so
d(1) = 0.

2.2: Let I be the augmentation ideal of ZG and let D : G → I be the derivation defined by g 7→ g−1 (this
is the principal derivation G → ZG corresponding to 1 ∈ ZG). Given any G-module A and any derivation
d : G → A, we can extend d to an additive map d¯ : ZG → A such that d(rs) ¯ ¯ · ε(s) + r · ds,
= dr ¯ where ε is
¯ ¯
the augmentation map and r, s ∈ ZG. This map is well-defined because d(rs · t) = d(rs) · ε(t) + rs · dt ¯ =
¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
dr · ε(s) · ε(t) + r · ds · ε(t) + rs · dt = dr · ε(st) + r · (ds · ε(t) + s · dt) = dr · ε(st) + r · d(st) = d(r · st),
and dg¯ = d(g1)
¯ = dg¯ · ε(1) + g · d1¯ = dg · 1 + g · d1 = dg + g · 0 = dg. The restriction f of d¯ to I is
a G-module homomorphism since f (r · s) = f (r) · ε(s) + r · f (s) = f (r) · 0 + r · f (s) = r · f (s), and
f (g − 1) = dg¯ − d1¯ = dg − d1 = dg − 0 = dg, so f is the unique module map I → A such that d = f D.
This means D is the universal derivation on G, and Der(G, A) ∼ = HomZG (I, A).

2.3(a): Let F = F (S) be the free group generated by the set S, and consider the F -module A with
a family of elements (as )s∈S . For the set map S → A o F defined by s 7→ (as , s), there is a unique
extension to a homomorphism ϕ : F → A o F by the universal mapping property of F . This is a splitting
π
of 1 → A → A o F → F → 1 because πϕ(f ) = π(df, f ) = f , where d : F → A is some function which
maps s ∈ S to as ∈ A. Since derivations F → A correspond to splittings of the above group extension,
d is the unique derivation such that ds = as ∀ s ∈ S.

2.3(b): Given any function φ : S → A where A is a ZF -module, there is a unique map d : F → A by

part(b) above, hence a unique ZF -module homomorphism f : I → A such that f D = d by Exercise
IV.2.2 above. In particular, I satisfies the universal property of free modules, φ = f D|S , and so the
augmentation ideal I of ZF is a free ZF -module with basis (Ds)s∈S = (s − 1)s∈S .
Note that this reproves Exercise II.5.3(b).
P
2.3(c): By part(b) above the universal derivation D : F → I satisfies Df = s∈S (∂f /∂s)ds, where
I is the augmentation ideal of ZF . By Exercise IV.2.2 any derivation d : F → M (where M is an
P-module) correspondsPto a unique F -module map ϕ : I → M and hence satisfies df = ϕ(Df ) =
F
s∈S (∂f /∂s)ϕ(Ds) = s∈S (∂f /∂s)ds, where ∂f /∂s lies in ZF .
Note that this reproves Exercise II.5.3(c).

2.4(a): Let G = F/R where F = F (S) and R is the normal closure of some subset T ⊆ F . For
any G-module A, derivations d : G → A correspond to splittings of 1 → A → A o G → G → 1; they are
of the form s(g) 7→ (dg, g) ∈ A o G. Consider the homomorphism ϕ : F → A o G given by f 7→ (df, g)
where g is the image of f under the projection map p : F → F/R, which is the extension of the set map
S → A o G, s 7→ (ds, p(s)), by the universal mapping property of F . Now r = f tf −1 ∈ R is mapped
to ϕ(r) = df + f · (dt + t · df −1 ) = df + f · dt − f tf −1 · df = (1 − r) · df + f · dt = dt, so ϕ induces a
homomorphism G → A o G iff dt = 0 ∀ t ∈ T [note: 1 − r = 1 − 1 = 0 when computing the G-action on
df ∈ A]. This homomorphism is a splitting iff d is a derivation, and so derivations G → A correspond to
derivations d : F → A such that d(T ) = 0.

2.4(b): From part(a) above and Exercise P IV.2.3(c) we see that derivations G → A correspond to
derivations d : F → A such that dt = s∈S (∂t/∂s)ds = 0 for all t ∈ T , where ∂t/∂s is the image of
∂t/∂s under ZF → ZG due to the G-action on A (restriction P of scalars). By Exercise IV.2.3(a) these
correspond to families (as )s∈S of elements of A such that s∈S (∂t/∂s)as = 0 for all t ∈ T .

2.4(c): The identity map idI : I → I [where I is the augmentation ideal of ZG] corresponds to a
derivation d : G → I such that ds̄ = idI D(s̄) = idI (s̄ − 1) P
= s̄ − 1 by Exercise IV.2.2, and this cor-
responds to a family (s̄ − 1)s∈S of elements of I such that s∈S (∂t/∂s)(s̄ − 1) = 0 for all t ∈ T by
∂ ∂
part(b) above. Thus there is an exact sequence ZG(T ) →2 ZG(S) →1 I where ∂2 et = s∈S (∂t/∂s)es and
P
∂1 es = s̄ − 1 (et and es are basis elements of their respective groups). Now ∂1 is surjective because

31
the elements s̄ − 1 generate I as a left ZG-ideal by Exercise I.2.1(b), where S̄ is a set of generators for
∂ ∂
G = F (S)/R, and so we have the desired exact sequence ZG(T ) →2 ZG(S) →1 I → 0.
Note that this reproves the second part of Exercise II.5.3(d).

2.4(d): Since the augmentation ideal I of ZF is free (by Exercise IV.2.3(b)), we have a free resolution
ε
0 → I → ZF → Z → 0 of Z over ZF . Taking R-coinvariants and noting that (ZF )R ∼ = Z[F/R] = ZG by
Exercise II.2.1, we obtain a complex IR → ZG → Z → 0 whose homology is H∗ R because H1 (R, Z) ∼ =
Ker(IR → ZG) by the dimension-shifting technique; ZF is an H∗ -acyclic module by Proposition III.6.1[1]
since it is free as a ZR-module by Exercise I.8.2, and it maps onto Z with kernel I. Since I is free with
ϕ ∂1
basis (s − 1)s∈S , I ∼= ZF (S) and we have the exact sequence 0 → H1 R → ZG(S) → ZG → Z → 0, where
we note that taking coinvariants is a right-exact functor.
Now we can map the standard (bar) resolution of Z over ZR to the aforementioned free resolution:
··· / F2 ∂2
/ F1 ∂1
/ F0 ∂0
/Z /0

φ2 φ1 φ0 id
   
0 /I i /
ZF
ε /Z /0
where φn>1 = 0, φ1 [r] = r − 1 = Dr, and φ0 [ ] = 1. This is a commutative diagram because
φ0 ∂1 [r] = φ0 (r[ ] − 1) = r − 1 = i(r − 1) = iφ1 [r] and φ1 ∂2 [r1 |r2 ] = φ1 (r1 [r2 ] − [r1 r2 ] + [r1 ]) =
r1 (r2 − 1) − (r1 r2 − 1) + r1 − 1 = r1 r2 − r1 − r1 r2 + 1 + r1 − 1 = 0. By applying the coinvariants functor
and noting that φ1 |R = D|R and H1 R = Ker(∂1 )R /Im(∂2 )R ∼ = Rab , we have a commutative diagram
D|R
R /I

 
H1 R
ϕ
/ ZG(S)
D
where the vertical arrows are quotient maps. The composite F → I → ZG(S) is a derivation such that
s 7→ es because s1 s2 7→ (s1 − 1) + s1 · (s2 − 1) 7→ es1 + s̄1 es2 . Thus the map ϕ is given by r mod [R, R] 7→
P θ (S) ∂ ε
s∈S (∂r/∂s)es , and we have the desired exact sequence 0 → Rab → ZG → ZG → Z → 0 where
P
∂es = s̄ − 1 and θ(r mod [R, R]) = s∈S (∂r/∂s)es .
Note that this reproves the first part of Exercise II.5.3(d).
π
3.1(a): Let 0 → A → E → G → 1 be an extension and let α : G0 → G be a group homomorphism, and
consider the pull-back (fiber-product) E ×G G0 = {(e, g 0 ) ∈ E × G0 | π(e) = α(g 0 )}. The kernel of the
canonical projection p : E ×G G0 → G0 corresponds to g 0 = 0 ⇒ α(g 0 ) = 0 = π(e) ⇒ Kerπ ∼ = A, and
p
thus we have an extension 0 → A → E ×G G0 → G0 → 1 which by definition fits into the commutative
diagram
0 /A /E π /G /1
O O
α

0 /A / E ×G G0 p / G0 /1
This extension is classified up to equivalence (by fitting into the above commutative diagram) because
given another such extension [corresponding to E 0 ] of G0 by A, commutativity of the right-hand square
implies there is a unique map φ : E 0 → E ×G G0 by the universal property of the pull-back, and this
gives commutativity of the right-half of the diagram below:
0 /A i1
/E π /G /1
O OX
ϕ α

0 /A i2
/ E ×G G0 p
/ G0 /1
O
φ

0 /A / E0
i3
/ G0 /1
Note that α for the E -extension yields the identity map G0 → G0 . It suffices to show that the left-side
0

32
of the diagram also commutes, for then we can apply the Five-Lemma which states φ is an isomorphism
(E 0 ∼
= E ×G G0 ). Now φi3 (a) = i2 (b) for some b ∈ A because i3 (a) maps to 0 ∈ G0 by exactness of the
bottom row and hence lies in the kernel of E ×G G0 which is contained in i2 (A). Then ϕφi3 (a) = ϕi2 (b),
and ϕφi3 (a) = i1 (a) by commutativity of the outer left-hand square while ϕi2 (b) = i1 (b) by commuta-
tivity of the top left-hand square. Thus i1 (a) = i1 (b) ⇒ a = b by injectivity of the inclusion, and this
yields φi3 (a) = i2 (a) which gives commutativity of the bottom left-hand square and completes the proof.
Therefore, α induces a map E(G, A) → E(G0 , A), and this corresponds to H 2 (α, A) : H 2 (G, A) →
H 2 (G0 , A) under the bijection of Theorem IV.3.12[1].
i p
3.1(b): Let 0 → A → E → G → 1 be an extension and let f : A → A0 be a G-module homomor-
phism, and consider the largest quotient E 0 of A0 o E such that the left-hand square in the following
diagram commutes:
0 /A i /E p /G /1

f φ
 0  0
0 / A0 i / E 0 p / G /1
Explicitly, E 0 = A0 o E/∼ with the equivalence relation (a01 , e1 ) ∼ (a02 , e2 ) iff a01 + f (a1 ) = a02 + f (a2 ) and
e1 − i(a1 ) = e2 − i(a2 ) for some a1 , a2 ∈ A; this relation is obviously reflexive and symmetric. It is tran-
sitive because if a01 + f (a1 ) = a02 + f (a2 ) and a02 + f (c) = a03 + f (a3 ), then a01 + f (a1 ) = a03 + f (a3 + a2 − c)
and e1 − i(a1 ) = [e3 − i(a3 ) + i(c)] − i(a2 ) = e3 − i(a3 + a2 − c). Define i0 by i0 (a0 ) = (a0 , 0) and define p0
by p0 (a0 , e) = p(e) and define φ by φ(e) = (0, e). The map p0 is well-defined because for (a01 , e1 ) ∼ (a02 , e2 )
we have p0 (a02 , e2 ) = p(e2 ) = p(e1 ) + p[i(a2 − a1 )] = p(e1 ) + 0 = p(e1 ) = p0 (a01 , e1 ). Now p0 (a0 , e) = 0 ⇒
p(e) = 0 ⇒ ∃ a | i(a) = e ⇒ (a0 , e) = (a0 , i(a)) ∼ (a0 + f (a), 0) = i0 (a0 + f (a)) ⇒ Kerp0 ⊆ Imi0 , and
p0 [i0 (a0 )] = p0 (a0 , 0) = p(0) = 0 ⇒ Imi0 ⊆ Kerp0 , so Kerp0 = Imi0 and the bottom row is an E 0 -extension.
This extension is classified up to equivalence (by fitting into the above commutative diagram) because
given another such extension [corresponding to E 00 ] of G by A0 , we get a diagram
0 /A i /E p /G /1

f φ
 0  p0
0 / A0 i / E0 /G /1
ϕ
 00   00
0 / A0 i / E 00 p / G /1
where E → E is the map Φ and the identity map A0 → A0 is induced from the f for the E 00 -extension.
00

We obtain an induced map ϕ : E 0 → E 00 given by ϕ(a0 , e) = i00 (a0 ) + Φ(e) which is well-defined be-
cause if (a01 , e1 ) ∼ (a02 , e2 ) then ϕ(a02 , e2 ) = i00 (a01 ) + i00 f (a1 ) − i00 f (a2 ) + Φ(e1 ) − Φi(a1 ) + Φi(a2 ) =
[i00 (a01 ) + Φ(e1 )] + i00 f (a1 ) − i00 f (a2 ) − i00 f (a1 ) + i00 f (a2 ) = ϕ(a02 , e2 ) + 0 + 0 = ϕ(a01 , e1 ), noting that
Φi = i00 f by commutativity of the outer left-hand square. The bottom right-hand square is commutative
because p00 ϕ(a0 , e) = p00 i00 (a0 ) + p00 Φ(e) = 0 + p(e) = p(e) = p0 (a0 , e). The bottom left-hand square [hence
the whole diagram] is also commutative because ϕi0 (a0 ) = ϕ(a0 , 0) = i00 (a0 ) + Φ(0) = i00 (a0 ) + 0 = i00 (a0 ).
We can now apply the Five-Lemma which states ϕ is an isomorphism (E 00 ∼ = E 0 ).
Therefore, f induces a map E(G, A) → E(G, A0 ), and this corresponds to H 2 (G, f ) : H 2 (G, A) →
H 2 (G, A0 ) under the bijection of Theorem IV.3.12[1].
i p
3.2(a): Let 0 → A0 → A → A00 → 0 be a short exact sequence of G-modules and let d : G → A00
be a derivation, and consider the set-theoretic pull-back E = {(a, g) ∈ A o G | p(a) = d(g)} where we
note that A o G = A × G as sets. If d¯ : E → A and π : E → G are the canonical projections, then
dπ[(a1 , g1 )(a2 , g2 )] = d(g1 g2 ) = d(g1 ) + g1 · d(g2 ) = p(a1 ) + g1 · p(a2 ) and the group law on E can be that
of the semi-direct product due to the agreement pd(a ¯ 1 +g1 ·a2 , g1 g2 ) = p(a1 +g1 ·a2 ) = p(a1 )+g1 ·p(a2 ) =
dπ(a1 + g1 · a2 , g1 g2 ). Thus E can be regarded as a subgroup of A o G, and d¯ is a derivation because
¯ 1 , g1 )(a2 , g2 )] = d(a
d[(a ¯ 1 + g1 · a2 , g1 g2 ) = a1 + g1 · a2 = d(a¯ 1 , g1 ) + g1 · d(a
¯ 2 , g2 ). Mimicking the proof
of Exercise IV.3.1(a) verbatim, for each derivation d there is an extension 0 → A0 → E → G → 1
characterized by the fact that it fits into a commutative diagram with derivation d¯

33
 0 / A0 i /A p
/ A00 /0
O O
d¯ d

0 / A0 j
/E π /G /1
This construction gives a map Der(G, A ) → E(G, A0 ).
00

3.2(b): What follows will be set-theoretic, and we use the same notation/maps as in part(a). A lifting of
d : G → A00 to a function l : G → A is given by l(g) = p−1 d(g), where p−1 (x) = 0 if x ∈ / Imp. This yields
a cross-section s : G → E of π given by s(g) = d¯−1 l(g), because πs = π d¯−1 l = π d¯−1 p−1 d = π(pd)
¯ −1 d =
−1 −1 −1
π(dπ) d = ππ d d = idG . Consider the diagram
Der(G, A00 ) / E(G, A0 )

∼
=
 
H 1 (G, A00 )
δ / H 2 (G, A0 )
where the top-horizontal map is the constructed map of part(a) and the left-vertical map is the natural
quotient; the isomorphism (right-vertical map) sends an extension to a factor set f : G × G → A0 . The
connecting homomorphism δ([d]) = [w2 ] is defined by lifting the element d of Z 1 (G, A00 ) = Der(G, A00 )
[equality comes from Exercise III.1.2] to an element w1 in C 1 (G, A) and then taking its image under
the coboundary map ∂ ∗ w2 and noting that it must be the image of some w2 ∈ Z 2 (G, A0 ) under the
injection i∗ : C 2 (G, A0 ) → C 2 (G, A); all of this is well-defined. From above, w1 in this scenario is l
and so ∂ ∗ l = L : G × G → A is defined by L(g, h) = gl(h) − l(gh) + l(g) via the equation on pg59[1].
Then we have i∗ (w2 ) = L, and w2 : G × G → A0 is a factor set which corresponds to a section S via
S(g)S(h) = j(w2 (g, h))S(gh) by the normalization condition on pg91[1]. I claim that S is the section
s defined above and that w2 is the factor set corresponding to s. Assuming this claim holds for the
moment, the connecting homomorphism sends the [class of the] derivation d to the factor set w2 which
is precisely the image of the extension under the vertical-isomorphism, and that extension is the image
of d under the top-horizontal map, so commutativity of the diagram is satisfied. Therefore, it suffices to
prove the claim, i.e. we must check the validity of the equation s(g)s(h) = j(w2 (g, h))s(gh). Note that
j ◦ w2 = d¯−1 L by commutativity of the left-side of the diagram in part(a), and s = d¯−1 l, so the equation
reduces to l(g) + [d¯−1 l(g)] · l(h) = L(g, h) + [d¯−1 L(g, h)] · l(gh) after we apply the derivation d¯ to both
sides. The action of E on A is given by e · a = π(e)a, so [d¯−1 l(g)] · l(h) = π[s(g)]l(h) = gl(h). Thus the
equation becomes l(g) + gl(h) = [gl(h) − l(gh) + l(g)] + [d¯−1 L(g, h)] · l(gh) ⇒ l(gh) = [d¯−1 L(g, h)] · l(gh).
It now suffices to show that the element π[d¯−1 L(g, h)] fixes l(gh), i.e. d¯−1 L(g, h) ∈ A0 ⊂ E. But this is
immediate because d¯−1 L(g, h) = j(w2 (g, h)) ∈ A0 where A0 has lazily been identified with j(A0 ).

3.3: Let G be a finite group which acts trivially on Z. For any homomorphism G → Q/Z we can
construct a central extension of G by Z by pulling back the canonical extension (the top row)
0 /Z /Q / Q/Z /0
O O

0 /Z /E /G /1
Thus we have a map ϕ : Hom(G, Q/Z) → E(G, Z), which by Exercise IV.3.2 can be identified with
the boundary map δ : H 1 (G, Q/Z) → H 2 (G, Z) since H 1 (G, Q/Z) ∼ = Hom(G, Q/Z) by Exercise III.1.2.
Now H1 G = Gab is finite, and so is H2 G by Exercise AE.16 [G is finite and Z is finitely generated];
thus Hom(H2 G, Q) = 0 = Hom(H1 G, Q) because there are no nontrivial homomorphisms from a finite
group into the infinite group Q. Also, Q is Z-injective and hence Ext1Z (H0 G, Q) = 0 = Ext1Z (H1 G, Q)
by Proposition 17.1.9[2]. The universal coefficient sequence of Exercise III.1.3 now yields H 1 (G, Q) =
0 = H 2 (G, Q); alternatively, we could have simply noted that H n (G, Q) = 0 ∀ n > 0 by Corollary
III.10.2[1] since |G| is invertible in Q. By the long exact cohomology sequence (Proposition III.6.1[1]), δ
is an isomorphism. Thus ϕ is a bijection.

3.4(a): Let E be a group which contains a central subgroup C ⊆ Z(E) of finite index n.
Method 1 : We have a central extension 1 → C → E → G → 1 by taking G = E/C. The nth
power map C → C (where C is considered as an abelian group) induces multiplication by n on H 2 (G, C)

34
because the cochain f : G2 → C is sent to nf : G2 → nC ⊆ C, and this is the zero-map because n = |G|
annihilates H 2 (G, C) by Corollary III.10.2[1]. Now Exercise IV.3.1(b) implies the above central extension
fits uniquely into a commutative diagram with the split extension [the trivial element in E(G, C)]
1 /C /E /G /1

n φ
 
1 /C / C ×G /G /1
∼
where we note that C o G = C × G because C is central. Looking at the left-hand square with E
restricted to C, commutativity implies φ(c) = (cn , 1), so the C-component of E → C × G gives us the
desired homomorphism E → C whose restriction to C is the nth power map.
Method 2 : The abelianization mapQρ : E → Eab composed with the transfer map tr : Eab → Cab = C
is given by ϕ : e 7→ e mod [E, E] 7→ g∈C\E ge(ge)−1 . Now |C\E| = n and c ∈ C commutes with all
elements of E, so ϕ(c) = g∈C\E gc(gc)−1 = cn g(cg)−1 = cn gg −1 = cn . Thus ϕ = tr ◦ ρ : E → C
Q Q Q
is the desired homomorphism whose restriction to C is the nth power map.

3.4(b): Given a finitely generated group E, suppose the commutator subgroup [E, E] of E is finite.
Then there are finitely many nontrivial elements g −1 (ege−1 ) where g is a generator of E (and e ∈ E
is arbitrary), so there are only finitely many conjugates ege−1 of each generator g of E. Consider the
inner automorphism group Inn(E) ∼ = E/C, where C = Z(E) is the center of E. An arbitrary element is
a function fe (x) = exe−1 [with e ∈ E fixed] which is determined by where it sends the generators of E.
Since there are finitely many generators and finitely many conjugates of each generator, there are only
finitely many non-identity maps fe ∈ Inn(G). Thus |Inn(G)| is finite, and so the center C of E has finite
index.
Conversely, suppose the center C of the finitely generated group E has finite index. Then part(a) above
n
gives us a homomorphism ϕ : E → C such that ϕ|C is the nth power map C → C. It suffices to show
that Kerϕ is finite, for then E/Kerϕ is abelian (being isomorphic to a subgroup of the abelian group C
by the First Isomorphism Theorem) and hence [E, E] ⊆ Kerϕ by Proposition 5.4.7[2], so the commutator
subgroup [E, E] of E is finite. Now ϕ is the C-component of the map φ : E → C × G given in part(a),
and the kernel of φ is contained in the kernel of the projection π : E → G by commutativity of that
n
diagram in part(a), so Kerϕ ⊆ Kerφ ⊆ Kerπ = C and hence Kerϕ ⊆ Ker(C → C). It suffices to show
n
that C ⊆ E is a finitely generated abelian group, for then Ker(C → C) is a subgroup of finite exponent
in a finitely generated abelian group (hence finite by the Fundamental Theorem of Finitely Generated
Abelian Groups). The Nielsen-Schreier Theorem (Theorem 85.1[6]) states that every subgroup of a free
group is free. The Schreier Index Formula (Theorem 85.3[6]) states that for a free group F of finite
rank with a subgroup H of finite index, rkZ H = |F : H|(rkZ F − 1) + 1. The finitely generated group
E has the presentation F/R with F free of finite rank, and C ⊆ F corresponds bijectively to some
H ⊆ F with H/R = C by the Lattice [4th] Isomorphism Theorem; H is free and finitely generated
by the Nielsen-Schreier Theorem and the Schreier index formula. Since quotients of finitely generated
groups are finitely generated (the generators of the quotient are the images of the generators under the
projection), C is finitely generated (and abelian since it lies in the center of E).

3.5(a): Let E be a group which contains an infinite cyclic central subgroup of finite index.
Method 1 : The group E of an extension in E(G, Z) gives us a homomorphism ϕ : G → Q/Z by Exercise
IV.3.3 (where G = E/Z), along with a map φ : E → Q which is injective on Z ⊆ E. Now the nontrivial
finitely generated subgroups of Q are of the form m m ∼
n Z with m, n ∈ N, so Imφ = n Z = Z [note: E is finitely
generated because both Z and E/Z are]. Thus we have a surjective map φ̃ : E → Imφ ∼ = Z which is simply
φ rephrased. Its kernel F := Kerφ̃ = Kerφ is finite because |F ∩Z| = 1 and |F Z/Z| = |F : F ∩Z| = |F | by
the 2nd Isomorphism Theorem, where we note that F Z/Z ⊆ G is finite [F Z is a subgroup of E because
φ̃
F / E]. We then have an extension 0 → Kerφ̃ → E → Z → 0 which must split (by Exercise AE.27)
because Z is free. Thus E ∼ = F o Z with F finite.
Method 2 : By Exercise IV.3.4(a) we have a map φ : E → Z which has a finite kernel as proved in Ex-
ercise IV.3.4(b). Letting φ̃ : E → φ(E) be the surjective map induced from φ, we have φ(E) = mZ ∼ =Z
for some m ∈ N and so φ̃ : E → Z is a surjective map with finite kernel. We then have an extension
0 → Kerφ̃ → E → Z → 0 which must split (by Exercise AE.27) because Z is free. Thus E ∼ = F o Z with

35
F finite (taking F = Kerφ̃ = Kerφ).

3.5(b): Let E be a torsion-free group which has an infinite cyclic subgroup Z of finite index. Let
E act by left multiplication on the finite set T of left cosets of Z in E and let πZ : E → S|T | be the asso-
ciated permutation representation
T afforded by this action. By Theorem 4.2.3[2], the kernel of the action
is the core subgroup KerπZ = e∈E eZe−1 ≡ A and so A ⊆ Z is a normal subgroup of E of finite index
(A is necessarily nontrivial and hence infinite cyclic). The group action of E on A ∼ = Z by conjugation
is a homomorphism E → Aut(Z) = {±1}, and the kernel is a subgroup E 0 ⊆ E of index 1 or 2 such
that A ⊆ Z(E 0 ). By part(a) we have E 0 ∼= F o A with F finite, but since E is torsion-free, F = {1} and
hence E 0 ∼
= Z. We therefore have an extension 0 → Z → E → G → 1 with |G| ≤ 2 (take G = E/E 0 ). If
G = {1} we are done. If G = Z2 acts non-trivially on Z then H 2 (G = Z2 , Z) = ZZ2 /N Z = 0/N Z = 0,
where N ∈ ZG is the norm element. In view of Theorem IV.3.12[1] we see that the extension splits (so
E∼ = Z o G), contradicting the assumption that E is torsion-free. Hence G acts trivially, so Z is central
in E and E ∼ = Z by part(a).

3.6: Let E be a finitely generated torsion-free group which contains an abelian subgroup of finite
index. This subgroup is isomorphic to Zn for some n by the Fundamental Theorem of Finitely Gen-
erated Abelian Groups. Note that Zn is a polycyclic group because it is solvable (it has the series
1 / Z / Zn with Zn /Z = Zn−1 abelian) and every subgroup is finitely generated; an equivalent defi-
nition of a polycyclic group is that it has a subnormal series with each quotient cyclic (so for Zn we
have the series 1 / Z / Z2 / · · · / Zn−1 / Zn with each quotient Zi /Zi−1 = Z cyclic). Thus E is a
virtually polycyclic group because it has a polycyclic subgroup of finite index [note: E is also virtu-
ally abelian]. Let E act by left multiplication on the finite set T of left cosets of Zn in E and let
πZn : E → S|T | be the associated permutation T representation afforded by this action. By Theorem
4.2.3[2], the kernel of the action is KerπZn = e∈E eZn e−1 =: A and so A ⊆ Zn is a normal subgroup
of E of finite index; A is necessarily nontrivial and hence isomorphic to Zn because it is a subgroup
of Zn of finite index (|E : Zn | = |(E/A)/(Zn /A)| is finite and |E : A| is finite). We therefore have an
p
extension 0 → Zn → E → G → 1 with G finite. The group action of G on Zn is a homomorphism
n ∼
ρ : G → Aut(Z ) = GLn (Z), and GLn (Z) contains only integral matrices with determinant ±1 as de-
duced from the surjective map det : GLn (Z) → Z× = {±1} or from the fact that an integral matrix
is invertible iff its determinant is a unit in Z. Consider the finite kernel K := Kerρ and its preimage
E 0 := p−1 (K) ⊆ E under p, and note that we have E 0 / E by the Lattice Isomorphism Theorem because
E 0 /Zn = K / G = E/Zn . Now this torsion-free group E 0 is finitely generated because its subgroup Zn
and its quotient K are both finitely generated [if e ∈ E 0 − Zn then it can be written as a finite sum with
generators of K, and if e ∈ Zn then it can be written as a finite sum with generators of Zn ]. Also, the
corresponding K-action on Zn is trivial (as K is the kernel of the G-action) and hence Zn lies in the
center of E 0 , so by Exercise IV.3.4(b) the commutator subgroup [E 0 , E 0 ] is finite. But the only finite
subgroup of a torsion-free group is the trivial group, so [E 0 , E 0 ] = 0 and Eab
0
= E 0 /[E 0 , E 0 ] = E 0 (i.e. E 0
n
is abelian, hence isomorphic to Z by the Fundamental Theorem of Finitely Generated Abelian Groups,
noting that |E 0 : Zn | = |K| is finite). Letting F = E/E 0 ∼
= (E/Zn )/(E 0 /Zn ) = G/K where the isomor-
phism follows from the 3 Isomorphism Theorem, we have a group extension 0 → Zn → E → F → 1
rd

coupled with the faithful group action F ,→ GLn (Z); this action is faithful because we modded the map
ρ by its kernel and injected A ∼ = Zn into E 0 ∼= Zn .
Since |F | = r annihilates H (F, Z ) by Corollary III.10.2[1], the rth power map Zn → Zn induces the
2 n

zero-map on H 2 (F, Zn ). Exercise IV.3.1(b) then implies the above extension fits uniquely into a com-
mutative diagram with the split extension [the trivial element in E(F, Zn )]
0 / Zn i1
/E p1
/F /1

r φ
 
0 / Zn i2
/ Zn o F p2 / F /1
The map φ is injective because if φ(e) = 0 then p1 e = 0 ⇒ ∃ z | i1 z = e ⇒ i2 (rz) = φ(i1 z) = 0 ⇒
rz = 0 ⇒ z = 0 ⇒ e = i1 z = i1 0 = 0. Alternatively, we could simply note that since the r-map is
injective, Kerφ ∩ Zn = {1} and hence Kerφ injects into F , which means Kerφ is trivial by commutativity
of the right-side diagram.
A crystallographic group is a discrete cocompact subgroup of the group Rn o On of isometries of some

36
Euclidean space; in general, V ⊆ W is cocompact if W/V is compact. Note that Rn has the usual
topology (its basis consists of the open n-balls), and the relative topology for the subspace Zn is the
discrete topology (any point x ∈ Zn is equal to the intersection Zn ∩ Bx where Bx is the ball of radius
1 T
3 centered at x). The orthogonal group On = {M ∈ GLn (R) | M M = 1} has the relative topology
induced from the matrix group Mn (R) ∼
2
= Rn .
It is a fact that a faithful action F ,→ GLn (R) can also be considered as a faithful action F ,→ On ,
so in our case we have an injection F ,→ On since GLn (Z) ⊂ GLn (R). The product topology on
Zn o F ⊂ Rn o On is the discrete topology [thus Zn o F is discrete] because Zn has the discrete topology
as mentioned above and the relative topology on F is discrete since F is finite. Since a subgroup of a
discrete group is discrete (the relative topology induced from the discrete topology is discrete), E can
be embedded (by φ) as a discrete subgroup of Rn o On .
The Zn -action on Rn given by translation x 7→ x + z is properly discontinuous, so the quotient map
p : Rn → Rn /Zn ∼ = Tn is a regular covering map (see Exercise I.4.2) where Tn ≡ S 1 × · · · × S 1 is the
compact n-torus. Now On is a compact space by the Heine-Borel Theorem because it is a closed bounded
2
subspace of Rn (see pg292[3]), so Zn is a cocompact subgroup of Isom(Rn ) := Rn o On because the
quotient T o On is compact. But Zn is a subgroup of E of finite index, so E is also cocompact. This
n

completes the proof that E is a crystallographic group.

3.7(a): Let G be a perfect group (so H1 G = 0) and let A be an abelian group with trivial G-action.
The universal coefficient sequence of Exercise III.1.3 then implies H 2 (G, A) ∼
= Hom(H2 G, A).

3.7(b): Let G be a perfect group and let A be any abelian group with trivial G-action. Yoneda’s lemma
from Exercise I.7.3(a) states that a natural transformation ϕ : Hom(H2 G, −) → H 2 (G, −) is determined
by where it sends idH2 G ∈ Hom(H2 G, H2 G), and so for the isomorphism ϕ of part(a) there is an element
u ∈ H 2 (G, H2 G) such that ϕ(idH2 G ) = u. Now for any v ∈ H 2 (G, A) there is a unique map f : H2 G → A
such that ϕ(f ) = v because ϕ is an isomorphism, and Yoneda’s lemma gives ϕ(f ) = H 2 (G, f )u. Thus
u is the “universal” cohomology class of H 2 (G, H2 G), in the sense that for any v ∈ H 2 (G, A) there is a
unique map f : H2 G → A such that v = H 2 (G, f )u.

3.7(c): In view of Theorem IV.3.12[1], part(b) can be reinterpreted as saying that the [perfect] group
π
G admits a “universal central extension” 0 → H2 G → E → G → 1 characterized by a certain property.
π0
This property states that given any abelian group A and any central extension 0 → A → E 0 → G → 1,
there is a unique map f : H2 G → A such that the extension is the image of the universal extension under
H 2 (G, f ) = E(G, f ). By Exercise IV.3.1(b) this latter part means there is a map E → E 0 making the
following diagram commute
E
π /G
>

 π0
0
E
and we assert that this map is unique. Indeed, two such maps h1 , h2 : E ⇒ E 0 which induce the same
map f must differ by a homomorphism φ : E → A because π 0 h1 (e) = π(e) = π 0 h2 (e) ⇒ h1 e = h2 e · a
with a ∈ Kerπ 0 = A, giving φ(e) = a [it is obviously a homomorphism since h1 and h2 are]; φ must
also factor through G (i.e. φ is equal to E → E/H2 G ∼ = G → A) because h1 and h2 must agree on
where it sends H2 G by commutativity of the diagram in Exercise IV.3.1(b). Now any homomorphism
ψ : G → A satisfies [G, G] ⊆ Kerψ by Proposition 5.4.7[2]; but G is perfect, so G = Kerψ and there are
no non-trivial maps G → A (hence φ = 0 ⇒ h1 = h2 ).
Note: It happens to be true that E in the universal central extension is necessarily perfect, so there
is another way to show uniqueness of the map E → E 0 [this is presented in John Milnor’s Introduction to
Algebraic K-Theory]. For any y, z ∈ E we have h1 y = h2 y·c and h1 z = h2 z·c0 with c, c0 ∈ Kerπ 0 ⊆ Z(E 0 ).
Thus h1 (yzy −1 z −1 ) = h2 (yzy −1 z −1 ) by basic rearrangements of the two previous equations, noting that
we can move c and c0 around as they lie in the center of E 0 . Since E = [E, E], it is generated by
commutators and hence h1 = h2 .
i π
3.8(a): Let 0 → A → E → G → 1 be a central extension with G abelian. The commutator pairing associ-
ated to the extension is the map c : G×G → A defined by c(g, h) = i−1 ([g̃, h̃]) = i−1 (g̃ h̃g̃ −1 h̃−1 ), where g̃

37
and h̃ are lifts of g and h to E. Since A/E and G = E/A is abelian, [E, E] ⊆ A by Proposition 5.4.7[2] and
so i−1 ([g̃, h̃]) is defined. Any other lift of g to E is of the form g̃a = ag̃, and [ag̃, a0 h̃] = [g̃, h̃] because A lies
in the center of E. Thus c is well-defined, and it is alternating because c(g, g) = i−1 ([g̃, g̃]) = i−1 (1) = 0.
Now g̃ h̃ is a lifting of g + h because π(g̃ h̃) = π(g̃) + π(h̃) = g + h, and [g̃ h̃, k̃] = g̃ h̃k̃ h̃−1 g̃ −1 k̃ −1 =
g̃[h̃, k̃]g̃ −1 [g̃, k̃] = [g̃, k̃][h̃, k̃] where the last equality follows from [E, E] ⊆ A ⊆ Z(E). We then have
c(g + h, k) = c(g, k) + c(h, k) because i is injective, giving i−1 ([g̃, k̃][h̃, k̃]) = i−1 ([g̃, k̃]) + i−1 ([h̃, k̃]). An
analogous computation gives c(k, g + h) = c(k, g) + c(k, h), so c is Z-bilinear. Since c is alternating and
V2 V2
bilinear, c(g, h) = −c(h, g) and hence c can be viewed as a map G → A, where G = G⊗G/h{g⊗g}i
is the second exterior power of G.

3.8(b): Let f be a factor set to the central extension in part(a). To show that c(g, h) = f (g, h) − f (h, g)
it suffices to show that [g̃, h̃] = i[f (g, h)]i[f (h, g)]−1 because i is injective. Given the section s : G → E,
f satisfies s(g)s(h) = i[f (g, h)]s(gh) and s(h)s(g) = i[f (h, g)]s(gh). Thus we have [s(g), s(h)] =
s(g)s(h)s(g)−1 s(h)−1 = i[f (g, h)]i[f (h, g)]−1 . Since s(g) is a lifting of g for all g ∈ G, we have the
desired [s(g), s(h)] = [g̃, h̃].
V2
3.8(c): Let θ : H 2 (G, A) → Hom( G, A) be the map which sends the class of a cocycle f to the
alternating map f (g, h) − f (h, g). This map is well-defined because [δc](g, h) − [δc](h, g) = c(g) + gc(h) −
c(gh) − c(h) − hc(g) + c(hg) = c(g) + c(h) − c(gh) − c(h) − c(g) + c(gh) = 0, where we note that G is
abelian and the G-action on the cochain c is trivial since A is central. In view of Theorem IV.3.12[1],
part(b) implies this image is the commutator pairing c(g, h), and f ∈ Kerθ iff θf = c(g, h) = 0. Now
c(g, h) = i−1 ([g̃, h̃]) = 0 iff [g̃, h̃] = 0 which is equivalent to E being abelian, i.e. [E, E] = 0. Thus there
is a bijection Kerθ ∼ = Eab (G, A), where Eab (G, A) is the set of equivalence classes of abelian extensions of
G by A.

4.1: Let Q2n be a generalized quaternion group. It is a fact that Q2n has a unique element of or-
der 2, hence a unique subgroup of order 2. Any subgroup of Q2n is also a 2-group (by Lagrange’s
Theorem) and so it has an element of order 2 by Cauchy’s Theorem; this element is then unique in each
subgroup (giving a unique Z2 subgroup). Therefore, by Theorem IV.4.3[1] every subgroup of Q2n is
either a cyclic group or a generalized quaternion group.
n−1
Alternatively, Q2n = hx, y | x2 = y 4 = 1, yxy −1 = x−1 i has the property that any subgroup G is
a 2-group with a unique Z2 subgroup (as mentioned above). If G is abelian, then by the Fundamental
Theorem of Finitely Generated Abelian Groups, G ∼ = Z2α1 × · · · × Z2αi which has more than one Z2 sub-
group if i > 1, and so G ∼ = Z2α is a cyclic subgroup. Suppose G is nonabelian, which means G contains
elements of the form xk (they necessarily form a cyclic subgroup H ⊂ hxi) and elements of the form xk y
[note: each element of Q2n can be written uniquely in the form xi y j for 0 ≤ i ≤ 2n−1 − 1 and 0 ≤ j ≤ 1,
and any element in Q2n − hxi has order 4 because (xk y)4 = (xk yxk y)2 = (xk x−k yy)2 = (y 2 )2 = y 4 = 1].
Let X = {xk1 y, xk2 y} ⊂ G be the elements of the form xk y and let xi be the generator of H. Now
n−2
xk2 y · xk1 y = xk2 −k1 y 2 = xMk+2 and (xk1 y)2 = y 2 , so xMk y 2 · y 2 = xMk and xMk · xk1 y = xk2 y. Thus
x y is a generator of G (of order 4) and the other generator is xr = xgcd(i,Mk) which is cyclic [note: if X
k1

includes other elements (up to xkm y) then the above still applies with r = gcd(i, km −k1 , . . . , km −km−1 ),
and if either i = 0 or k1 = · · · = km = 0 then omit those integers/differences in the gcd-term]. Since
(xk1 y)xr (xk1 y)−1 = xk1 (yxr y −1 )x−k1 = xk1 x−r x−k1 = x−r , the presentation is complete and G is a
generalized quaternion group.
n−1
4.2: The dihedral group D2n = Z2n−1 o Z2 = hx, y | x2 = y 2 = 1, xy = yx−1 i has the prop-
erty that each element can be written uniquely in the form y x for 0 ≤ i ≤ 1 and 0 ≤ j ≤ 2n−1 − 1.
i j

Any element not in Z2n−1 = hxi has order 2 because (yxk )2 = yxk yxk = (yxk y)xk = x−k xk = 1.
If a subgroup contains only elements of hxi then it is cyclic, so we consider the only other subgroups
G, and these contain elements of the form yxk and elements in hxi (which necessarily form a sub-
group H ⊆ hxi). Let X = {yxk1 , . . . , yxkm } ⊂ G be the elements of the form yxk and let xi be the
generator of H. The given subset X implies G contains the elements xkm −k1 , . . . , xkm −km−1 . Then
xr = xgcd(i,km −k1 ,...,km −km−1 ) generates the elements in G of the form xk (because k would be a multiple
of r), and yxk1 · x(km −k1 )−(km −kj ) = yxkj for any j ≤ m [note: if i = 0 (meaning the only elements of
G are of the form yxk ) then r = gcd(km − k1 , . . . , km − km−1 ), and if k1 = · · · = km = 0 (meaning y is

38
the only element in G of the form yxk ) then r = i]. Thus G = hxr , yxk1 i where yxk1 is of order 2 and
(yxk1 )xr (yxk1 ) = (yxk1 +r y)xk1 = x−k1 −r xk1 = x−r , i.e. G is a dihedral group.
The non-cyclic abelian subgroups of D2n are isomorphic to Z2 × Z2 = D4 . The group D8 contains two
non-cyclic abelian normal subgroups, hx2 , yi and hx2 , yxi. We assert that such subgroups are not normal
if n > 3 (i.e. if |D2n | > 8). From the subgroup presentations above we see that the non-cyclic abelian
n−2
subgroups are Hi = hx2 , yxi i for all i ∈ {0, . . . , 2n−1 − 1}. It suffices to show that there is an integer
j such that the element xj (yxi )x−j = yxi−2j does not lie in Hi (i.e. i − 2j is neither i nor i + 2n−2
modulo 2n−1 ). The first condition implies 2j 6≡ 0 mod 2n−1 ⇒ j 6= 2n−2 m for any m ∈ Z, and the latter
condition implies 2j + 2n−2 6≡ 0 mod 2n−1 ⇒ j 6= 2n−2 m − 2n−3 . Thus we can take j = 2n−2 − 1, and
n−2 n−2 n−2
x2 −1 ∈ D2n will yield the non-normality (x2 −1 )Hi (x1−2 ) * Hi [note: if n = 3 then j cannot be
an even integer 2m nor can it be an odd integer 2m − 1, which means j does not exist].

4.3: Let G = Zq o Z2 with q = 2n (n ≥ 3) and let A ⊂ Zq be the subgroup of order 2. Note that
A o Z2 = A × Z2 is a non-cyclic abelian subgroup of G since Z2 acts trivially on A [multiplication by
−1 + 2n−1 is the action, and (−1 + 2n−1 )2n−1 = −2n−1 + 2n−2 2n = 2n−1 + 0 = 2n−1 for the generator
2n−1 ∈ A]. If H ⊂ G is a non-cyclic proper subgroup, then H * Zq and HZq is a subgroup of G by
Corollary 3.2.15[2], so HZq = G. Thus 2q = |G| = |HZq | = |H|q/|H ∩ Zq | ⇒ |H : H ∩ Zq | = 2, and
since H ∩ Zq / H we have H/(H ∩ Zq ) ∼ = Z2 and this yields the extension 0 → H ∩ Zq → H → Z2 → 0.
The generator of Z2 acts as multiplication by −1 on H ∩ Zq because −1 + 2n−1 ≡ −1 mod 2m with
m < n [note: 2m = |H ∩ Zq | with m < n because H ∩ Zq is a subgroup of Zq and hence has
order 2m , and if m = n then Zq ⊆ H which means H = G, contradicting the assumption that
H is proper]. If H were abelian then H ∩ Zq would be central in H, and if H ∩ Zq were cen-
tral in H then H would be abelian since H/(H ∩ Zq ) ∼ = Z2 is cyclic; this latter fact holds because
H/(H ∩ Zq ) = {1(H ∩ Zq ), x(H ∩ Zq )} and so any h ∈ H has a representation h = xa z for z ∈ H ∩ Zq ,
giving h1 h2 = xa1 z1 xa2 z2 = xa1 xa2 z1 z2 = xa2 xa1 z2 z1 = xa2 z2 xa1 z1 = h2 h1 . Now H ∩ Zq is central in H
iff the Z2 -action is trivial (−z = z) iff H ∩ Zq = A, and so H is abelian iff H ∩ Zq = A.
We have H 1 (Z2 , Zq ) = KerN where N : (Zq )Z2 → (Zq )Z2 is the norm map. But (Zq )Z2 is a quotient
of a cyclic group (it is then necessarily cyclic of order m = 2r where m|q), and in particular we must
have (−1 + 2n−1 )1 = 1 mod m ⇒ 2n−2 − 1 = ms = 2(2r−1 s) which is impossible because an even
number cannot equal an odd number, so (Zq )Z2 = 0 and the kernel of the norm map is trivial. Thus
H 1 (Z2 , Zq ) = 0, and by Proposition IV.2.3[1] this means that the extension 0 → Zq → G → Z2 → 0 has
a unique splitting (up to conjugacy) and hence that G contains only two conjugacy classes of subgroups
of order 2 (specifically, 0 × Z2 lies in one class, and A × 0 is the other class because the number of con-
jugates of A is |G : NG (A)| = |G : G| = 1 by Proposition 4.3.6[2]). As an abelian non-cyclic subgroup,
H contains at least two subgroups of order 2 (one necessarily being A) and hence H can be A × Z2 and
its conjugates.
We assert that H = A × Z2 is not normal in G. Indeed, the element (1, x) ∈ G where x is the genera-
tor of Z2 can be used with (2n−1 , 1) ∈ H to obtain (1, x)(2n−1 , 1)(1, x)−1 = (1 + x · 2n−1 , x)(−1, x) =
(1 + 2n−1 + x · (−1), x2 ) = (2 + 2n−1 , 1) which is not in H since 2 + 2n−1 is neither 0 nor 2n−1 modulo
2n for n ≥ 3.

4.4: Let G be a p-group such that every abelian normal subgroup is cyclic and choose a maximal abelian
normal subgroup Zq ⊂ G with q = pn , so we have the corresponding extension 0 → Zq → G → H → 1.
If |H| = 1 then G ∼ = Zq is cyclic, hence of type (A). If |H| = p then Theorem IV.4.1[1] states G is of
type (A),(D),(E), or (F) because groups of type (C) have a non-cyclic abelian subgroup of index p by
Proposition IV.4.4[1] (such subgroups are necessarily normal by Corollary 4.2.5[2]) and groups of type
(B) are non-cyclic abelian normal subgroups of themselves [note: if G is of type (D) then we must have
|G| ≥ 16 = 24 because G = D8 contains two non-cyclic abelian normal subgroups and G = D4 ∼ = Z2 × Z2
is a non-cyclic abelian normal subgroup of itself]. Suppose, then, that |H| ≥ p2 , and consider the normal
subgroups H 0 ⊆ H of order p. If such an H 0 acted trivially on Zq , then the inverse image G0 ⊂ G
of H 0 = G0 /Zq would be a central extension of H 0 ∼ = Zp by Zq , hence an abelian subgroup of G [as
explained in Exercise IV.4.3] bigger than Zq . But G0 / G by the Lattice Isomorphism Theorem (since
H 0 / H), so it contradicts the maximality of Zq and hence H 0 cannot act trivially on Zq . Now G0 is
a p-group with a cyclic subgroup Zq of index p, so by Theorem IV.4.1[1] it is of type (C),(D),(E), or
(F) because types (A) and (B) were eliminated by the previous statement. Also, if H 0 acted as in (C)

39
then we would have a unique non-cyclic abelian subgroup G00 of G0 of index p by Proposition IV.4.4[1];
this applies to all p and n ≥ 2 except for the case p = 2 = n. Now G acts on G0 by conjugation (since
G0 / G) and hence maps G00 to another non-cyclic p-index subgroup. But G00 is the only such subgroup,
so conjugation sends G00 to itself (i.e. G00 / G) and G is a non-cyclic abelian normal subgroup of G,
contradicting the hypothesis. Thus the only possibility is that p = 2 since groups of type (D),(E), and
(F) are 2-groups, and the non-trivial element of H 0 acts as −1 [corresponding to (D),(E), and (C)] or
−1 + 2n−1 with n ≥ 3 [corresponding to (F)]. This means that H 0 embeds in Z∗2n ∼ = Z2 × Z2n−2 as
either Z2 × {0} or {(0, 0), (1, 2n−3 )}. But the composite H → Z∗2n → {±1} [where the first map is the
action-representation and the latter map is the projection] has a non-trivial kernel, and we can simply
take H 0 to be contained in the kernel; this implies H 0 is embedded as {0} × 2n−3 Z/2n−2 Z ⊂ Z∗2n which
is not any of the aforementioned subgroups. Thus we do not have |H| ≥ p2 and the proof is complete.

6.1: Suppose N is a group with trivial center.

Method 1 : The center of N is C = {1}, and so H 2 (G, C) = 0 = H 3 (G, C) for any group G. Then any
homomorphism ψ : G → Out(N ) gives rise to an obstruction in H 3 (G, C) = 0 which necessarily vanishes,
thus E(G, N, ψ) 6= ∅ by Theorem IV.6.7[1]. Therefore, by Theorem IV.6.6[1], E(G, N, ψ) ∼= H 2 (G, C) = 0
and hence there is exactly one extension of G by N (up to equivalence) corresponding to any homomor-
phism G → Out(N ).
Method 2 : Note that N is an Aut(N )-crossed module via the canonical map α : N → Aut(N ) and
the canonical action of Aut(N ) on N , and Kerα = Z(N ) = {1}. Thus any extension of G by N fits into
a commutative diagram with exact rows
1 /N i /E π /G /1

ψ
 
1 /N α /
Aut(N ) / Out(N ) /1
where ψ is determined by the E-extension. By Exercise IV.3.1(a), this is a pull-back diagram and hence
all such extensions for a given ψ are equivalent (under the above diagram). Thus there is exactly one
extension of G by N (up to equivalence) corresponding to any homomorphism G → Out(N ).

6.2: Let 1 → N → E → G → 1 be an extension of finite groups such that gcd(|N |, |G|) = 1. Corol-
lary IV.3.13[1] states that such an extension must split if N is abelian, and it is indeed true that this
result could be generalized to the non-abelian case. Now one might hope to deduce this generalization
directly from Theorem IV.6.6[1] in view of the vanishing of H 2 (G, C) [note that H 2 (G, C) = 0 by the
cohomology analogue of Exercise AE.16 because gcd(|C|, |G|) = 1, where C is the center of N ]. However,
this does not work because E(G, N, ψ) doesn’t contain the semidirect product (hence the split extension)
N o G unless ψ : G → Out(N ) [which is determined by the above extension] lifts to a homomorphism
ϕ : G → Aut(N ).

40
5 Chapter V: Products
t−1
1.1: Let G = hti be a finite cyclic group of order m and let F be the periodic resolution · · · → ZG →
N t−1 ε Pm−1
ZG → ZG → ZG → Z → 0, where N = i=0 ti is the norm element (note that Fn = ZG for all
n ≥ 0). Let ∆ : F → F ⊗ F be the map whose (p, q)-component ∆pq : Fp+q → Fp ⊗ Fq is given by


 1⊗1 p even
∆pq (1) = 1⊗t p odd, q even

P i j
0≤i<j≤m−1 t ⊗ t p odd, q odd

Now ∆ is a G-module map of degree 0, where ∆(g · x) = (g, g) · ∆(x) with the action given by re-
striction of scalars with respect to the diagonal embedding G → G × G, so in order to prove that it is
a diagonal approximation it suffices to show that it commutes with L the boundary maps (making it a
chain map) and is augmentation-preserving. Note that (F ⊗ F )n = p+q=n Fp ⊗ Fq with differential
D(f ⊗ f 0 ) = df ⊗ f 0 + (−1)degf f ⊗ df 0 , where d is the boundary operator of F . In dimension zero we have
(ε ⊗ ε)[∆00 (1)] = ε(1) ⊗ ε(1) = 1 ⊗ 1 ≡ 1 = ε(1), so ∆ is augmentation-preserving [the equivalence equa-
∼
=
tion is from Z⊗Z Z → Z]. Moving up a dimension, we have ∆00 [(t−1)1] = (t, t)[1⊗1]−[1⊗1] = t⊗t−1⊗1
and D1 [∆10 (1) + ∆01 (1)] = D1 (1 ⊗ t) + D1 (1 ⊗ 1) = (t − 1) ⊗ t + (−1)1 1 ⊗ 0 + 0 ⊗ 1 + (−1)0 1 ⊗ (t − 1) =
t ⊗ t − 1 ⊗ t − 0 + 0 + 1 ⊗ t − 1 ⊗ 1 = tP⊗ t − 1 ⊗ 1, so commutativity is satisfied.
P k Moving up
P another
dimension, we have (∆10 + ∆01 )[N 1] = k (tk , tk )[1
P k k k+1 k k
⊗ t] + k (t , t )[1 ⊗ 1] = k t ⊗ t + kt ⊗t
i j 0
P
and D2 [∆02 (1) + ∆11 (1) + ∆20 (1)] = D2 (1 ⊗ 1) + i<j D2 (t ⊗ t ) + D2 (1 ⊗ 1) = 0 ⊗ 1 + (−1) 1 ⊗ N 1 +
[(t − 1)ti ⊗ tj + (−1)1 ti ⊗ (t − 1)tj ] + N 1 ⊗ 1 + (−1)2 1 ⊗ 0 = 0 + k 1 ⊗ tk + i<j ti+1 ⊗ tj − i<j ti ⊗
P P P P
i<j
tj − i<j ti ⊗ tj+1 + i<j ti ⊗ tj + k tk ⊗ 1 + 0 = k 1 ⊗ tk + ( k tk ⊗ tk − 1 ⊗ 1) − ( k 1 ⊗ tk − 1 ⊗
P P P P P P

t) + ( k t ⊗ tk+1 − 1 ⊗ t − tm−1 ⊗ 1) − ( k tk ⊗ 1 − 1 ⊗ 1 − tm−1 ⊗ 1) = k tk ⊗ tk+1 + k tk ⊗ tk , so

P k P P P
commutativity is satisfied. Thus we can proceed by induction.
For even integers p+q = 2c (c ∈ N) there are c+1 (1⊗1)-elements, 0 (1⊗t)-elements, and c ( i<j ti ⊗tj )-
P

elements. For odd integers p+q = 2c+1 (c ∈ N) there are c+1 (1⊗1)-elements, 0 ( i<j ti ⊗tj )-elements,
P

and c + 1 (1 ⊗ t)-elements. This then gives D2c [∆2c (1)] = c[N ⊗ 1 + 1 ⊗ N ] + c i<j [ti+1 ⊗ tj − ti ⊗ tj+1 ]
P

and ∆2c−1 (N 1) = c k [tk ⊗ tk+1 + tk ⊗ tk ], and these are equal because they are just a multi-
P
ple of the low-dimensional case; the c[N ⊗ 1 + 1 ⊗ N ] came from (c + 1)[N ⊗ 1 + 1 ⊗ N ] minus
N ⊗ 1 + 1 ⊗ N which accounts for the two zero-dimensional tensor components which give trivial bound-
ary, i.e. D(1 ⊗ 1 + 1 ⊗ 1) = (0 ⊗ 1 + N ⊗ 1) + (N ⊗ 1 + 1 ⊗ 0). Commutativity in the next dimension is
also satisfied [similar calculation], and the proof is complete.

2.1: Let M (resp. M 0 ) be an arbitrary G-module (resp. G0 -module), let F (resp. F 0 ) be a projective reso-
lution of Z over ZG (resp. ZG0 ), and consider the map (F ⊗G M )⊗(F 0 ⊗G0 M 0 ) → (F ⊗F 0 )⊗G×G0 (M ⊗M 0 )
given by (x ⊗ m) ⊗ (x0 ⊗ m0 ) 7→ (x ⊗ x0 ) ⊗ (m ⊗ m0 ). Note that (F ⊗G M ) ⊗ (F 0 ⊗G0 M 0 ) =
(F ⊗ M )G ⊗ (F 0 ⊗ M 0 )G0 , which is the quotient of F ⊗ M ⊗ F 0 ⊗ M 0 by the subgroup generated
∼
=
by elements of the form gx ⊗ gm ⊗ g 0 x0 ⊗ g 0 m0 . The isomorphism F ⊗ M → M ⊗ F of chain complexes
degm·degx
(M in dimension 0) is given by x ⊗ m 7→ (−1) m ⊗ x = m ⊗ x, and so the aforementioned
quotient is isomorphic to F ⊗ F 0 ⊗ M ⊗ M 0 modulo the subgroup generated by elements of the form
(gx⊗g 0 x0 ⊗gm⊗g 0 m0 ) = (g, g 0 )·(x⊗x0 ⊗m⊗m0 ) where this latter action is the diagonal (G×G0 )-action.
Now this is precisely (F ⊗ F 0 ⊗ M ⊗ M 0 )G×G0 = (F ⊗ F 0 ) ⊗G×G0 (M ⊗ M 0 ) and hence the considered
map is an isomorphism.
Assuming
Ln now that either M or M 0 is Z-free, we have a corresponding
Ln−1 Künneth formula
0 0 0 0 Z 0 0
p=0 Hp (G, M )⊗Hn−p (G , M ) ,→ Hn (G×G , M ⊗M )  p=0 Tor1 (Hp (G, M ), Hn−p−1 (G , M ))
by Proposition I.0.8[2]. Note that in order to apply the proposition we needed one of the chain complexes
(say, F ⊗G M ) to be dimension-wise Z-free (and so with a free resolution F this means we needed M to
be Z-free). Actually, the general Künneth theorem has a more relaxed condition and it suffices to choose
M (or M 0 ) as a Z-torsion-free module.
*Must require some conditions on groups/actions: We have assumed G acts trivially on M 0 while G0
acts trivially on M . Since these actions don’t mix, we cannot deduce a Universal Coefficients theorem
(UCT) with nontrivial actions.

41
2.2: [[no proofs, just notes]] (for more info, see topological analog in §60[4])
Let M (resp. M 0 ) be an arbitrary G-module (resp. G0 -module), let F (resp. F 0 ) be a projective resolu-
tion of Z over ZG (resp. ZG0 ), and consider the cochain cross-product HomG (F, M )⊗HomG0 (F 0 , M 0 ) →
HomG×G0 (F ⊗ F 0 , M ⊗ M 0 ) which maps the cochains u and u0 to u × u0 given by hu × u0 , x ⊗ x0 i =
0
(−1)degu ·degx hu, xi ⊗ hu0 , x0 i. This map is an isomorphism under the hypothesis that either H∗ (G, M ) or
H∗ (G0 , M 0 ) is of finite type, that is, the ith-homology group is finitely generated for all i (alternatively,
we could simply require the projective resolution F or F 0 to be finitely generated). For example, if
M = Z then HomG (−, Z) commutes with finite direct sums, so we need only consider the case F = ZG.
0 0
An inverse to the above map is given by t 7→ ε ⊗ φ, where ε is the augmentation map and L∞φ : F → M
0 0
is given by φ(f Q ) = t(1 ⊗ f ). Note that this does not hold for infinitely generated P = ZG because
∞
HomG (P, Z) ∼ = Z is not Z-projective (i.e. free abelian).
Assuming
Ln now that either M or M 0 is Z-free, we have a corresponding
Ln+1 Künneth formula
0 0 0 0
p=0 H p
(G, M )⊗H n−p
(G , M ) ,→ H n
(G×G , M ⊗M )  p=0 TorZ
1 (H p
(G, M ), H n−p+1 (G0 , M 0 ))
by Proposition I.0.8[2].
*Must require some conditions on groups/actions: We have assumed G acts trivially on M 0 while G0
acts trivially on M . Since these actions don’t mix, we cannot deduce a Universal Coefficients theorem
(UCT) with nontrivial actions.

3.1: Let m ∈ H 0 (G, M ) = M G and u ∈ H q (G, N ), and let fm : H ∗ (G, N ) → H ∗ (G, M ⊗ N ) de-
note the map induced by the coefficient homomorphism n 7→ m ⊗ n. This homomorphism is also given
∼
=
by n 7→ 1⊗n 7→ m⊗n where the former map in the composition is the canonical isomorphism N → Z⊗N ,
the latter map in the composition is Fm ⊗ idN : Z ⊗ N → M ⊗ N , and Fm : Z → M is given by F (1) = m.
Then using two properties of the cup product (existence of identity element and naturality with respect
to coefficient homomorphisms) we obtain fm (u) = (Fm ⊗ idN )∗ (1 ` u) = Fm ∗
(1) ` id∗N (u) = m ` u.
0 G
Now let m ∈ H (G, M ) = M and z ∈ Hq (G, N ), and let fm : H∗ (G, N ) → H∗ (G, M ⊗ N ) denote
the map induced by the same coefficient homomorphism n 7→ m ⊗ n. Then using two properties of the
cap product (existence of identity element and naturality with respect to coefficient homomorphisms)
we obtain fm (z) = (Fm ⊗ idN )∗ (1 a z) = Fm∗
(1) a id∗N (z) = m a z.

3.2: Consider the diagonal transformation ∆ presented in Exercise V.1.1, along with the cohomol-
ogy groups H 2r (G, M ) ∼ = M G /N M and H 2r+1 (G, M 0 ) ∼ = Ker(N : M 0 → M 0 )/IM 0 of the finite cyclic
group G = hσi of order n, where I = hσ − 1i is the augmentation ideal of G. The cup product in
H ∗ (G, M ⊗ M 0 ) is given by u ` v = (u × v) ◦ ∆ with hu × v, x ⊗ x0 i = (−1)degv·degx hu, xi ⊗ hv, x0 i. Choose
representatives hu, xi = m ∈ M of H i (G, M ) with m ∈ M G for i even and m ∈ Ker(N : M → M )
for i odd, and choose representatives hv, x0 i = m0 ∈ M 0 of H j (G, M 0 ) with m0 ∈ M 0G for j even and
m0 ∈ Ker(N : M 0 → M 0 ) for j odd. If i is even then (−1)degv·degx = (−1)degv·i = 1, and if j is even then
(−1)degv·degx = (−1)−j·degx = 1, and if both i and j are odd then (−1)degv·degx = (−1)−j·i = −1. Thus
the cup
P product element of H i+j (G, M ⊗ M 0 ) is represented by m ⊗ m0 for i or j even and is represented
by − 0≤p<q≤n−1 t m ⊗ tq m0 when i and j are both odd.
p

3.3(a): Let G be a finite group which acts freely on S 2k−1 , and consider the exact sequence of G-
modules from pg20[1], 0 → Z → C2k−1 → · · · → C1 → C0 → Z → 0 where each Ci = Ci (S 2k−1 ) is
free. Tensoring the sequence with an arbitrary G-module M gives an exact sequence (as explained on
∂2k−1 ∂ ∂
pg61[1]) 0 → M → C2k−1 ⊗ M → · · · → C1 ⊗ M →1 C0 ⊗ M →0 M → 0. We can break this up into
short exact sequences 0 → Ker∂0 → C0 ⊗ M → M → 0 and 0 → Ker∂i → Ci ⊗ M → Ker∂i−1 → 0
for all i, and we can then apply the H ∗ (G, −) functor to obtain corresponding long exact cohomol-
ogy sequences. First note that HomG (F, C∗ ⊗ M ) → HomG (F, M ) is a weak equivalence by Theorem
I.8.5[1], so H n (G, Ci ⊗ M ) ∼= H n (HomG (F, M )) = 0 for all i with n > 0 (M is considered a chain
complex concentrated in dimension 0). Thus we can apply the dimension-shifting argument to the above
short exact sequences and obtain H i (G, M ) ∼= H i+1 (G, Ker∂0 ) ∼
= H i+2 (G, Ker∂1 ) ∼
= ··· ∼
= H i+2k (G, M )
for i > 0 where we note that Ker∂2k−1 = M . For i = 0 we use the first short exact sequence men-
tioned above to obtain the exact sequence H 0 (G, M ) → H 1 (G, Ker∂0 ) → H 1 (G, C0 ⊗ M ) = 0, with
H 1 (G, Ker∂0 ) ∼
= H 2k (G, M ). Thus there is an iterated coboundary map d : H i (G, M ) → H i+2k (G, M )
which is an isomorphism for i > 0 and an epimorphism for i = 0.

42
3.3(b): Consider the “periodicity map” d from part(a) above with the finite group G which acts freely
on the sphere S 2k−1 . Since d is simply an iteration of coboundary maps δ, we can use Property V.3.3[1]
which states d(w ` v) = d(w) ` v for any w ∈ H ∗ (G, Z) and v ∈ H ∗ (G, M ). Choosing w = 1 ∈ H 0 (G, Z)
and noting that 1 ` v = v by Property V.3.4[1], d(v) = d(1 ` v) = d(1) ` v. But d is an isomorphism
from H 0 (G, Z) to H 2k (G, Z) by part(a) above, so there exists an element u ∈ H 2k (G, Z) satisfying
d(1) = u. Thus there is an element u ∈ H 2k (G, Z) such that the “periodicity map” d of H ∗ (G, M ) is
given by d(v) = u ` v for all v ∈ H ∗ (G, M ).

3.3(c): Let G be a finite cyclic group of order |G| = n, and note that it acts freely on the circle
S 1 by rotations. By part(b) above, the periodicity isomorphism d maps 1 ∈ H 0 (G, Z) ∼
= Z to a generator
α ∈ H 2 (G, Z) ∼= Zn . For the ring structure on H ∗ (G, Z) with multiplication being the cup product,
α2 is a generator of H 4 (G, Z) ∼
= Zn since d is an isomorphism and d(α) = α ` α = α2 . General-
izing, α is a generator of H (G, Z) ∼
m 2m
= Zn and hence the cohomology ring is the polynomial ring
H ∗ (G, Z) ∼
= Z[α]/(nα) with |α| = 2.
Alternatively, we could note that the infinite-dimensional lens space L∞ is a K(G, 1)-complex, so
H ∗ (G, Z) ∼
= H ∗ (L∞ ) by the cohomological analog of Proposition II.4.1[1]. The ring structure was
calculated in Example 3.41 on pg251[3], giving H ∗ (L∞ ) ∼
= Z[α]/(nα) with |α| = 2.

3.4: Let G be cyclic of order n and consider the endomorpism α(m) of G given by α(m)g = g m ,
∗ ∗ ∗ ∗
P anyi m ∈ Zn . Since α(m) (u ` v) = α(m) u ` α(m) v and H (G, Z) consists of
for
∗
elements of the form
i zi β with zi ∈ Z and |β| = 2 by Exercise V.3.3(c), it suffices to calculate α(m) on H 2 (G, Z) in order
∗ ∗ ∗
to calculate α(m) : H (G, Z) → H (G, Z). By Exercise III.1.3 we have the universal coefficient isomor-
phism H 2 (G, Z) ∼= Ext(H1 G, Z) since Hom(H1 G, Z) = Hom(Zn , Z) = 0. The map α(m) is multiplication
by m on Gab = G and hence is multiplication by m on H1 G by Exercise II.6.3(a). Now f (mg) = mf (g)
for any group homomorphism f , so given a free resolution F of the abelian group Gab = H1 G = G
(namely, its presentation) we have the commutative diagram
0o Hom(F1 , Z) o Hom(F0 , Z) o
φ
Hom(G, Z)
m m m
  
0o Hom(F1 , Z) o Hom(F0 , Z) o
φ
Hom(G, Z)
Thus α(m) induces multiplication by m on Cokerφ ≡ Ext(H1 G, Z) and hence α(m)∗ is multiplication by
m on H 2 (G, Z) by naturality of the universal coefficient formula. Alternatively, we could have used the
interpretation of H 2 in terms of group extensions (Exercise IV.3.1(a)) which states that α(m) induces
a map E(G, Z) → E(G, Z), sending the extension E to the fiber-product E ×G G which correponds to
mE (i.e. the elements em ∈ E) and hence gives the m-multiplication map in cohomology by Theorem
∗ i th
IV.3.12[1].
P Referring P back to∗the cohomology ring, P α(m) is∗multiplication by m for the
P (2i) -dimension
i ∗
since iP zi β 7→ i zi α(m) (β ` · · · ` β) = i zi [α(m) β] ` · · · ` [α(m) β] = i zi (mβ) ` · · · `
(mβ) = i zi mi β i .

3.5: The symmetric group S3 on three letters has a semi-direct product representation S3 = Z3 o Z2
where Z2 acts on Z3 by conjugation. Thus H ∗ (S3 ) = H ∗ (S3 )(2) ⊕ H ∗ (S3 )(3) ∼= H ∗ (S3 )(2) ⊕ H ∗ (Z3 )Z2 by
Theorem III.10.3[1], with H (S3 )(2) isomorphic to the set of S3 -invariant elements of H ∗ (Z2 ). Exercise
∗

III.10.1 showed that H ∗ (S3 )(2) ∼

= Z2 , so it suffices to compute H ∗ (Z3 )Z2 where we know that Z2 acts by
conjugation on Z3 (1 7→ 1, x 7→ x2 , x2 7→ x). But this action can be considered as the endomorphism
α(2) from Exercise V.3.4 above since (1)2 = 1 and (x)2 = x2 and (x2 )2 = x4 = x, and that exercise
implies that the induced map on the (2i)th -cohomology is multiplication by 2i [we know that the co-
homology is trivial in odd dimensions]. Now 21 ≡ 2 mod3 and 22 ≡ 1 mod3, so by multiplying both of
those statements by 2 repeatedly we see that 2i ≡ 1 mod3 for i even and 2i ≡ 2 mod3 for i odd. Thus
the largest Z2 -submodule of H 2i (Z3 ) ∼ = Z3 on which Z2 acts trivially is Z3 (itself) for i even, and is 0
for i odd. It now follows that the integral cohomology H ∗ (S3 ) is the same as that which was deduced in
Exercise III.10.1, namely, it is Z2 in the 2 mod4 dimensions and is Z6 in the 0 mod4 dimensions and is 0
otherwise (besides the 0th -dimension in which it is Z).

4.1:

43
4.2:

5.1(a): Let G be an abelian group so that ZG is a commutative ring, and let M ⊗ZG N be the tensor
product with ZG-module structure defined by r ·(m⊗n) = rm⊗n = m⊗rn, where r ∈ ZG. There exists
× µ∗
a Pontryagin product given by H∗ (G, M ) ⊗ H∗ (G, N ) → H∗ (G × G, M ⊗ N ) → H∗ (G, M ⊗ZG N ). The
former map is the homology cross product (see pg109[1]), where G×G acts on M ⊗N via (g, g 0 )·(m⊗n) =
gm⊗g 0 n. The latter map in the composition is induced by µ = (α, f ) : (G×G, M ⊗N ) → (G, M ⊗ZG N ),
where α(g, g 0 ) = gg 0 and f (m ⊗ n) = m ⊗ n. As explained on pg79[1], µ∗ is a well-defined map as long
as f (rm) = α(r)f (m) for r ∈ ZG. This is indeed the case because α(g, g 0 )f (m ⊗ n) = gg 0 · (m ⊗ n) =
g · (m ⊗ g 0 n) = gm ⊗ g 0 n = f (gm ⊗ g 0 n) = f [(g, g 0 ) · (m ⊗ n)].

5.1(b): Take G = Z3 = hgi and take the modules M = N = Z7 with nontrivial Z3 -action, i.e. g ·xi = x2i
and g 2 · xi = x4i with g 3 = 1 [consistency check: g 3 · xi = x8i = (x8 )i = xi ]. Note that there is only one
nontrivial action in this scenario because an action is a homomorphism Z3 → Aut(Z7 ) ∼ = Z6 and there
are two such maps, the trivial map and the inclusion. The tensor product Z7 ⊗Z3 Z7 can be interpreted
as the group (Z7 ⊗ Z7 )Z3 where Z3 acts diagonally, g · (m ⊗ n) = gm ⊗ gn. This group contains only
idempotent elements since (xi ⊗ xj )2 = x2i ⊗ x2j = gxi ⊗ gxj = g · (xi ⊗ xj ) ∼ xi ⊗ xj . Now the
tensor product Z7 ⊗Z[Z3 ] Z7 can be interpreted as the group (Z7 ⊗ Z7 )Z3 where Z3 acts anti-diagonally,
g · (m ⊗ n) = g −1 m ⊗ gn = g 2 m ⊗ gn. This group, however, does contain elements which are not idem-
potent. For example, (x ⊗ x)2 = x2 ⊗ x2 = g 2 x4 ⊗ gx ∼ x4 ⊗ x = (x ⊗ x)(x3 ⊗ 1) and (x3 ⊗ 1)  1 ⊗ 1.
Thus Z7 ⊗Z[Z3 ] Z7  Z7 ⊗Z3 Z7 .

5.2: If G and G0 are abelian and k is a commutative ring, then the cohomology cross-product H ∗ (G, k)⊗k
×
H ∗ (G0 , k) → H ∗ (G × G0 , k ⊗ k) is a k-algebra homomorphism. Moreover, z × z 0 = p∗ z ` p0∗ z 0 for any
z ∈ H (G, k) and z 0 ∈ H ∗ (G0 , k), where p : G × G0 → G and p0 : G × G0 → G0 are the projections.
∗
0
To prove these two statements, note first that the cross-product is given by hz×z 0 , x⊗x0 i = (−1)|z ||x| z(x)⊗
z 0 (x0 ) for x ∈ F and x0 ∈ F 0 , with F and F 0 being the resolutions for G and G0 , respectively. Then
0 0 0 0
h(z1 ⊗ z10 )(z2 ⊗ z20 ), x ⊗ x0 i = h(−1)|z1 ||z2 | z1 z2 × z10 z20 , x ⊗ x0 i = (−1)|z1 ||z2 | (−1)|z1 z2 ||x| z1 (x)z2 (x) ⊗
0 0
0
z1 (x0 )z20 (x0 ) = (−1)|z1 ||x| (−1)|z1 ||x| [z1 (x) ⊗ z10 (x0 )][z2 (x) ⊗ z20 (x0 )] = hz1 × z10 , x ⊗ x0 i · hz2 × z20 , x ⊗ x0 i
and so the cross-product is a k-algebra homomorphism (it obviously commutes with addition). Since
z ⊗ z 0 = (z ⊗ 1) · (1 ⊗ z 0 ) with 1 ∈ H 0 (G, Z) = Z, it follows that z × z 0 = (z × 1) ` (1 × z 0 ), where we
make the identifications Z ⊗ k = k = k ⊗ Z to switch from tensor-multiplication to cup product. I claim
that z × 1 = p∗ z; indeed, by naturality of the cross-product it suffices to check this for G0 = 1. But
this is obvious because p∗ : H ∗ (G, k) → H ∗ (G × 1, k) maps z to z × 1, noting that F 0 = 1. Similarly,
1 × z 0 = p0∗ z 0 , whence the proposition.

5.3(a): A directed set D is a partially-ordered set having the property that for each pair α, β ∈ D
there exists γ ∈ D such that α, β ≤ γ. A directed system of groups is a family of groups {Gα }
indexed by a directed set D along with a family of homomorphisms {fαβ : Gα → Gβ } such that
fαα = idGα and ` fαγ = fβγ ◦ fαβ for α ≤ β ≤ γ. The direct limit G = lim Gα of this directed system
−→
is defined to be α Gα /∼ where gα ∼ gβ if there exists some γ ∈ D such that fαγ (gα ) = fβγ (gβ ).
Now for any G-module M , we have a compatible family of maps H∗ (Gα , M ) → H∗ (G, M ), hence
a map ϕ : lim H∗ (Gα , M ) → H∗ (G, M ). It is true that homology commutes with direct limits of
−→
chain complexes (see Albrecht Dold’s Lectures on Algebraic Topology, Proposition VIII.5.20), so to
prove that ϕ is an isomorphism it suffices to show that lim(Fα ⊗Gα M ) = F ⊗G M where Fα is the
−→
standard resolution for Gα and F is the standard resolution for G. The obvious maps Fα → Fβ
are given by`(g1 , . . . , gn ) 7→ (fαβ (g1 ), . . . , fαβ (gn )), and F is obviously the direct limit of Fα . Thus
F`⊗G M = ( α Fα /∼) ` ⊗G M , and since we can switch actions from G to Gα via restriction of scalars,
( α Fα /∼)⊗` G M = α (Fα ⊗Gα M )/∼ where ∼ is simply altered by tensoring each tuple with an element
of M . But α (Fα ⊗Gα M )/∼ is by definition the direct limit of Fα ⊗Gα M , so lim(Fα ⊗Gα M ) = F ⊗G M
−→
and hence group homology commutes with direct limits of groups.

5.3(b): It is a fact that any group is the direct limit of its finitely generated subgroups. So for any abelian

44
group G = lim Gα and commutative ring k, the homology ring H∗ (G, k) is isomorphic to lim H∗ (Gα , k)
−→ −→
by part(a) above. If each of those rings H∗ (Gα , k) is strictly anti-commutative, then H∗ (G, k) will obvi-
ously be strictly anti-commutative since it is a quotient of the direct sum of those rings. Thus we can
reduce to the case where G is a finitely generated abelian group, hence isomorphic to a finite product
of cyclic groups.VLet’s argue by induction on the number of cyclic factors. The infinite cyclic group has
resolution F = (x) which is strictly anti-commutative
V as explained in subsection 5.2 on pg118[1], and
the finite cyclic group has resolution F = (e1 )⊗ZG Γ(e2 ) which is strictly anti-commutative by property
(iii) of subsection 5.3 on pg119[1]. Since the admissible product on F induces a k-bilinear product on
F ⊗G k via (f ⊗ k)(f 0 ⊗ k 0 ) = f f 0 ⊗ kk 0 , the complex F ⊗G k is strictly anti-commutative; thus H∗ (G, k) is
strictly anti-commutative for G cyclic. Applying the inductive hypothesis, we can attach another cyclic
factor by the method of subsection 5.4 on pg119[1]: F and F 0 are resolutions with admissible product
for G (cyclic) and G0 (inductive group), so F ⊗ F 0 is a resolution with admissible product for G × G0 .
This resolution is strictly anti-commutative because (x ⊗ y)(x ⊗ y) = (−1)|x|·|y| x2 ⊗ y 2 = 0 if either x2
or y 2 is 0, so for x (resp. y) of odd degree and y (resp. x) of even degree we have x ⊗ y of odd degree
which satisfies (x ⊗ y)2 = 0. Then (F ⊗ F 0 ) ⊗G×G0 k is strictly anti-commutative and so is H∗ (G × G0 , k),
completing the inductive process. Thus the ring H∗ (G, k) is strictly anti-commutative for any abelian
group G and commutative ring k.

5.4: Let n = p + q + r, and let σ ∈ Sn be a permutation with signature sgn(σ) being the number
of inversions of σ; an inversion of σ is a pair of elements (i, j) such that i < j and σ(i) > σ(j) [it
also indicates the number of swaps needed to give the original sequence ordering]. A permutation σ
is called a (p, q, r)-shuffle if σ(i) < σ(j) for 1 ≤ i < j ≤ p and for p + 1 ≤ i < j ≤ p + q and for
p + q + 1 ≤ i < j ≤ p + q + r. But this permutation is clearly the composition of a (p, q)-shuffle τ1
and a (p + q, r)-shuffle τ2 since the former shuffle will give τ1 (i) < τ1 (j) for 1 ≤ i < j ≤ p and for
p + 1 ≤ i < j ≤ p + q, and the latter shuffle will give τ2 (i) < τ2 (j) for p + q + 1 ≤ i < j ≤ p + q + r and
will preserve the ordering of the former shuffle via 1 ≤ i < j ≤ p + q. Then sgn(σ) = sgn(τ1 ) + sgn(τ2 )
because the inversions of τ2 give the original ordering of the sequence up to changes in {1, . . . , p + q} and
the inversions of τ1 give the original ordering of that set. Therefore (in the bar resolution),
P
[g1 | · · · |gp ] · [gp+1 | · · · |gp+q ] · [gp+q+1 | · · · |gp+q+r ] = σ σ[g1 | · · · |gp+q+r ]
where σ ranges over the (p, q, r)-shuffles. P
The notation is [g1 | · · · |gn ] · [gn+1 | · · · |gn+m ] = τ τ [g1 | · · · |gn+m ] where τ ranges over all (p, q)-shuffles,
and σ[g1 | · · · |gn ] = (−1)sgnσ [gσ−1 (1) | · · · |gσ−1 (n) ].
Generalizing, we have
P
[g1 | · · · |ga1 ] · [ga1 +1 | · · · |ga1 +a2 ] · · · [ga1 +···+an−1 +1 | · · · |ga1 +···+an ] = σ σ[g1 | · · · |ga1 +···+an ]
where σ ranges over the (a1 , . . . , an )-shuffles.

6.1: Let G be an abelian group (written additively) with n ∈ Z, and consider the endomorphism g 7→ ng
of G. To see what it induces on the rational homology
V∗ ring H∗ (G, Q) it suffices to figure out what the
endomorphism induces on the exterior algebra (G ⊗ Q) by Theorem V.6.4[1]V(they’re isomorphic),
∗
noting that the isomorphism in said theorem is natural. Now the induced map on (G ⊗ Q) is uniquely
V1
determined by the induced map ϕ : G ⊗ Q → (G ⊗ Q) = G ⊗ Q by the universal mapping property of
exterior algebras (pg122[1]), and this map ϕ is given by g ⊗ q 7→ ng ⊗ q = n(g ⊗ q), i.e. multiplication
by n. Then on the i-fold tensor product of G ⊗ Q with itself (hence the exterior algebra) we have the
induced map as f1 ⊗ · · · ⊗ fi 7→ nf1 ⊗ · · · ⊗ nfi = ni (f1 ⊗ · · · ⊗ fi ) where f1 , . . . , fi ∈ G ⊗ Q. Thus the
original endomorphism on G induces multiplication by ni on H∗ (G, Q) for the ith -dimension.

6.2: Let A and B be strictly anti-commutative graded k-algebras, where k is a commutative ring. We as-
sert that A⊗k B is the sum (i.e. coproduct) of A and B in the category of strictly anti-commutative graded
k-algebras, via the maps fA : A → A⊗k B and fB : B → A⊗k B with a 7→ a⊗1 and b 7→ 1⊗b. The coprod-
uct refers to the pair (A⊗k B, {fA , fB }) satisfying the universal property that given a family of algebra ho-
momorphisms {gA : A → C, gB : B → C}, there exists a unique algebra homomorphism h : A ⊗k B → C
such that hfA = gA and hfB = gB . We define a k-bilinear map A × B → C given by a × b 7→ gA (a)gB (b).
Then Corollary 10.4.16[2] gives us a unique homomorphism h : A⊗k B → C given by a⊗b 7→ gA (a)gB (b),

45
and this is clearly an algebra homomorphism since h[(a1 ⊗ b1 )(a2 ⊗ b2 )] = h[(−1)|a2 ||b1 | a1 a2 ⊗ b1 b2 ] =
(−1)|a2 ||b1 | gA (a1 a2 )gB (b1 b2 ) = (−1)|a2 ||b1 | gA (a1 )[(−1)|a2 ||b1 | gB (b1 )gA (a2 )]gB (b2 ) = h(a1 ⊗ b1 )h(a2 ⊗ b2 ).
Now hfA (a) = h(a ⊗ 1) = gA (a)gB (1) = gA (a), with a similar calculation for B, and so the universal
property of tensor products (Theorem 10.4.10[2]) guarantees that A ⊗k B is the categorical sum A + B.

6.3(a): Let A be a strictly anti-commutative graded ring with a differential ∂ and a system of di-
vided powers, so that there is a family of functions ϕi : A2n → A2ni denoted x 7→ x(i) satisfying the
properties on pg124[1]. Note that x(i) is a cycle if x is a cycle because ∂x(i) = x(i−1) ∂x = x(i−1) · 0 = 0.
Thus ϕi restricted to the kernel Z2n ⊆ A2n gives a function Z2n → Z2ni . These functions inherit the
same properties associated with A because every term in all of the properties are cycles xi and x(j) ,
assuming x is a cycle. Now suppose x(i) is a boundary whenever x is a boundary. Then ϕi induces a
function H2n A → H2ni A because x = x + ∂y 7→ x(i) + (∂y)(i) = x(i) + ∂w = x(i) , and the properties
remain untouched. Therefore, we have an induced system of divided powers on H∗ A.

6.3(b): Consider the divided polynomial algebra Γ(y) with degy = 2, and assume that y is a bound-
ary, y = ∂x for
P some x. I claim that y (i) is not a boundary. PIndeed, supposeP y (i) = ∂f for some
(j) i j j−1
element f = j zj y ∈ Γ(y), where zj ∈ Z. Then y /i! = j zj ∂(y )/j! = j zj [jy ∂y]/j! =
j−1 2 i
P
∂y j zj y /(j − 1)! = ∂ x · w = 0 · w = 0. But this implies y = 0, a contradiction.
V∗
6.4(a): By Theorem V.6.4[1] we have an injection ψ : (G ⊗ k) → H∗ (G, k) for G abelian and k
a PID. This k-algebra map was the unique extension of the isomorphism G ⊗ k → H1 (G, k) in dimen-
sion 1. Now ψ[(g ⊗ 1) ∧ (h ⊗ 1)] = ψ(g ⊗ 1) · ψ(h ⊗ 1) by definition of an algebra map, where · is
the Pontryagin product. On the bar resolution this product is given by the shuffle product, P and the
isomorphism ψ : G ⊗ k → H1 (G, k) sends g ⊗ 1 to [g]. Thus (g ⊗ 1) ∧ (h ⊗ 1) 7→ [g] · [h] = σ [g|h] =
[g|h] − [h|g], where σ ran over the two possible (1, 1)-shuffles. Remark: This map is well-defined because
V2
for (g ⊗ 1) ∧ (g ⊗ 1) = 0 ∈ (G ⊗ k), the image is [g|g] − [g|g] = 0.

6.4(b): LetVk be a PID in which 2 is invertible, let G be an abelian group, and considerVthe map
∗ ∗
C2 (G, k) → (G⊗k) given by [g|h] 7→ (g ⊗1)∧(h⊗1)/2. This induces a map ϕ : H2 (G, k) → (G⊗k)
because any 3-coboundary ∂[r|s|t] = [s|t] − [rs|t] + [r|st] − [r|s] is mapped to the trivial element
(s ⊗ 1) ∧ (t ⊗ 1)/2 − (r + s ⊗ 1) ∧ (t ⊗ 1)/2 + (r ⊗ 1) ∧ (s + t ⊗ 1)/2 − (r ⊗ 1) ∧ (s ⊗ 1)/2 = [(s ⊗ 1) ∧ (t ⊗ 1)/2] −
[(r ⊗ 1) ∧ (t ⊗ 1)/2] − [(s ⊗ 1) ∧ (t ⊗ 1)/2] + [(r ⊗ 1) ∧ (s ⊗ 1)/2] + [(r ⊗ 1) ∧ (t ⊗ 1)/2] − [(r ⊗ 1) ∧ (s ⊗ 1)/2] = 0.
Using ψ from part(a) above, ϕ is its left-inverse because (g ⊗ 1) ∧ (h ⊗ 1) 7→ [g|h] − [h|g] 7→ (g ⊗ 1) ∧ (h ⊗
1)/2 − (h ⊗ 1) ∧ (g ⊗ 1)/2 = (g ⊗ 1) ∧ (h ⊗ 1)/2 + (g ⊗ 1) ∧ (h ⊗ 1)/2 = (g ⊗ 1) ∧ (h ⊗ 1), where we note
in the last equality that the exterior algebra is strictly anti-commutative.

6.5: Let VG be abelian and let A be a G-module with trivial G-action. In view of the isomorphism
H2 G ∼
2
= G, the universal coefficient theorem gives us a split exact sequence 0 → Ext(G, A) →
2 θ V2 V2
H (G, A) → Hom( G, A) → 0. The isomorphism ψ : G → H2 G is given by g ∧ h 7→ [g|h] − [h|g]
by Exercise V.6.4(a). Now the map β : H 2 (G, A) → Hom(H2 G, A) in the universal coefficient sequence
sends the class kf k of the cocycle f to (βkf k)([g1 |g2 ]) = f (g1 , g2 ). Then θ is given by kf k 7→ βkf k 7→
(βkf k ◦ ψ)(g ∧ h) = (βkf k)([g|h] − [h|g]) = f (g, h) − f (h, g), and this element is an alternating map.
Thus θ coincides with the map θ in Exercise IV.3.8(c), and so we see that every alternating map comes
from a 2-cocycle. It also follows that Ext(G, A) ∼= Eab (G, A), whence the name “Ext.”

46
6 Chapter VI: Cohomology Theory
of Finite Groups
2.1: Suppose |G : H| < ∞ and that M is a ZG-module with a relative injective resolution Q, and η : M →
Q0 is the canonical admissible injection (i.e. Q0 = CoindG G
H ResH M ). If M is free as a ZH-module then Q
0
η 0
is ZG-free by Corollary VI.2.2[1]. Since η is H-split, the exact sequence 0 → M → Q → Cokerη → 0 is
split-exact and hence Q0 ∼ = M ⊕Cokerη. Since M and Q0 are both ZH-free (a free ZG-module is ZH-free
by Exercise I.3.1), Cokerη is by definition stably free. Let us find this cokernel explicitly. The canonical
injection is given by M ,→ HomH (ZG, ResG ∼ G G G
H M ) = IndH ResH M = ZG ⊗ZH ResH M , m 7→ [s 7→ sm] 7→
−1
m. But IndH ResH M ∼
G G
P
g∈G/H g ⊗ g = Z[G/H] ⊗ M with the mapping g ⊗ m 7→ g ⊗ gm, by Proposition
P
III.5.6[1]. Thus the canonical injection is given by m 7→ g∈G/H g ⊗ m. We can now consider M as a
free ZH-module, and without loss of generalization we can L assume M = ZH (since direct sums commute
with tensor products). We can also regard Z[G/H] = g∈G/H Z[g] as a direct sum of integers via the
L|G:H| L|G:H|
isomorphism Z[g] ∼ = Z, g 7→ 1. Thus the H-split injection is η : ZH ,→ Z ⊗ ZH ∼ = ZH
with the mapping m 7→ (g1 , . . . , g|G:H| ) ⊗ m 7→ (1, . . . , 1) ⊗ m 7→ (m, . . . , m). The cokernel of this map
L|G:H| L|G:H|−1
ZH/ZH ∼
Ln
is = ZH, a free ZH-module. Indeed, For any finite direct sum X, the
Ln−1
quotient of this group by its diagonal subgroup X = {(x, . . . , x)} is the direct sum X because
any element (x1 , . . . , xn−1 , xn )X is equivalent in the quotient group to (x1 x−1 −1
n , . . . , xn−1 xn , 1)X via the
−1 −1 −1
element (xn , . . . , xn , xn ). Since Cokerη is free as a ZH-module, we can apply Corollary VI.2.2[1] to
get an admissible injection Cokerη ,→ Q1 with Q1 free as a ZG-module. Continuing in this way, we
obtain the resolution Q as a complex of free ZG-modules.

3.1: Let G act freely on S 2k−1 , as on pg20[1]. From this we have a free resolution of Z over ZG
which is periodic of period 2k:
ε
· · · → C1 → C0 → C2k−1 → · · · → C1 → C0 → Z → 0
where C∗ = C∗ (X) and X ≈ S 2k−1 is a free G-complex that makes each Ci finitely generated. Then by
Proposition VI.3.5[1] we can take the above resolution ε : C → Z, form the dual (backwards resolution)
of its suspension ε∗ : Z∗ = Z → ΣC, and then splice together C and ΣC to form a complete resolution
F for G. This resolution is obviously periodic of period 2k because ε is periodic of period 2k and the
suspension just shifts the resolution (leaving the period unaltered) and the dual functor forms a periodic
resolution of the same period.
If G = hti is finite cyclic of order n, and k = 1, then G acts by rotations on the circle (n vertices/edges)
and we have a periodic resolution of period 2:
t−1 N t−1 ε
· · · → ZG → ZG → ZG → ZG → Z → 0
n−1
where N = 1 + t + · · · + t is the norm element. Now ε(1) = 1, so by Proposition VI.3.4[1] the dual
∗ ∗
P t−1 N
ε : Z → ZG is given by ε (1) = g∈G g = N . The maps ZG → ZG and ZG → ZG are invariant under
the dual functor because HomG (ZG, ZG) ∼ = ZG. Therefore, the explicit complete resolution is:
t−1 N t−1 N t−1 N t−1
· · · → ZG → ZG → ZG → ZG → ZG → ZG → ZG → ZG → · · · .

5.1: We have a natural map H ∗ → H b ∗ which is an isomorphism in positive dimensions and an

epimorphism in dimension 0. The cup product for both functors agrees in dimension 0 because `:
H 0 (G, M ) ⊗ H 0 (G, N ) → H 0 (G, M ⊗ N ) is the map M G ⊗ N G → (M ⊗ N )G induced by the in-
clusions M G ,→ M and N G ,→ N , and `: H b 0 (G, M ) ⊗ H
b 0 (G, N ) → H
b 0 (G, M ⊗ N ) is induced by
G G G 0 0
M ⊗ N → (M ⊗ N ) via the surjection H  H b . From this compatibilism in dimension 0 we can
deduce that the diagram
H p (G, M ) ⊗ H q (G, M )
`
/ H p+q (G, M ⊗ N )

 
H b q (G, M ) `
b p (G, M ) ⊗ H /H
b p+q (G, M ⊗ N )
commutes for all p, q ∈ Z. Indeed, embed M in the (co)induced module M̄ = Hom(ZG, M ) ∼ = Z⊗M
(noting that G is finite for our purposes) and let 0 → M → M̄ → C → 0 be the canonical Z-split exact
sequence (see Exercise III.7.3). For any G-module N the sequence 0 → M ⊗N → M̄ ⊗N → C ⊗N → 0 is

47
exact, and the module M̄ ⊗ N is induced (see Exercise III.5.2(b)). We therefore have dimension-shifting
isomorphisms δ : H p (G, C) → H p+1 (G, M ) and δ : H p (G, C ⊗ N ) → H p+1 (G, M ⊗ N ) which commutes
with the cup product (see pg110[1]). The compatibilism for p = 0, q = 0 allows us to prove by ascending
induction on p that the above diagram is commutative for p ≥ 0 and q = 0. Embedding N in an induced
module, we then see that it commutes for q ≥ 0. The scenario p, q < 0 is trivial because H p = 0 for
p < 0. Thus the cup product on H b ∗ is compatible with that defined originally on H ∗ , and the natural
∗ ∗
map H → H preserves products.
b

5.2(a): Let H b ∗ (G)(p) be the p-primary component of Hb ∗ (G) = Hb ∗ (G, Z), so that we have H b ∗ (G) =
L b∗ b∗ b∗ b∗
p | |G| H (G)(p) . Since H (G)(p) is a subgroup of H (G), in order to show that it is an ideal in H (G)
∗ ∗ ∗
it suffices to show that α ` β ∈ H b (G)(p) for α ∈ H
b (G)(p) and β ∈ H b (G). But this is trivial because
s s s
if p α = 0 then p (α ` β) = (p α) ` β = 0 ` β = 0. Now
L
Hb ∗ (G)(q) is also an ideal because
q6=p
a sum of ideals is an ideal. Consequently, H b ∗ (G)(p) is a quotient ring of Hb ∗ (G) via the projection
∗ ∗
Hb (G)  H b (G)(p) .
Note: The p-primary ring H b ∗ (G)(p) is not a subring of H b ∗ (G) because the inclusion Hb ∗ (G)(p) ,→
∗ ∗
b (G)(p) is 1 ∈ Zpr = H 0
Hb (G) does not preserve identity elements. The identity of H b (G)(p) while the
∗ 0
identity of H (G) is 1 = (1, . . . , 1) ∈ Z|G| = H (G), where we have used the factorization of Z|G| into the
b b
direct sum of its p-primary components and |G| = pr m with p - m. The inclusion will send the identity
1 to (1, 0, 0, . . . , 0) 6= 1, where we take the first summand of Z|G| to be Zpr .
∼
= Q
5.2(b): Consider the group isomorphism ϕ : H b ∗ (G) → b∗
p | |G| H (G)(p) , where each factor on the
right is a ring via part(a), and multiplication in the product is done componentwise [note: we switch
from direct sum to direct product notation in order to emphasize the fact thatPwe are dealing cat-
egorically with rings]. The map is given by ϕ(α) = (. . . , αp , . . .), where α = p αp in the decom-
∗ ∗
L
position H (G) =
b H (G)(p) . In order to show that this map is a ring isomorphism it suf-
b
p | |G|
fices to show that (αβ)p = αp βp , where αβ = α ` β, for then ϕ(α P ` β) = (. . . , (αβ)
Pp , . . .) =
(. . . , αp βp , . .P
.) = (. . . , αp , .P
. .)(. . . , βp , . . .) = ϕ(α) ` ϕ(β). Writing α = p αp and β = p0 βp0 we
have αβ = p,p0 αp βp0 = p αp βp , because αp βp0 6=p is annihilated by both p and p0 6= p, and hence
annihilated by 1 = gcd(p, p0 ) [note that gcd(p, 0 0
P p ) = mp + np for some m, n ∈ Z by the Euclidean
Algorithm]. It is now obvious that (αβ)p = ( q αq βq )p = αp βp .

6.1:

6.2: Let R be a ring, let C be a chain complex of finitely generated projective R-modules, and
let C̄ be the dual complex HomR (C, R) of finitely generated projective right R-modules. For any
z ∈ (C̄ ⊗ b R C)n any any chain complex C 0 , there is a graded map ψz : HomR (C, C 0 ) → C̄ ⊗ b R C 0 of
degree n, given by ψz (u) = (idC̄ ⊗u)(z).
b Let z ∈ (C̄ ⊗ b R C)0 correspond to idC under the isomor-
phism ϕ : C̄ ⊗ b R C → HomR (C, C) from Exercise VI.6.1. This element is indeed a cycle, because
ϕ−1 (∂z) = D0 ϕ0 (z) = D0 (idC ) = d ◦ idC − (−1)0 idC ◦ d = d − d = 0 implies that ∂z = 0 since ϕ
is an isomorphism. Then z = (zp )p∈Z , where zp ∈ C̄−p ⊗R Cp = (Cp )∗ ⊗R Cp corresponds to (−1)p idCp
under the canonical isomorphism φ : (Cp )∗ ⊗R Cp ∼ = HomR (Cp , Cp ) of Proposition I.8.3[1] given by
c̄ ⊗ c 7→ hc̄, xic. Indeed, denoting zp = c̄ ⊗ c, the canonical isomorphism gives hc̄, xic = (−1)p idCp and the
2
isomorphism ϕ0 = (ϕ−pp )p∈Z then becomes (−1)−p·p hc̄, xic = (−1)p (−1)p idCp = (−1)p(p+1) idCp = idCp ,
agreeing with our choice for z. Now ψz is induced by maps ψpq : HomR (C−p , Cq0 ) → C̄p ⊗R Cq0 . Since
u ∈ HomR (C−p , Cq0 ) is of degree p + q and idC̄p is in dimension p, these maps are clearly given by
ψpq (u) = (−1)(q+p)p (idC̄p ⊗ u)(z−p ); see the definition of a map between completed tensor products on
pg137[1]. Then Exercise I.8.7 states that ϕ−1 0
pq = ψϕ−1 (idCp ) = ψr , where ψr : HomR (C−p , Cq ) → C̄p ⊗R Cq
0

−1 pq −1 pq p
is defined by ψr (u) = (idC̄p ⊗R u)(r). Since r = ϕ (idCp ) = (−1) φ (idCp ) = (−1) (−1) z−p , we
2
have ϕ−1 pq p
pq (u) = ψr (u) = (idC̄p ⊗R u)((−1) (−1) z−p ) = (−1)
(pq+p )
(idC̄p ⊗ u)(z−p ) = ψpq (u), noting
2
p p
that (−1) = (−1) . Therefore, ψz is the inverse of the isomorphism ϕ : C̄ ⊗ b R C 0 → HomR (C, C 0 ) of
Exercise VI.6.1.

48
6.3: Let G be a finite group, let F be an acyclic chain complex of projective ZG-modules, and let
ε0 : F 0 → Z be a complete resolution. Part(b) of Proposition VI.6.1[1] states that if F is of finite type
then ε0 ⊗ M induces a weak equivalence F ⊗ b G (F 0 ⊗ M ) → F ⊗b G M = F ⊗G M . The proof of the propo-
sition considers the dual F̄ = HomG (F, ZG) and utilizes the fact that it is projective. However, if we
did not impose the finiteness hypothesis on F then its dual would not necessarily
L∞ by projective.Q∞Indeed,
the dual of an infinite direct sum is an infinite direct product, and HomG ( ZG, ZG) ∼ = ZG is
not ZG-projective. If it were ZG-projective, then by Exercise I.8.2 it would also be Z-projective Q∞ (i.e.
free abelian). ButQ∞any subgroup of a free abelian group is free abelian by Theorem I.7.3[5], and Z
is a subgroup of ZG which is not free abelian, giving the desired contradiction. Thus the finiteness
hypothesis is necessary for the given proof – this does not guarantee that the finiteness hypothesis is
necessary for the statement of the proposition.

6.4:
∼
= b
7.1(a): Let ϕ : H b i (G, M ) → H−1−i (G, M ) be the isomorphism established in the proof of Proposi-
tion VI.7.2[1], which on the chain level has the inverse ϕ−1 : HomG (F, ZG) ⊗G M → HomG (F, M ) given
by u ⊗ m 7→ [x 7→ u(x) · m], and let z = ϕ(1) ∈ H b −1 (G, Z). For an arbitrary G-module coefficient
homomorphism h : M → N we have a commutative diagram
ϕ−1
HomG (F, ZG) ⊗G M / HomG (F, M )

α β
 ϕ−1

HomG (F, ZG) ⊗G N / HomG (F, N )
where α(u ⊗ m) = u ⊗ h(m) and β(f ) = h ◦ f , because βϕ−1 (u ⊗ m) = β[u(x) · m] = h(u(x) · m) =
u(x) · h(m) = ϕ−1 (u ⊗ h(m)) = ϕ−1 α(u ⊗ m). Thus ϕ−1 is natural and hence so is ϕ. Since ϕ is natural
the following diagram with short exact rows is commutative (suppressing the end 0’s)
HomG (F, M 0 ) / HomG (F, M ) / HomG (F, M 00 )

ϕ ϕ ϕ
  
HomG (F, ZG) ⊗G M 0 / HomG (F, ZG) ⊗G M / HomG (F, ZG) ⊗G M 00
and so ϕ is compatible with connecting homomorphisms in long exact sequences by Proposition I.0.4[1].

7.1(b): By definition of z, ϕ and a z agree on 1 ∈ H b 0 (G, Z) since 1 a z = z. If now M and

0
u ∈ H (G, M ) are arbitrary, there is a coefficient homomorphism Z → M such that 1 7→ u un-
b
der the induced map α : Hb 0 (G, Z) → H
b 0 (G, M ), noting that this cohomology map is induced from
0 G G 0
H (G, Z) = Z → M = H (G, M ). Since the cap product is natural with respect to coefficient homo-
morphisms we have a commutative diagram
b 0 (G, Z)
H
az
/H
b −1 (G, Z)

α β
 
Hb 0 (G, M ) az / H
b −1 (G, M )
which defines β(z) = β(1 a z) = α(1) a z = u a z. Thus by naturality of ϕ from part(a) we have
an analogous commutative diagram as above (replacing a z with ϕ), and this yields ϕ(u) = ϕα(1) =
βϕ(1) = β(z) = u a z.

7.1(c): The maps ϕ and a z agree in dimension 0 (referring to the domain) by part(b), and ϕ is
δ-compatible by part(a). Thus we can use dimension-shifting [the δ boundary isomorphisms] to deduce
that ϕ and a z agree in all dimensions, up to sign. Indeed, we have the commutative diagram
Hb 0 (G, M ) ϕ
/Hb −1 (G, M )

∼
= δ ∼
= δ
 
b n (G, K)
H
ϕ
/H
b −1−n (G, K)

49
where the vertical maps are due to iterations of the dimension-shifting technique 5.4 on pg136[1]. These
isomorphisms provide ambiguity in the sign, so ϕ(u) = ±u a z in any dimension and hence Proposition
VI.7.2[1] has been reproved (the isomorphism is given by the cap product with the fundamental class z).

7.2: Let A be an abelian torsion group, and consider the injective resolution 0 → Z → Q → Q/Z → 0 of
δ0 δ1
Z. Applying Hom(A, −) gives the cochain complex 0 → Hom(A, Z) → Hom(A, Q) → Hom(A, Q/Z) → 0,
and we have Ext(A, Z) ≡ Ext1Z (A, Z) = Kerδ 1 /Imδ 0 = Hom(A, Q/Z)/Imδ 0 = A0 /Imδ 0 . But Q is torsion-
free, so Hom(A, Q) = 0 and Imδ 0 = 0. Thus Ext(A, Z) = A0 .
Equivalently, Theorem 17.1.10[2] provides us with a long exact sequence 0 → Hom(A, Z) → Hom(A, Q) →
Hom(A, Q/Z) → Ext(A, Z) → Ext(A, Q). But Hom(A, Q) = 0 as mentioned above, and Ext(A, Q) = 0
∼
=
by Proposition 17.1.9[2] since Q is Z-injective. Thus we have a desired isomorphism A0 = Hom(A, Q/Z) →
Ext(A, Z).

7.3: Let G be a finite group, let M be a G-module which is free as an abelian group, and let F be
a projective resolution of Z over ZG. Note that M ∗ = Hom(M, Z) by Proposition VI.3.4[1]. Consider
the split exact coefficient sequence 0 → Z → Q → Q/Z → 0. Since M is Z-free, applying the functor
Hom(M, −) yields the exact sequence 0 → M ∗ → Hom(M, Q) → M 0 → 0 where M 0 = Hom(M, Q/Z);
this sequence is Z-split exact because the original coefficient sequence is split exact (Hom commutes
with direct sums). I claim that H b ∗ (G, Hom(M, Q)) = 0 and H b ∗ (G, Hom(M, Q) ⊗ M ) = 0. Assuming
this for the moment, we then have dimension-shifting isomorphisms δ : H b j (G, M 0 ) → H
b j+1 (G, M ∗ ) for
all j. It is a fact that the tensor product of a G-module with a Z-split exact sequence is exact, so
0 → M ∗ ⊗ M → Hom(M, Q) ⊗ M → M 0 ⊗ M → 0 is an exact sequence. Thus we also have dimension-
shifting isomorphisms δ : H b j (G, M 0 ⊗ M ) → Hb j+1 (G, M ∗ ⊗ M ). Moreover, we have a commutative
diagram
b i−1 (G, M 0 ) ⊗ H
H b −i (G, M ) `
/H
b −1 (G, M 0 ⊗ M ) α /H
b −1 (G, Q/Z)

∼
= δ⊗id ∼
= δ ∼
= δ
  
b i (G, M ∗ ) ⊗ H
H b −i (G, M ) `
/H
b 0 (G, M ∗ ⊗ M ) β /H
b 0 (G, Z)
where the left-side square follows from compatibility with δ (see pg110[1]) and the right-side square
follows from naturality of the long exact cohomology sequence (see pg72[1]); α and β are induced by the
evaluation maps. Since the top row is a duality pairing by Corollary VI.7.3[1], so is the bottom row.
It suffices to prove the claim. The analog of Proposition III.10.1[1] for Tate cohomology states that if
Hb n (H, M ) = 0 for some n with H ⊆ G, then H n (G, M ) is annihilated by |G : H|. Taking H = {1}
and M a rational vector space, the norm map N : MH → M H is an isomorphism (N = idM ). Then
Hb −1 ({1}, M ) = KerN = 0 = CokerN = H b −1 (G, M ) and H
b 0 ({1}, M ), so H b 0 (G, M ) are annihilated by
|G| and are thus trivial groups since |G| is invertible in M . The claim is now justified since Hom(M, Q)
and Hom(M, Q) ⊗ M are both rational vector spaces.

7.4: Let k be an arbitrary commutative ring and Q an injective k-module. Let A0 = Homk (A, Q) for any
`
k-module A. If M is a kG-module, then the pairing H b i (G, M 0 ) ⊗ H
b −1−i (G, M ) → Hb −1 (G, M 0 ⊗ M ) →
Hb −1 (G, Q) ,→ Q induces an isomorphism H b i (G, M 0 ) ∼
=H b −1−i (G, M )0 . Indeed, this is simply the analog
of Corollary VI.7.3[1], and the proof of that corollary goes through untouched if we replace Q/Z by Q
(as both are injective) and Z by k (as both are commutative ring coefficients).

8.1: Let G be a group and M a ZG-module such that H b ∗ (G, M ) = 0 but M is not cohomologically
trivial. If G is cyclic, then by Theorem VI.8.7[1] it cannot be a p-group, so G = Zp ×Zq ∼ = Zpq for distinct
n
primes p and q. The complete resolution from Exercise VI.3.1 then implies that H (G, M ) = CokerN
b
for n even and H b n (G, M ) = KerN for n odd (see pg58[1]), so N : MG → M G is an isomorphism. Also,
Proposition VI.8.8[1] implies that either H b i (Zp , M ) 6= 0 for some i or H b j (Zq , M ) 6= 0 for some j (or
both). As mentioned on pg150[1] as a consequence of Theorem VI.8.5[1], H b i (Zp , Zp ) 6= 0 for all i > 0.
Therefore, let us consider M = Zp . It suffices to find a Zpq -action on Zp such that N : (Zp )Zpq → (Zp )Zpq
is an isomorphism. But Zp is simple, so either Zpq acts trivially on Zp or (Zp )Zpq = (Zp )Zpq = 0. But if

50
Zpq acts trivially on Zp then N : Zp → Zp is the zero map (hence has nontrivial kernel) since multipli-
cation by |Zpq | = pq annihilates Zp . Thus we must have (Zp )Zpq = (Zp )Zpq = 0. Taking p to be an odd
prime, this condition is satisfied by the Zpq -action x · mi = m2i , where Zp = hmi and Zpq = hxi, as long
pq
as 2pq ≡ 1 mod p (because we must have m = xpq · m = m2 ). For example, {p = 3, q = 2} works, as
does {p = 7, q = 3}.
As a result, a desired example is the group G = Z6 = hxi and the Z6 -module M = Z3 = hmi coupled
with the action x · mi = m2i .

8.2: Suppose M is a G-module which is Z-free and cohomologically trivial (G is of course finite).
Then M is ZG-projective by Theorem VI.8.10[1], so F = M ⊕ N for some projective module N and free
module F. For any G-module K, the module Hom(F, K) is an induced module by Exercise III.5.2(b)
(since F ∼
= ZG ⊗ F0 = IndG 0 0
{1} F where F is a free Z-module of the same rank) and hence cohomologically
trivial. Since the Hom-functor commutes with direct sums, Hom(M, K) is also cohomologically trivial
for any G-module K.
Alternatively, for any G-module K choose an exact sequence 0 → L → F → K → 0 with F free (such
sequences exist because every module is a quotient of a free module). Since M is Z-free, we can apply
Hom(M, −) to get the exact sequence 0 → Hom(M, L) → Hom(M, F ) → Hom(M, K) → 0. Since F
and L are also Z-free, Hom(M, L) and Hom(M, F ) are cohomologically trivial by Lemma VI.8.11[1]; L
is free because it embeds in the free Z-module F and any submodule of a Z-free module is free. Thus
Hom(M, K) is cohomologically trivial by the long exact Tate cohomology sequence.

8.3(a): Let M and P be ZG-modules such that M is Z-free and P is ZP -projective, and consider
i ϕ
any exact sequence 0 → P → E → M → 0. The obstruction to splitting the sequence lies in
H 1 (G, Hom(M, P )) ∼ 1
= ExtZG (M, P ), where the isomorphism follows from Proposition III.2.2[1]. More
precisely, we have a short exact sequence of G-modules 0 → Hom(M, P ) → Hom(M, E) → Hom(M, M ) →
δ
0 since M is Z-free, and this yields the sequence HomG (M, E) → HomG (M, M ) → H 1 (G, Hom(M, P ))
via the long exact cohomology sequence, where we recall that HomG (·, ·) = Hom(·, ·)G = H 0 (G, Hom(·, ·)).
Hence the extension splits iff δ(idM ) = 0, because if the extension splits then there is a section s : M → E
which maps onto idM (i.e. s 7→ ϕ ◦ s = idM ) and so idM ∈ Kerδ by exactness, and if δ(idM ) = 0 then
by exactness there exists a map M → E which maps onto idM and that map is then the desired section.
It suffices to show that Hom(M, P ) is cohomologically trivial, for then H 1 (G, Hom(M, P )) = 0 and
δ = 0. By additivity [Hom commutes with direct sums], it suffices to show that Hom(M, F) is cohomo-
logically trivial for any free ZG-module F, since the projective P is a direct summand of some F. But
F∼= CoindG 0 0
{1} F where F is a free Z-module of the same rank (by Corollary VI.2.3[1]), so Hom(M, F) is
induced (by Exercise III.5.2(b)) and hence cohomologically trivial.
Alternatively, since M is Z-free the original exact sequence in consideration is Z-split (see Exercise
AE.27), so the injection i : P → E of G-modules is a Z-split injection, hence admissible. Since P is
G-projective, it is relatively injective by Corollary VI.2.3[1] and so the mapping problem
P
i /E

idP
  f
P
can be solved (i.e. there exists a map f : E → P such that f ◦ i = idP ). But this just means that
i ϕ
f : E → P is a ZG-splitting homomorphism for the sequence 0 → P → E → M → 0, and so this
sequence splits.

8.3(b): Let M be a ZG-module such that proj dim M < ∞, and consider the projective resolution
0 → Pn → · · · → P0 → M → 0. We can break this up into short exact sequences Zi → Pi → Zi−1 → 0,
where Zi is the kernel of Pi → Pi−1 . Now Zi (i ≥ 0) is Z-free because it is a submodule of a
ZG-projective module which is a submodule of a ZG-free module F, and F is also necessarily Z-
free and any subgroup (in particular, Zi ) of a Z-free group is Z-free. Therefore, for the sequence
0 → Pn = Zn−1 → Pn−1 → Zn−2 → 0 with Pn G-projective and Zn−2 Z-free, part(a) above implies that
this sequence splits and hence Pn−1 ∼
= Pn ⊕ Zn−2 . Since Pn−1 is G-projective, so is Zn−2 . One now sees
by descending induction on i that Zi is G-projective for i ≥ 0, so that 0 → Z0 → P0 → Z−1 = M → 0

51
is a projective resolution of length 1. Thus proj dim M ≤ 1.

8.4: Let G be a group such that there exists a free, finite G-CW-complex X with H∗ (X) ∼ = H∗ (S 2k−1 ).
2k−1
Since Hi (S ) = 0 for i 6= 0 and i 6= 2k − 1, the augmented cellular chain complex C∗ = C∗ (X) is a free
∂2k−1 ∂2k−2
resolution of Z over ZG up to dimension 2k − 1. From the chain sequence C2k−1 → C2k−2 → C2k−3
we have Ker∂2k−2 = Im∂2k−1 ∼ = C2k−1 /Ker∂2k−1 [the isomorphism is due to the 1st Isomorphism
∂2k−2
Theorem] and hence we have an exact sequence 0 → C2k−1 /Ker∂2k−1 ,→ C2k−2 → C2k−3 → · · · .
Now Im∂2k = B is the module of (2k − 1)-boundaries of C∗ , and we have a surjection C2k−1 /B 
C2k−1 /Ker∂2k−1 with kernel Ker∂2k−1 /B ∼
= Z [note: this isomorphism comes from the fact that Z ∼
=
∂2k−2
H2k−1 (S 2k−1 )]. Thus we have an exact sequence S of G-modules 0 → Z → C2k−1 /B → C2k−2 →
ε
C2k−3 → · · · → C0 → Z → 0 where each Ci is G-free. I claim that C2k−1 /B is Z-free and has finite projec-
tive dimension. Assuming this claim holds, C2k−1 /B is cohomologically trivial by Theorem VI.8.12[1] and
hence is ZG-projective by Theorem VI.8.10[1]. Thus we can splice together an infinite number of copies of
S (which forms an acyclic complex of projective G-modules) and we can then apply Proposition VI.3.5[1]
to obtain a complete resolution which is periodic of period 2k. It suffices to prove the claim. Since C2k−1
is ZG-free, it is necessarily Z-free and hence any subgroup (in particular, B) must also be Z-free; thus
C2k−1 /B is Z-free. As X is a finite complex, C∗ (X) stops after Cn (X) for some 2k − 1 < n < ∞.
n ∂ 2k ∂
Thus we have a finite projective resolution 0 → Cn → · · · → C2k → C2k−1 → C2k−1 /B → 0 and
proj dim C2k−1 /B < ∞.

9.1: Let G be a nontrivial finite group which has periodic cohomology of period d. Then there is
an element u ∈ H b ∗ (G), so cup product with u gives a peri-
b d (G) which is invertible in the ring H
odicity isomorphism v 7→ u ` v. Taking v = u and using anti-commutativity of the cup product,
2
u ` u = (−1)d (u ` u). If d is not even, then 2u2 = 0. If |G| = 2 then G is cyclic (of order 2) and hence
has period d = 2 (which is even), so we must have |G| ≥ 3. But then 2u2 = 0 implies that u ` u = u2 = 0
and hence u = 0 by the periodicity isomorphism. This contradicts the fact that u is nontrivial (it is
invertible), so d must be even.

9.2: If G is cyclic then Hb ∗ (G) is periodic of period 2 because G admits a 2-dimensional fixed-point-
free representation as a group of rotations (see pg154[1]); we could also just note that Hb i (G) is Z|G|
for i even and is 0 for i odd. Conversely, if a group G has periodic cohomology of period 2, then
Gab = H1 G = H b −2 (G) ∼
=H b 0 (G) = Z|G| . Now |G| = |Gab | = |G|/|[G, G]| ⇒ |[G, G]| = 1 and hence
∼
G = Gab = Z|G| , which is cyclic.

9.3: Suppose H b ∗ (G) is periodic of period 4. We have H b −1 (G) = 0 and Hb 0 (G) = Z|G| , as explained on
pg135[1]. We also have H b −2 (G) = H1 G = Gab , and Hb 1 (G) = H 1 G = Hom(G, Z) = 0 by Exercise III.1.2
(noting that G is finite by hypothesis). Thus, since H b n (G) ∼=Hb n+4 (G) for all n,
 

 Z n=0 
 Z n=0
 
G n ≡ 1 mod 4  Z|G| n ≡ 0 mod 4 , n 6= 0
Hn (G) ∼
ab
= H n (G) ∼
=

 Z
 |G| n ≡ 3 mod 4 G
 ab

 n ≡ 2 mod 4
0 otherwise 0 otherwise
 

Note that this reproves Exercise II.5.7(a), because the finite subgroup G ⊂ S 3 ⊂ H∗ has H b ∗ (G) peri-
odic of period 4 (see pg155[1]) and hence H2 G = 0.

9.4: Suppose the finite group G has periodic cohomology. Now H b ∗ (G) ∼ Q b ∗ (G)(p) by Exer-
= p | |G| H
cise VI.5.2(b), and each factor H b ∗ (G)(p) embeds in Hb ∗ (Sylp (G)) by the Tate cohomology version of
i i
Theorem III.10.3[1], so H b (G) = 0 for i odd if H b (Sylp (G)) = 0 for each prime p. Thus to show that
H (G) = 0 for i odd it suffices to do this when G is a p-group. For the p-group G (of order pr ) with
b i

periodic cohomology, G is either a cyclic group Zpr or a generalized quaternion group Q2r by Proposition
VI.9.3[1]. If G is cyclic then it has period 2 by Exercise VI.9.2, so since H b −1 (G) = 0, Hb i (G) = 0 for i
odd. If G is generalized quaternion then it has period 4 as explained on pg155[1] (it is a finite subgroup
of H∗ ), so by Exercise VI.9.3, Hb i (G) = 0 for i odd.

52
9.5: Suppose G is a p-group which has a unique subgroup C of order p; note that C is necessarily
cyclic. Choose a fixed-point-free representation of C on a 2-dimensional vector space W (as a group of
L the induced module V = ZG ⊗ZC W . Since C is unique, it is normal in G and
rotations), and form
∼
hence ResGC V = g∈G/C gW by Proposition III.5.6[1]; each gW is clearly a fixed-point-free representa-
tion of C (c · gw = (cg)w 6= gw). Consequently, V is a fixed-point-free represenation of C. But then
V is also a fixed-point-free representation of G, for a nontrivial isotropy group Gv (v ∈ V − {0}) would
contain an element x of order p (by Cauchy’s Theorem, Theorem 3.2.11[2]) and hence would contain
C (uniqueness implies C = hxi), contradicting the fact that C acts freely on V − {0}. Thus G admits
a periodic complete resolution of period 2|G : C| as explained on pg154[1], so G has periodic cohomology.

9.6: Let G = Zm o Zn , where m and n are relatively prime and Zn acts on Zm via a homomor-
phism Zn → Z∗m whose image has order k. If a prime q divides n, then a Sylow q-subgroup H lies
in Zn and is necessarily central (in Zn ) because cyclic groups are abelian. By Theorem III.10.3[1] we
have H b ∗ (Zn )(q) ∼= H b ∗ (H)Zn /H and by Exercise III.8.1 we know that Zn /H acts trivially on H b ∗ (H),
∗ ∼ ∗ ∗ ∗
so H (H) = H (Zn )(q) . By Theorem III.10.3[1] we also have H (G)(q) ⊆ H (H) = H (Zn )(q) . Since
b b b b ∼ b ∗

Hb ∗ (Zn )(q) ⊆ H b ∗ (G)(q) by Exercise AE.55, we must have H b ∗ (G)(q) ∼ b ∗ (Zn )(q) . Now if a prime p divides
=H
m, then the same argument yields H ∗
b (H) = H∼ ∗
b (Zm )(p) where H is now a Sylow p-subgroup of Zm ⊂ G.
We have H /G because H is the unique subgroup of Zm /G (hence gHg −1 ∼ = H for all g ∈ G), so Theorem
∗
III.10.3[1] implies H (G)(p) = H (H)
b ∼ b ∗ G/H ∼ b ∗ G/H b ∗
= H (Zm )(p) = H (Zm )(p)
(Zm /H)oZn
. We have a Zn -action
on Zm given by ϕ : Zn → Aut(Zm ) = Z∗m , and we have a trivial Zm /H-action ψ : Zm /H → Aut(Zm )
given by ψ(g) = idZm ≡ 1. We then have a (Zm /H) o Zn -action ψ o ϕ which is precisely the action
ϕ : Zn → Z∗m because (ψ o ϕ)(g, z) = ψ(g) · ϕ(z) = 1 · ϕ(z) = ϕ(z). Thus we can consider the G/H-action
on Zm (hence on its Tate cohomology) as the Zn -action, so H b ∗ (G)(p) ∼
=H b ∗ (Zm )Zn . Since H b ∗ (G) is the
(p)
direct sum of its primary components and p (resp. q) ranges over prime divisors of m (resp. n), we have
the isomorphism H b ∗ (G) ∼=Hb ∗ (Zn ) ⊕ H b ∗ (Zm )Zn .
Let us examine the Zn -action a little more carefully. The image of Zn under the action-homomorphism
consists solely of automorphisms f : Zm → Zm such that f k = idZm . Such a map induces f∗ = H b 2 (f ) on
Hb (Zm ) ∼
2 nd
= Zm . If α generates the 2 -dimension cohomology, then in the cohomology ring, f∗ (α) = λα
for some λ ∈ Zm . But α = id∗ (α) = f∗k (α) = λk α, so λk ≡ 1 mod m. Noting that H b i (Zm ) = 0 for i
i
odd, but is nontrivial for i even. Now f∗ (α ) = f∗ (α ` · · · ` α) = f∗ (α) ` · · · ` f∗ (α) = (λα) ` · · · `
(λα) = λi αi which is the identity only when i is a multiple of 2k. Thus H b i (Zm )Zn is nontrivial only
when i ∈ 2kZ (in which case it is Zm since the action is trivial). Therefore,

Zmn
 i ≡ 0 mod 2k
i ∼ i
H (Zm o Zn ) = H (Zn ) ⊕ H (Zm ) = 0
b b b i Zn ∼
i odd

otherwise

Zn

for the case gcd(m, n) = 1. This means that the period of H ∗ (G) is 2k.

9.7: Let Fq be a field with q elements, where q is a prime power. The special linear group G = SLn (Fq ) is
the kernel of the surjective determinant homomorphism det : GLn (Fq ) → F∗q , i.e. it is the group of matri-
ces with determinant 1. Let us first assume that n ≥ 3. Then the cyclic groups A = hdiag(a, a−1 , 1, . . . , 1)i
and B = hdiag(1, . . . , 1, b, b−1 )i form a non-cyclic abelian subgroup A × B ⊆ G, noting that Fq is commu-

tative. If we now let n = 2 then we will assume that q is not prime. Then the cyclic groups A = h 1a i




0 1
and B = h 10 1b i form a non-cyclic abelian subgroup A × B ⊆ G, since 10 a1 10 1b = 10 a+b 1 with b not
equal to any multiple of a (and vice versa). By Theorem VI.9.5[1], G = SLn (Fq ) does not have periodic
cohomology if n ≥ 3 or if q is not prime.
Note that if n = 2 and q is prime then SL2 (Fq ) does have periodic cohomology, as explained on pg157[1].

9.8:

9.9: Suppose that G has p-periodic cohomology. Let P ⊆ G be a subgroup of order p, let N (P )
(resp. C(P )) be the normalizer (resp. centralizer) of P in G, and let W = N (P )/C(P ); note that

53
if C(P ) = P then W is called the Weyl group. Choose a Sylow p-subgroup H containing P . Since
G has p-periodic cohomology, H is either cyclic or generalized quaternion by Theorem VI.9.7[1]. I
assert that H ⊆ C(P ) ⊆ N (P ). Indeed, if H is cyclic then it is necessarily abelian so the result fol-
lows, and if H is generalized quaternion (must have p = 2) then it has a unique element of order 2
(as stated on pg98[1]) and this is then the generator for P ∼= Z2 which must be central in H, so the
result follows. Denote by XG the G-invariant elements of H b ∗ (H, M ) and similarly for XN (P ) , where
b ∗ (H, M ) is G-invariant if resH gHg −1
an element z ∈ H −1 z = res
H∩gHg −1 gz for all g ∈ G.
H∩gHg Theorem
III.10.3[1] states that H b ∗ (G, M )(p) ∼
= XG and H b ∗ (N (P ), M )(p) ∼
= XN (P ) . Now trivially, XG ⊆ XN (P ) ,
so it suffices to show that XN (P ) ⊆ XG , for then H (G, M )(p) ∼
b ∗
= XG = XN (P ) ∼ = H b ∗ (N (P ), M )(p) .
Note that P is the unique subgroup in H of order p because if H is generalized quaternion then the
reasoning is as stated above and if H is cyclic then every subgroup has unique order (by Theorem
2.3.7[2]). If H ∩ gHg −1 is trivial then every element in H ∗ (H, M ) is clearly invariant for such g ∈ G.
If H ∩ gHg −1 is not trivial then its order is at least p and the intersection contains P . This implies
P ⊆ gHg −1 ⇒ g −1 P g ⊆ H ⇒ g −1 P g = P and hence g ∈ N (P ), so the question of G-invariance
reduces to the question of N (P )-invariance, i.e. XN (P ) ⊆ XG .
Also, XN (P ) ⊆ XC(P ) trivially, so by Theorem III.10.3[1] we have the inclusion (up to isomorphism)
Hb ∗ (N (P ), M )(p) ⊆ H b ∗ (C(P ), M )(p) . Since H is a Sylow p-subgroup contained in C(P ), |W | and
N (P ) N (P )
p are relatively prime and hence corC(P ) resC(P ) = |N (P ) : C(P )| = |W | is an isomorphism on
Hb ∗ (N (P ), M )(p) , where the equality is due to Proposition III.9.5[1]. Thus the restriction map induces
a monomorphism H b ∗ (N (P ), M )(p) ,→ Hb ∗ (C(P ), M ). But as explained on pg84[1], if z = resN (P ) u
C(P )
∗ N (P )
then z is N (P )-invariant; let the N (P )-invariants be denoted by Y ⊆ H (C(P ), M ). Thus resC(P )
b
maps Hb ∗ (N (P ), M )(p) monomorphically into Y . Since C(P ) / N (P ), Y = H b ∗ (C(P ), M )W as noted
∗
b (N (P ), M )(p) ⊆ H ∗ W
on pg84[1]. Thus H b (C(P ), M ) , and so combining the two inclusions we see that
∗ ∗ b ∗ (C(P ), M )W = Hb ∗ (C(P ), M )W . For the other direction, if
H (N (P ), M )(p) ⊆ H (C(P ), M )(p) ∩ H
b b
(p)
z ∈ Y(p) = Hb ∗ (C(P ), M )W then consider the element w = corN (P ) z. Since H b ∗ (C(P ), M )(p) is annihi-
(p) C(P )
lated by a power of p, w ∈ H b ∗ (N (P ), M )(p) . Regurgitating the proof of Theorem III.10.3[1] using the
N (P ) 0
double-coset formula, we deduce that z = res C(P ) w where w0 = w/|W | ∈ H
b ∗ (N (P ), M )(p) . This means
that H(C(P ), M )W ⊆H b ∗ (N (P ), M )(p) because resN (P ) maps H b ∗ (N (P ), M )(p) monomorphically into
(p) C(P )
the N (P )-invariants. Thus H b ∗ (G, M )(p) ∼
=Hb ∗ (N (P ), M )(p) ∼
=Hb ∗ (C(P ), M )W .
(p)

9.10: For any finite group G, the augmentation ideal I ⊂ ZG is a cyclic G-module if G is cyclic
group by Exercise I.2.1(b); we could also just note that if G = hsi then I = ZG · (s − 1) because I
consists of elements of the form sk − 1 [we then form the elements si − sj ∈ I via summation], and
sk − 1 = N · (s − 1) where N = sk−1 + · · · + s + 1 ∈ ZG. Conversely, if I is cyclic as a G-module, so that
I = ZG · x, then I claim that G admits a periodic resolution of period 2. Assuming this for the moment,
G then has periodic cohomology (of period 2) by Theorem VI.9.1[1] and hence G is cyclic by Exercise
VI.9.2. It suffices to prove the claim. The multiplication map ZG → ZG given by r 7→ rx has image I
x ε
and kernel K, so we can form the exact sequence 0 → K → ZG → ZG → Z → 0. Under the category of
abelian groups, the Rank-Nullity Theorem gives |G| = dimZ ZG = dimZ I + dimZ Z = dimZ I + 1 for the
augmentation map ε, and gives |G| = dimZ ZG = dimZ K + dimZ I for the x-multiplication map. The first
equation implies dimZ I = |G| − 1 and the second equation then implies dimZ K = 1, i.e. K ∼ = Z as an
abelian group. So with K = hki, G acts on K via gk0 = zk0 (for z ∈ Z). But then k0 = g |G| k0 = z |G| k0
x ε
and hence z = 1, i.e. the G-action is trivial. Our exact sequence is now 0 → Z → ZG → ZG → Z → 0.
Splicing together this sequence infinitely many times, we obtain the desired periodic resolution.

54
7 Chapter VII: Equivariant Homology and
Spectral Sequences
2.1: Let 0 → C 0 → C → C 00 → 0 be a short exact sequence of chain complexes, and let {Fp C} be
the filtration such that F0 C = 0, F1 C = C 0 , and F2 C = C. Then E1q
1
= H1+q (C 0 /0) = H1+q (C 0 ) and
0 00
E2q = H2+q (C/C ) = H2+q (C ) and Epq = 0 for p ≤ 0 and p > 2. The differential d1 : Epq
1 1 1
→ Ep−1,q1
00 0 1
gives maps ∂ : Hn (C ) → Hn−1 (C ) with n = 2 + q and ∂ = d2q .
d1 ∂ d1
Now E 1 is given by · · · → E3q
1 1
= 0 →3 E2q 1
→ E1q 1
→1 E0q = 0 → · · · , and from E 2 = H(E 1 ) we see that
0
E1q = Kerd1 /Im∂ = H(C )/Im∂ = Coker∂ and E2q = Ker∂/Imd13 = Ker∂ and Epq
2 1 2 2
= 0 for p ≤ 0 and
d22q d21q
p > 2. In E 2 we have Ker∂ → 0 and Coker∂ → 0 and trivial maps for p 6= 1 and p 6= 2, because the
differential d2 : Epq
2 2
→ Ep−2,q+1 is of bidegree (−2, 1). Thus H(E 2 ) = E 2 , and so the spectral sequence
“collapses” at E (i.e. E = E ) because E r+1 ∼
2 2 ∼ ∞
= H(E r ) in general.
∞ ϕ
Since Ep = Grp H(C) := Fp H(C)/Fp−1 H(C) where Fp H(C) = Im{H(Fp C) → H(C)}, we see that
Ep∞ = 0 for p ≤ 0 and p > 2 and Coker∂ = E 2 ∼ = E1∞ = F1 H(C) and Ker∂ = E 2 ∼ = E2∞ =
F2 H(C)/F1 H(C). Since F2 H(C) = H(C) we have a surjection Hn (C)  F2 H(C)/F1 H(C) ∼ = Ker∂,
ϕ
and since F1 H(C) = Im{H(C 0 ) → H(C)} we have a commutative diagram
Hn (C 0 ) / Hn (C)
O
ϕ
%%  ?
F1 Hn (C)
hence an injection Coker∂ = F1 Hn (C) ,→ Hn (C). We can now rewrite the E ∞ sequence 0 → F1 Hn (C) →
F2 Hn (C)/F1 Hn (C) → 0 as a short exact sequence 0 → Coker∂ ,→ Hn (C)  Ker∂ → 0, and we then
have a commutative diagram
Hn (C 0 )
i / Hn (C) j
/ Hn (C 00 )
9 :
ϕ
%% + $$ ,
Coker∂ Ker∂
The top row is exact at Hn (C) because Kerj = Ker{Hn (C)  Ker∂} = Coker∂ = Imϕ = Imi. Since
Imj = Ker∂ and Keri = Kerϕ = Im∂, the top row of this commutative diagram extends to a long exact
homology sequence.
We have thus deduced the familiar long exact homology sequence from this spectral sequence, and the
spectral sequence of a filtered complex can be regarded as a generalization of the long exact sequence
associated to a chain complex and a subcomplex.

3.1(a): Let C be a first-quadrant double complex such that the associated spectral sequence to Fp (T C)n =
1 1 1
L
i≤p C i,n−i has Epq = 0 for q 6= 0, andLlet D be the chain
L complex E∗,0 withLdifferential d . We
have the isomorphisms Hn (T C) = Ker[ p≤n Cp,n−p → p≤n−1 Cp,n−1−p ]/Im[ p≤n+1 Cp,n+1−p →
1
L L L
C
p≤n p,n−p ] = [ H
p≤n−1 n−p (C p,∗ )]⊕X = [ E
p≤n−1 p,n−p ]⊕X = 0⊕X, where X = Cn,0 /(Im[Cn+1,0 →
1
Cn,0 ] ⊕ Im[Cn,1 → Cn,0 ]) = Hn (C∗,0 /Im[C∗,1 → C∗,0 ]) = Hn (H0 (C∗,∗ )) = Hn (E∗,0 ) = Hn (D). Thus
Hn (T C) ∼ = Hn (D).

3.1(b): Take τ : T C → D to be the canonical surjection, and note that this can be viewed as a
map of double complexes C → D (where D is regarded as a double complex concentrated on the line
1 1 1
q = 0); this is obviously a filtration-preserving chain map. Now Epq (D) = Hq (Ep,∗ ≡ Ep,0 ) which is 0 if
1 1
q 6= 0, and is Ker∂0 ≡ Ep,0 if q = 0 (since Ep,−1 = 0). This is precisely the spectral sequence associated to
C, so the induced map on spectral sequences from τ is an isomorphism at the E 1 -level. Thus by Propo-
sition VIII.2.6, τ induces an isomorphism H∗ (T C) → H∗ (T D) ∼ = H∗ (D) and hence is a weak equivalence.

4.1: Suppose X is the union of subcomplexes Xα such that every non-empty intersection Xα0 ∩ · · · ∩ Xαp
(p ≥ 0)Lis acyclic, and let K be the nerve of the covering {Xα }. Let C be the double complex
Cpq = σ∈K (p) Cq (Xσ ), and let T be the [total] chain complex T C. As shown on pg167[1], we have
1
a spectral sequence with Epq equal to 0 if q 6= 0 and equal to Cp (X) if q = 0. Then Exercise 3.1(b)

55
implies that we have a weak equivalence T → C(X). Moreover, we have another spectral sequence with
1 1
Epq = Cp (K, Hq ) where Hq ≡ {Hq (Xσ )} is a coefficient system on K. Since each Xσ is acyclic, Epq is
0 for q 6= 0 and is Cp (X, Z) for q = 0. Then Exercise 3.1(b) implies that we have a weak equivalence
T → C(K). Thus we have an isomorphism H∗ (X) ∼ = H∗ (K).

7.2: Let X be a G-complex such that for each cell σ of X, the isotropy group Gσ fixes σ pointwise; in this
case the orbit space X/G
L inherits a CW-structure. Assume further that each Gσ is finite. We have a spec-
1 G
tral sequence Epq = σ∈Σp Hq (Gσ , Mσ ) ⇒ Hp+q (X, M ), where Σp is a set of representatives for Xp /G
(Xp is the set of p-cells of X) and Mσ = Zσ ⊗ M (Zσ is the Gσ -module additively isomorphic to Z, on
which Gσ acts by the orientation character χσ : Gσ → {±1}). Since Gσ fixes σ pointwise, χσ (Gσ ) = {1}
and hence Zσ = Z and hence Qσ = Z ⊗ Q = Q (where we now take rational coefficients M = Q). Now
for all q > 0, Hq (Gσ , Q) = Hq (Gσ ) ⊗ Q = 0 where the first equality is proved in Exercise AE.6 and the
latter equality follows from the fact that Hq (Gσ ) is finite (proved in Exercise AE.16). The E 1 term is
then concentrated on the line q = 0, and the spectral sequence therefore collapses at E 2 = H∗ (E 1 ) to
∗
yield H∗G (X, Q) ∼ = H∗ ( σ∈Σp H0 (Gσ , Qσ )) = H∗ (H0 (G, σ∈Σp IndG
L L
Gσ Qσ )) = H∗ (H0 (G, Cp (X; Q))) =
H∗ (Cp (X; Q)G ) = H∗ (X/G; Q), where the starred equality is the result of Shapiro’s lemma and the last
equality is given by Proposition II.2.4[1].
If X is also contractible, then X is necessarily acyclic. Thus the above result and Proposition VII.7.3[1]
imply H∗ (G, Q) ∼ = H∗G (X, Q) ∼
= H∗ (X/G; Q).
Note: The hypothesis that Gσ fixes σ pointwise is not very restrictive in practice. In the case of a
simplicial action, it can always be achieved by passage to the barycentric subdivision X̃. Indeed, for
σ 0 ⊂ X̃ which lies in σ ⊂ X, Gσ0 ⊆ Gσ . If Gσ0 did not fix σ 0 pointwise then this would break continuity
of the G-action on σ (consider two such simplices of X̃ which lie in σ and have a common face).

7.3: Let X be a G-complex and E a free contractible G-complex. There is a G-map X × E → X

which is a homotopy equivalence, so Proposition VII.7.3 implies that H∗G (X × E) → H∗G (X) is an iso-
morphism. As G acts freely on X × E, H∗G (X × E) ∼ = H∗ ((X × E)/G). Thus we have the equivalence
H∗G (X) ∼
= H∗ (X ×G E).

7.5: Let X be a G-complex and let N be a normal subgroup of G which acts freely on X. Let Zσ
be the orientation module and let Xp denote the set of p-cells of X. Let Σp be a set of representatives
for Xp /G and let Σ0p be a set of representatives for (X/N )p /(G/N ); it is easy to see that both sets
) ∼
G G/N
are in bijective correspondence. To prove L that H∗ (X, M ∼ LH∗ (X/N, M ) with any G/N -module
=
coefficient M , it suffices to show that σ∈Σp Hq (Gσ , Mσ ) = σ0 ∈Σ0p Hq ((G/N )σ0 , Mσ0 ); this is because
1 G
L
we have a spectral sequence Epq = σ∈Σp Hq (Gσ , Mσ ) ⇒ Hp+q (X, M ) and so the said isomorphism
will give isomorphic associated graded modules (since E r = H(E r−1 ) and E ∞ is the associated graded
module), and this will give H∗G (X, M ) ∼
G/N
= H∗ (X/N, M ) by Lemma VII.2.1[1]. In view of the bi-
jection Σp → Σp , σ 7→ σ , it suffices to show that Hq (Gσ , Mσ ) ∼
0 0
= Hq ((G/N )σ0 , Mσ0 ). First note
that Mσ ≡ Zσ ⊗ M = Zσ0 ⊗ M ≡ Mσ0 because the Gσ -action and the (G/N )σ0 -action on Z coin-
cide (if g ∈ Gσ preserves the orientation of σ ∈ X, gσ = +σ, then gN preserves the orientation of
σ 0 = σ = nσ ∈ X/N because +σ 0 = +σ = gσ = g · nσ = (gn)σ = (gn)σ 0 ∈ X/N for all n ∈ N )
and because the two actions coincide on M by definition (since M is a G/N -module). Thus it suffices
to show (due to the Künneth formula) that Hq (Gσ ) ∼ = Hq ((G/N )σ0 ) where the homology is now us-
ing integer coefficients, and in turn it suffices to show that Gσ ∼ = (G/N )σ0 where σ 0 is the image of
σ under the quotient X → X/N . Consider the obvious monomorphism ϕ : Gσ → (G/N )σ0 given by
g 7→ gN ; it suffices to show that ϕ is surjective. But this is immediate, because if gN ∈ (G/N )σ0
then gN σ 0 = σ 0 which implies gn1 σ = n2 σ for some n1 , n2 ∈ N which implies n−1 2 gn1 ∈ Gσ , and then
ϕ(n−1
2 gn 1 ) = n −1
2 gn1 N = n −1
2 gN = n −1
2 g(g −1
n 2 g)N = gN where we note that N / G.

10.1: The proof of Theorem VII.10.5[1] used the assumption that |G| = p (p prime) to state that
dimZp Hb n (G, Z)p = 1 for all n ∈ Z, because G ∼
= Zp has Tate cohomology group Zp in every dimen-
sion. The proof also used that assumption in order to apply Proposition VII.10.1[1] to the G-invariant
subcomplex X G of X; the isotropy group Gσ for every cell σ ∈ X − Y cannot equal G (because X G is
the largest subcomplex on which G acts trivially) and hence must be the trivial group (the only proper

56
subgroup of G ∼
= Zp ).

10.2: The extended theorem holds for |G| = p1 by the original Theorem VII.10.5[1], so we proceed
by induction, assuming the extended theorem holds for |G| = pr . Let |G| = pr+1 and remember the
hypothesis that X H is a subcomplex for all H ⊆ G. Since G is a p-group we can choose a maximal
normal subgroup N of index p, so that X G = (X N )G/N . Via induction (since |N | = pr ) we can apply the
extended theorem to X with X N [which is subcomplex by hypothesis], so every condition of the theorem
on X is also satisfied on X N . Since |G/N | = p we can apply the original theorem to X N with (X N )G/N
[which is a subcomplex since it is equal to X G ], so every condition of the theorem on X N is also satisfied
on (X N )G/N . Thus every condition of the extended theorem on X is also satisfied on (X N )G/N = X G ,
and the proof is complete.

10.3: Let X be a finite-dimensional free G-complex (G finite) with H∗ (X) ∼ = H∗ (S 2k ). Proposition

G
VII.10.1[1] (with Y = ∅) implies that H b ∗ (X, M ) = 0, noting that Gσ is trivial for all σ since X is
2 b p (G, Hq (X, M )) ⇒ H G
G-free. On the other hand, we have a spectral sequence Epq = H b p+q (X, M ).
Since the spectral sequence is concentrated on the horizontal lines q = 0 and q = 2k (the only
nonzero homology terms of the 2k-sphere), it follows that the differential d2k+1 : H b p (G, H2k (X, M )) →
Hb p−(2k+1) (G, H2k+(2k+1)−1 (X, M )) = H b p−2k−1 (G, H4k (X, M )) = 0 is an isomorphism. To prove that
every nontrivial element of G acts nontrivially on H2k (X) it suffices to show that every cyclic sub-
group of G (generated by the elements of G) acts nontrivially on H2k (X), so we are immediately re-
duced to the case where G is cyclic and nontrivial. Using the isomorphism d2k+1 with M = Z and p
odd, we conclude that H b odd (G, H2k (X)) = 0. This means G acts nontrivially on H2k (X), otherwise
Hb odd (G, H2k (X)) = H
b odd (G, Z) = Z|G| which is not equal to 0. The proof is now complete. Note that
a nontrivial G-action on H2k (X) ∼ = Z means that G = Z2 , so having every nontrivial element of G act
nontrivially on H2k (X) means that |G| ≤ 2 (the case |G| = 1 is satisfied vacuously since there are no
nontrivial elements).

57
8 Chapter VIII: Finiteness Conditions
2.1: By definition, cd Γ = 0 iff Z admits a projective resolution 0 → P → Z → 0 of length 0, i.e. Z ∼=P
and Z is ZΓ-projective. But Exercise I.8.1 states that only the trivial group Γ = {1} makes Z a projective
module. Thus the trivial group is the only group of cohomological dimension zero.
L
2.2: Take Γ = Z2 . Then cd Γ = ∞ by Corollary 2.5. Now a free ZΓ-module F is a direct sum ZΓ
where the Γ-action would be a parity-permutation on all or some of the summands, so H n
(Γ, F ) ∼
=
[ H n (Γ, (ZΓ)2 )] ⊕ [ H n (Γ, ZΓ)] where the first collection of summands has the nontrivial Γ-action.
L L
But then that module (ZΓ)2 is an induced module IndΓ{1} ZΓ, and so are the other modules ZΓ (triv-
ially). So these modules are H ∗ -acyclic and hence H n (Γ, F ) = 0 for all n > 0, i.e. sup {n | H n (Γ, F ) 6=
0 for some F } = 0.

2.7(a): Induced Γ-modules ZΓ ⊗ A are cohomologically trivial (as noted on pg148[1]) and hence have
projective dimension ≤ 1 by Theorem VI.8.12[1].

2.7(b): If proj dimR M ≤ n, then Extn+1

R (M, −) = 0 by Lemma VII.2.1[1]. For any direct summand
M 0 of M , we must then have Extn+1
R (M 0
, −) = 0 since Extn+1
R (−, −) commutes with direct sums. Thus
0
proj dimR M ≤ n by Lemma VII.2.1[1].

2.7(c): Suppose Γ is finite and M is a Γ-module in which |Γ| is invertible. It suffices to show that
M is a direct summand of an induced module ZΓ ⊗ A, for then proj dimZΓ (ZΓ ⊗ A) ≤ 1 by part(a) and
hence proj dimZΓ M ≤ 1 by part(b). Take A = M0 , where M0 is the underlying abelian group of M . By
−1
Corollary III.5.7[1] there is a Γ-module isomorphism ϕ : ZΓ ⊗ M → ZΓ ⊗ M0 given by g ⊗ mP7→ g ⊗ g m,
where ZΓ ⊗ M has the diagonal Γ-action. The inclusion
P i:M →P ZΓ ⊗ M given byP m 7→ g∈Γ g ⊗ m is
a Γ-module homomorphism because γ · i(m) = γ · ( g g ⊗ m) = g γg ⊗ γm = g g ⊗ γm = i(γ · m).
1
The Γ-module map π : ZΓ ⊗ M0 → M defined by g ⊗ m 7→ |Γ| gm is a Γ-splitting to ϕi [note: the
action on ZΓ ⊗ M0 is γ · (g ⊗ m) = γg ⊗ m]. Indeed, π[ϕi](m) = π[ϕ( g g ⊗ m)] = π[ g g ⊗ g −1 m] =
P P
1 −1 1
P
|Γ| g gg m = |Γ| |Γ|m = m = idM (m). Thus the injection ϕi : M → ZΓ ⊗ M0 splits, and M is then
(by definition of a splitting homomorphism) a direct summand of ZΓ ⊗ M0 as a Γ-module.

4.1: If Z is finitely presented as a ZΓ-module then Z is finitely generated and every surjection P  Z
(with P finitely generated and projective) has a finitely generated kernel (Proposition VIII.4.1[1]). In
particular, the augmentation map ε : ZΓ → Z has kernel I which then must be finitely generated (ob-
viously noting that ZΓ is free of rank 1). Exercise I.2.1(d) then implies Γ is a finitely generated group.
Conversely, suppose Γ is a finitely generated abelian group, so that Γ = F (S)/R is a group presentation
for Γ with |S| < ∞. Then there is an exact sequence (ZΓ)|S| → ZΓ → Z → 0 by Exercise IV.2.4(d) and
hence Z is finitely presented as a ZΓ-module by Proposition VIII.4.1[1].

6.1: Let Γ be of type F L and cd Γ = n. Then Γ is of type F P and hence of type F P∞ by Propo-
sition VIII.6.1, and so there is a partial resolution Fm → · · · → F0 → Z → 0 with each Fi free of finite
rank by Proposition VIII.4.3 (for all m ≥ 0). Due to its cohomological dimension, we can make a finite
projective resolution 0 → P → Fn−1 → · · · → F0 → Z → 0 with each Fi free and P projective. Then
Proposition VIII.6.5 implies that P is stably free, and so there is some free module F of finite rank such
that P ⊕ F is free. Take the free resolution 0 → F → F → 0 → · · · → 0 and consider its direct sum with
the finite projective resolution. This gives us a finite free resolution for Z over ZΓ of length n.

6.3: Let Γ be of type F P and cd Γ = n. Then we have a finite projective resolution Pn → · · · →

P0 → Z → 0, and taking the Hom-dual we obtain the resolution for cohomology which behaves as
· · · → HomZΓ (Pn , ZΓ) → 0. Since Pn is a finitely generated projective module, so is its Hom-dual; thus
H n (Γ, ZΓ) is a finitely generated Γ-module.

58
9 Chapter IX: Euler Characteristics
1.1: For a ring A, suppose there is a Z-valued function r on finitely generated projective A-modules,
satisfying r(P ⊕ Q) = r(P ) + r(Q) and r(A) = 1 and r(P ) > 0 if P 6= 0. I claim that A is indecomposable,
i.e. that A cannot be decomposed as the direct sum of two non-zero left ideals. Indeed, a decomposition
A = I ⊕ J yields the equation 1 = r(I) + r(J) because I and J are projective A-modules (direct sum-
mands of the free module A). Since both ideals are non-zero, r(I) ≥ 1 and r(J) ≥ 1, and this yields the
desired contradiction 1 ≥ 1 + 1 = 2.

2.1: Let P be a finitely generated projective (left) A-module and let P ∗ = HomA (P, A) be its dual.
Then P ∗ is a right A-module and we have P ∗ ⊗A P ∼ = HomA (P, P ) by Proposition I.8.3. Consider the
diagram
∼
P ∗ ⊗A P
= / HomA (P, P )

ev tr
$ x
T (A)
where ev(u ⊗ x) = u(x) is the evaluation map. The isomorphism P is given by u ⊗ x 7→ [p 7→ Pu(p) · x], and
on basis elements ei ∈ P this image homomorphism is ei 7→ j u(ei )rj · ej , where x = j rj ej . Thus
P P P
the composition is u ⊗ x 7→ tr[p 7→ u(p) · x] = i u(ei )ri = i ri · u(ei ) = i u(ri ei ) = u(x) = ev(u ⊗ x),
and the diagram is commutative.

2.4: Let F be a finitely generated free module and e : F → F a projection operator of F onto a
direct summand isomorphic to P (this is idempotent since e2 = e). Then e = i ◦ idP ◦ π, where i and π
are the inclusion and projection maps between F and P , so tr(e) = tr(i ◦ idP ◦ π) = tr(idP ) = R(P ).
Thus R(P ) is equal to the trace of an idempotent matrix defining P .

2.5: Let Γ be a group and ϕ : ZΓ → Z the augmentation map, and let P be a finitely generated projective
Γ-module. Then Proposition 2.3 implies that trZ (Z ⊗Γ idP ) = T (ϕ)(trZΓ (idP )) = T (ϕ)(RZΓ (P )). As this
is an element of T (Z) = Z, it is immediate that trZ (Z ⊗Γ idP ) is precisely the Z-rank of PΓ = Z ⊗Γ P .
Thus εΓ (P ) = T (ϕ)(RZΓ (P )).

2.6(a): Let Γ be a finite group. From the definition, trZΓ/Z : T (ZΓ) → T (Z) lifts to the map
ZΓ → T (Z) = Z given by γ 7→ trZ (µγ ), where µγ : ZΓ → ZΓ is right-multiplication by γ. Taking
µγ as a matrix over Z, it is the identity matrix for γ = 1 and is a matrix with zeros on the diagonal for
1 6= γ ∈ Γ (γ permutes the basis elements). Thus trZΓ/Z (1̄) = |Γ| and trZΓ/Z (γ̄) = 0 for 1 6= γ ∈ Γ.
Consequently, there is a well-defined homomorphism τ : T (ZΓ) → Z such that τ (1̄) = 1 and τ (γ̄) = 0
for 1 6= γ ∈ Γ, and one has trZΓ/Z = |Γ| · τ .

2.6(b): Applying Proposition 2.4 to α = idP and ϕ : Z ,→ ZΓ, and following the same method as
in the proof of Exercise 2.5, and using the result of part(a), we have that rkZ (P ) = |Γ| · τ (RZΓ (P )) for
any finitely generated projective Γ-module P . Thus ρΓ (P ) = τ (RZΓ (P )).

2.7: The previous exercise shows that ρΓ (P ) ∈ Z, for Γ a finite group. It is obvious from the definition of
ρ that ρΓ (P ) > 0 if P 6= 0 (it must be greater than or equal to 1/|Γ|). Now for two finitely generated pro-
jective Γ-modules P and Q, they are necessarily free Z-modules and hence rkZ (P ⊕Q) = rkZ (P )+rkZ (Q).
Furthermore, ρΓ (ZΓ) = rkZ (ZΓ)/|Γ| = |Γ|/|Γ| = 1. Thus ZΓ is indecomposable by Exercise 1.1.

4.1: The proof of Theorem IX.4.4[1] used the assumption that Γ was finite in order to apply Proposition
IX.4.1[1] by replacing Γ by the cyclic subgroup Γ0 = hγi. The proposition requires Γ0 to be of finite index
(for all γ ∈ Γ), and this will hold in general if |Γ| < ∞.

4.3(b): It is a fact from representation theory that a finitely generated kΓ-module is determined up
to isomorphism by its character. By part(a), the character is in bijective correspondence with the
Hattori-Stallings rank. Thus two finitely generated kΓ-modules are isomorphic iff they have the same

59
Hattori-Stallings rank.

4.3(c): Take k = Q. If Γ is finite and P is a finitely generated projective ZΓ-module then Q ⊗Z P

is a finitely generated projective QΓ-module, hence a finitely generated projective ZΓ-module by Ex-
ercise I.8.2. Then by Theorem 4.4 there is an integer r such that RΓ (Q ⊗Z P ) = r · [1]. But this
Hattori-Stallings rank is precisely that of (QΓ)r , so Q ⊗Z P is a free QΓ-module by part(b).

4.4: Taking ρΓ (P ) = RΓ (P )(1) as a definition, the result is precisely Proposition 4.1 applied to γ = 1.

60
10 Additional Exercises
1: Find a counterexample to the statement MG ∼
= M G for a G-module M .

For an arbitrary group G we have (ZG)G ∼ = Z ⊗ZG ZG ∼ = Z. Alternatively, (ZG)G ∼ = ZG/[I · ZG] =
ZG/I ∼ = Z (where the latter isomorphism follows from application of the 1 st
Isomorphism Theorem on the
augmentation map). If G is finite, then the norm element N exists, and the integer multiples zN (z ∈ Z)
are the only elements of ZG fixed by all g ∈ G (assuming left-multiplication action), so (ZG)G = Z · N
[∼
= Z] is the ideal generated by N . But if G is infinite then thereP is no norm element, and (ZG) = 0.
G

To prove this last statement, take any nonzero element x = i ri gi of ZG (since it’s a finite sum we
can assume 0 ≤ i ≤ n and all ri ∈ Z are nonzero) and consider the set S = {g0 , . . . , gn }. To show that
x∈/ (ZG)G it suffices to show that there is at least one nontrivial g with gS 6= S. We can assume 1 ∈ / S,
otherwise for any g ∈/ S we have g · 1 = g ∈ / S. Suppose that gS = S ∀ g ∈ G. Then g0 · g0 = g02 ∈ S,
and so through trivial induction we see that g0i ∈ S (1 ≤ i ≤ n + 1) and g0 · g0n+1 = g0n+2 ∈ / S (otherwise
|S| > n + 1), noting that g0i 6= 1 since 1 ∈
/ S. Thus we have arrived at a contradiction, and so the choices
G = Z and M = ZG suffice.
Another solution uses the choices G = Z2 = hxi and M = Z where G acts by x · n = −n. Then
x · n = n ⇒ −n = n ⇒ 2n = 0, so the largest quotient on which G acts trivially is ZZ2 = Z2 , and
2n = 0 only holds for n = 0 ∈ Z, hence ZZ2 = 0.

2: Let Z2 = hxi act on the additive complex numbers C by x · z = z ∗ , where z ∗ = x − iy is the

complex conjugate of z = x + iy. [Note: CZ2 ∼
= CZ2 ∼
= R]. Find H 1 (Z2 , C).

For a derivation f : Z2 → C we have f (x0 ) = f (x2 ) = f (1) = 0 and f (x) determines f . Now
f (x2 ) = f (x) + x · f (x) = f (x) + f (x)∗ and so we must have f (x) = −f (x)∗ (i.e. a pure-imaginary
number). Since iy = −(iy)∗ and 0 + iy = (iy)/2 + (iy)/2 = (iy)/2 − (−iy)/2 = (iy)/2 − (iy)∗ /2, we have
f (x) = 21 f (x) − 12 f (x)∗ = − 12 f (x)∗ − x · [ 12 f (x)] = x · [ 12 f (x)∗ ] − [ 21 f (x)∗ ]. Therefore, f is a principal
derivation and hence H 1 (Z2 , C) = 0, using the result of Exercise III.1.2 above.

3: Let the multiplicative cyclic group C2k = hxi act on Z by x · n = (−1)k n. Calculate H 1 (C2k , Z)
under this action.

For k = 2m even, C2k acts trivially on Z, so H 1 (C2k , Z) ∼ = Hom(C2k , Z) = 0 where this equation follows
from Exercise III.1.2 above. For k = 2m + 1 odd, we start by viewing the 1-cocycles as Z 1 = Der(C2k , Z)
and the 1-coboundaries as B 1 = PDer(C2k , Z). For f ∈ Z 1 , f (x2i ) = f (x)+x·f (x2i−1 ) = f (x)−f (x2i−1 )
and f (x2i ) = f (x2i−1 ) + x2i−1 · f (x) = f (x2i−1 ) + (−1)(2i−1)(2m+1) f (x) = f (x2i−1 ) − f (x), so 2f (x) =
2f (x2i−1 ) ⇒ f (x) = f (x2i−1 ) and f (x2i ) = 0. So Z 1 consists of the derivations which are determined
by f (x) and satisfy the derived properties, hence Z 1 ∼ = Z. For f ∈ B 1 , f (xi ) = xi · n − n = (−1)i n − n
which satisfies f (x ) = 0 ∀ i = 2j and f (x ) = −2n ∀ i = 2j + 1, and so B 1 ∼
i i
= 2Z. Thus
H 1 (C2k , Z) ∼
= Z/2Z = Z2 when k is odd.
Alternatively (for k odd), we know that H 1 (C2k , Z) = KerN where the map N : ZC2k → ZC2k is the
norm map induced by multiplication on Z by the norm element N . Now N m = 1 · m + x · m + x2 ·
m + · · · + x2k−1 m = m − m + m − m + · · · + m − m = 0 and so N Z = 0 ⇒ KerN = ZC2k . Since
xi · n = (−1)i n, the action is trivial for all g ∈ C2k if −n = n, hence the largest quotient on which C2k
acts trivially is ZC2k = Z2 .

4: Noting that the only nontrivial map ϕ : Zi = hti → Zij = hsi is the canonical inclusion defined
by t 7→ sj , show that the induced map under H2n−1 is the same inclusion. This is the corestriction map
Z
corZij
i
: H∗ (Zi ) → H∗ (Zij ).

Considering the two periodic free resolutions of Z, there exists an augmentation-preserving chain map f
between them by Theorem I.7.5[1] and we look at the commutative diagram in low dimensions:

61
 Z[Zi ]
t−1
/ Z[Zi ] ε /Z /0

f1 f0 idZ
  
Z[Zij ]
s−1
/ Z[Zij ] ε / Z /0
The right square yields f0 (1) = 1, and the left square yields (s − 1)f1 (1) = f0 (t − 1) = t · f0 (1) − f0 (1) =
ϕ(t)1 − 1 = sj − 1 = (s − 1)(1 + s + · · · + sj−1 ) ⇒ f1 (1) = 1 + s + · · · + sj−1 . We assert that f is
given by fn (1) = 1 for n even and fn (1) = 1 + s + · · · + sj−1 for n odd. Indeed, assuming inductively
that the chain map is valid up to n, it suffices to check commutativity (1 + s + · · · + sij−1 )fn+1 (1) =
fn (1 + t + · · · + ti−1 ) for n odd and (s − 1)fn+1 (1) = fn (t − 1) for n even. For the latter case [n even]
we have (s − 1)(1 + s + · · · + sj−1 ) = (sj − 1) and fn (t − 1) = (sj − 1)fn (1) = sj − 1, so commutativity
is satisfied. For the former case [n odd] we have fn (1 + t + · · · + ti−1 ) = (1 + sj + · · · + sij−j )fn (1) =
(1+sj +· · ·+sij−j )(1+s+· · ·+sj−1 ) = 1+s+· · ·+sij−1 and (1+s+· · ·+sij−1 )fn+1 (1) = (1+s+· · ·+sij−1 ),
so commutativity is satisfied.
Using this chain map, and after moving to quotients, the cycle elements (for odd-dimensional homology)
are mapped via ϕ∗ (1) = 1 + 1 + · · · + 1 = j while the boundary elements are mapped via ϕ∗ (1) = 1; thus
the result follows.

xn −1
5: Let d : G → A be a derivation. Prove the relation d(xn ) = x−1 dx for x ∈ G.

1−1
For n = 0, d(1) = d(x0 ) = x−1 dx = 0, and we argue by induction on n. Assuming the relation at
k k+1 k+1
−1 −x
n = k holds, d(x ) = d(x · x ) = d(x) + x · d(xk ) = [1 + x xx−1
k+1 k
]dx = [ x−1+x
x−1 ]dx = x x−1−1 dx and
we are finished.
n
−1
Alternatively, we note that xx−1 = 1 + x + · · · + xn−2 + xn−1 , so the relation immediately follows by
successive calculations d(xn ) = d(x) + x · d(xn−1 ) = d(x) + x[d(x) + x · d(xn−2 )] = d(x) + x · d(x) +
x2 [d(x) + x · d(xn−3 )] = [1 + x + x2 ]d(x) + x3 · d(xn−3 ), etc..

6: What information do we obtain about the homology of a group G by computing its homology with
rational coefficients?

Assuming Q is an abelian group with trivial G-action, we can apply the result of Exercise III.1.2 to obtain
the short exact sequence 0 → Hn (G) ⊗ Q → Hn (G, Q) → TorZ1 (Hn−1 (G), Q) → 0. Since Q is torsion-free
we have the equality TorZ1 (Hn−1 (G), Q) = 0 and hence the isomorphism Hn (G, Q) ∼ = Hn (G) ⊗ Q. Now
A ⊗ Q = 0 for any torsion abelian group A because q ⊗ a = |a|q
|a| ⊗ a = q
|a| ⊗ |a|a = q
|a| ⊗ 0 = 0. Therefore,
dimQ (Hn (G, Q)) = rkZ (Hn G). Moreover, if Hn (G, Q) is nontrivial then Hn G is torsion-free.
Ln
7: Let the multiplicative cyclic group Cn = hxi act on M = j=1 Z2 by x · ai = ai+1 where aj
generates the j th Z2 -summand. Compute H 1 (Cn , M ).

For f ∈ Z 1 = Der(Cn , M ) we have 0 = f (xn ) = f (x) + xn−1 · f (x) ⇒ xn−1 · f (x) = −f (x) = f (x) [we
can drop the negative sign because the maximum order for elements is 2]. So f is determined by f (x) =
(z1 , . . . , zn ) and we must have (z1 , . . . , zn ) = xn−1 · (z1 , . . . , zn ) = (z2 , . . . , zn , z1 ) ⇒ z1 = z2 = · · · = zn .
Thus f (x) is 0 or (1, 1, . . . , 1), and Z 1 = Z2 . But f (xi ) = f (x) + xi−1 · f (x) = f (x) + f (x) = 2f (x) = 0,
and f (x) = (1, 1, . . . , 1) = x · n − n where n ∈ M is the element consisting of alternating 1’s and 0’s [the
case f (x) = 0 is trivial]. Thus f ∈ B 1 = PDer(Cn , M ), and so H 1 (Cn , M ) = 0.

8: Let GLn (Z) act on Zn by left matrix multiplication, where we consider Zn as an n × 1 column
vector with integral entries. Compute the induced map ψ : H∗ (GLn (Z), Zn ) → H∗ (GLn (Z), Zn ) under
the action of −[δij ] on z ∈ Zn .

The anti-identity matrix m = −[δij ] is in the center Z(GLn (Z)) and so the conjugation action on
GLn (Z) by m is the identity. Thus we can rewrite the map of the action m · z as (g 7→ mgm−1 =
g , z 7→ m · z = −z) ∈ (GLn (Z), Zn ). By Proposition III.8.1[1] this map induces the identity on the
respective homology with coefficients, hence ψ = id∗ . [But clearly ψ = −id∗ because it’s induced from
z 7→ −z ∈ Zn . Thus 2 · id∗ = 0 and H∗ (GLn (Z), Zn ) is all 2-torsion].

62
9: The cyclic group Cm is a normal subgroup of the dihedral group Dm = Cm o C2 (of symmetries
of the regular m-gon). There is a C2 -action on Cm = hσi given by σ 7→ σ −1 . Determine the action of
C2 on the homology H2i−1 (Cm , Z), noting that there is an element g ∈ Dm such that gσg −1 = σ −1 .

Letting c(g) : Cm → Cm be conjugation by g, we can apply Corollary III.8.2[1] to obtain the in-
duced action of Dm /Cm ∼ = C2 on H∗ (Cm , Z) given by z 7→ c(g)∗ z. It suffices to compute c(g)∗ on
the chain level, using the periodic free resolution P of Cm , and using the trivial action on Z. Us-
ing the condition τ (hx) = [c(g)](h)τ (x) = h−1 τ (x) on the chain map τ : P → P (for h ∈ Cm ), we
claim that τ2i−1 (x) = τ2i (x) = (−1)i σ m−i x for i ∈ N and τ0 (x) = x. Assuming this claim holds,
the chain map P ⊗Cm Z → P ⊗Cm Z [in odd dimensions] is given by x ⊗ y 7→ (−1)i σ m−i x ⊗ gy =
(−1)i σ m−i x ⊗ y = (−1)i x ⊗ σ i−m y = (−1)i x ⊗ y, and so c(g)∗ [hence the C2 -action] is multiplication
by (−1)i on H2i−1 (Cm , Z). It suffices to prove the claim. Seeing that N τ2i (1) = τ2i−1 (N ) = N τ2i−1 (1)
where N is the norm element, we can restrict our attention to τ2i−1 and use induction on i since
(σ − 1)τ1 (1) = (σ − 1)(−σ m−1 ) = σ m−1 − 1 = σ −1 − 1 = τ0 (σ − 1). This chain map must satisfy
commutativity (σ − 1)τ2i−1 (1) = τ2(i−1) (σ − 1), and this is indeed the case because (σ − 1)τ2i−1 (1) =
(−1)i (σ m−i+1 − σ m−i ) and τ2(i−1) (σ − 1) = (σ −1 − 1)(−1)i−1 σ m−i+1 = (−1)i (σ m−i+1 − σ m−i ).
P
10: Assuming |G : H| < ∞, we define tr : MG → MH by tr(m) = g∈H\G gm, and we then define
the transfer map resGH : H∗ (G, −) → H∗ (H, −) to be the unique extension of tr to a map of homological
functors (using Theorem III.7.3[1]). Show that this agrees with the map defined by applying H∗ (G, −)
to the canonical injection M → CoindG ∼ G
H M = IndH M and using Shapiro’s lemma. Assume that we know
this latter map is compatible with ∂.

By Theorem III.7.3[1] it suffices to verify that the latter map equals tr in dimension zero.
explicit isomorphism for CoindG ∼ G
P first want an
We H M = IndH M , asserting that it is given by ψ(f ) =
g∈G/H g⊗f (g
−1
). Following the proof of Proposition III.5.9[1], there is an H-map ϕ0 : M → CoindG HM
given by [ϕ0 (m)](g) = {gm ∀ g ∈ H , 0 otherwise}, and by the universal property of induction
this extends to a G-map ϕ P : IndG G
H M → CoindH M given by ϕ(g ⊗ m) = gϕ0 (m). Now we have
ψϕ : g 0 ⊗ m 7→ g 0 [ϕ0 (m)] 7→ g∈G/H g ⊗ [ϕ0 (m)](g −1 g 0 ) = g 00 ⊗ g 00−1 g 0 m = g 00 g 00−1 g 0 ⊗ m = g 0 ⊗ m,
noting that there is one and only one g 00 ∈ G/H such that g 00−1 g 0 ∈ H (otherwise g1−1 g 0 = h1 and
g2−1 g 0 = h2 gives P g1 h1 = g2 h2 which contradicts their coset representations). In the other direction we
have ϕψ : f 7→ g∈G/H g⊗f (g −1 ) 7→ g∈G/H g[ϕ0 (f (g −1 ))](g 0 ) = [ϕ0 (f (g 00−1 ))](g 0 g 00 ) = g 0 g 00 f (g 00−1 ) =
P

f (g 0 g 00 g 00−1 ) = f (g 0 ) = f , also noting that there is only one g 00 ∈ G/H such that g 0 g 00 ∈ H. Thus ϕ = ψ −1
and ψ is the desired isomorphism.
Applying the zeroth homology functor H0 (G, −) to the aforementioned map M → CoindG G
H M → IndH M
g ⊗ g −1 m, and using the isomorphism from Shapiro’s lemma,
P
given by m 7→ (g 7→ gm) 7→ g∈G/HP
x ⊗G (g ⊗H g −1 m) = g∈G/H xg ⊗G (1 ⊗H g −1 m) ∼
P
we obtain the chain map x ⊗G m 7→ g∈G/H =
−1 −1 −1
P P P P
g∈G/H xg ⊗ H g m = g∈G/H g x ⊗ H g m = g∈H\G gx ⊗ H gm = g∈H\G g · (x ⊗ H m). Using
the natural isomorphism H0 (G, −) ∼ = (−)G of Proposition III.6.1[1], this chain map yields the trace map
tr described above.

11: Prove that Hn (G, M ) ∼

= TorG
n−1 (I, M ) where I is the augmentation ideal of ZG.

ε
We have the short exact sequence of G-modules, 0 → I ,→ ZG → Z → 0. By an analogue of The-
orem 17.1.15[2] we have a long exact sequence of abelian groups · · · → TorG G
n (ZG, M ) → Torn (Z, M ) →
G G R
Torn−1 (I, M ) → Torn−1 (ZG, M ) → · · · . It is a fact that if P is R-projective then Torn (P, B) = 0 for
any R-module B (n ≥ 1). Therefore, TorG n (ZG, M ) = 0 since ZG is a free [hence projective] ZG-module,
and so we obtain the isomorphisms TorG G G
n (Z, M ) → Torn−1 (I, M ). Since Hn (G, −) = Torn (Z, −), the
result follows.

12: Given |G : H| < ∞ and z ∈ H(G, M ) where H(−, −) is either H∗ or H ∗ , show that corG G
H resH z =
|G : H|z.

63
Referring to Exercise AE.10 above, we have the map M → CoindG G
H M → IndH M which yields the
G
P
chain map x ⊗G m 7→ g∈H\G gx ⊗H gm, and this induces the transfer map resH in homology. Com-
posing this P with the corestrictionPmap on the chain level x ⊗H m 7→ x ⊗G m, we obtain the chain map
x ⊗G m 7→ g∈H\G gx ⊗G gm = g∈H\G x ⊗G m = |G : H|(x ⊗G m) which induces corG G
H resH as multi-
plication by |G : H|.
On the other hand, we have the chain map Hom(F, M )H → Hom(F, M )G given by f 7→ g∈G/H [x 7→
P

gf (g −1 x)] which induces the transfer map corG H in cohomology; G acts diagonally. Composing this with
the restriction map on the chain level, we obtain the chain map f 7→ f 7→ g∈G/H [x 7→ gf (g −1 x)] =
P
P
g∈G/H [x 7→ f (x)] = |G : H|[x 7→ f (x)], noting that the domain element f is a G-invariant homomor-
phism. This induces the aforementioned map corG G
H resH .

Mi ) ∼
Lm Lm
13: Give another proof that H(G, i = i H(G, Mi ) where H(−, −) is either H∗ or H ∗ .
α β
Applying Proposition III.6.1[1] to the short exact sequence of G-modules 0 → M1 → M1 ⊕M2 → M2 → 0,
we obtain the standard long exact sequence in (co)homology. Since α is an injection onto a direct sum-
mand, the induced map α∗ under the covariant functor H∗ (G, −) is an injection (refer to Exercise II.7.3);
similarly, the induced map β ∗ = H ∗ (G, β) is also an injection. Thus the long exact sequence breaks up
into short exact sequences 0 → Hn (G, M1 ) → Hn (G, M1 ⊕ M2 ) → Hn (G, M2 ) → 0 (similar for coho-
mology). These are split exact sequences because γ∗ α∗ = id∗ and Γ∗ β ∗ = id∗ , where γ is the projection
M1 ⊕ M2 → M1 and Γ is the inclusion M2 → M1 ⊕ M2 . The result now follows by trivial induction on
m.

14: Prove that if the G-module M has exponent p [prime] then H n (G, M ) and Hn (G, M ) are Zp -
vector spaces.

Showing that these groups are the specified modules is equivalent to showing that they are annihi-
lated by p (i.e. have exponent dividing p). Considering
P homology, a chain P z ∈ Cn (G, M ) is a finiteP sum
of terms of the form m ⊗ [g1 | · · · |gn ], so pz = p(m ⊗ [g1 | · · · |gn ]) = (pm) ⊗ [g1 | · · · |gn ] = 0 = 0.
Thus pHn (G, M ) = 0 ∀ n ≥ 0. Considering cohomology, a cochain f ∈ C 0 (G, M ) is an element of M
and hence pf = 0 trivially. If n ≥ 1 then f ∈ C n (G, M ) is a function Gn → M , so pf is a function
Gn → M → pM = 0 and hence pf = 0. Thus pH n (G, M ) = 0 ∀ n ≥ 0.

15: If G is a finite group, show that

L Hn (G, M ) is a Z|G| -module for n > 0. Thus we have the pri-
mary decomposition Hn (G, M ) = p Hn (G, M )(p) where p ranges over the primes dividing |G|.

By Proposition III.9.5[1], corG G

{1} res{1} z = |G : {1}|z = |G|z where z ∈ H∗ (G, M ). But this composition
factors through H∗ ({1}, M ) which is trivial in positive dimensions. Thus corG G
{1} res{1} z = 0 ⇒ |G|z = 0,
and so Hn (G, M ) is annihilated by |G| for n > 0.

16: Show that Hn (G, M ) = 0 ∀ n > 0 if M is a finite abelian group and G is finite with rela-
tively prime order, gcd(|G|, |M |) = 1. Furthermore, show that Hn (G, M ) is a finite group (for n > 0) if
G is finite and M is a finitely generated abelian group.

By Exercises AE.14+15 above, Hn (G, M ) is annihilated by both |G| and |M |, and hence must be anni-
hilated by a common factor. But gcd(|G|, |M |) = 1, so the common factor is 1 and Hn (G, M ) = 0 for
n > 0.
By Exercise AE.15 above (assuming n > 0), |G|Hn (G, M ) = 0 and so Hn (G, M ) is all torsion. Since
|G| < ∞, Fn is a free G-module of finite rank r = |G|n (F is the bar resolution of Z overLZG), where we
for Fn is the (n+1)-tuples whose first element is 1. Thus Fn ⊗G M ∼ = ( i ZG)⊗G M ∼
r
note that a G-basisL =
L r
(ZG ⊗ G M ) ∼
=
r
M is finitely generated since M is finitely generated, which implies that H n (G, M )
i i
is also finitely generated. A finitely generated torsion group is finite, so the result follows.

17: Why does resG n n

H induce a monomorphism H (G, M )(p) ,→ H (H, M ), where H is the Sylow p-
subgroup of G?

64
Consider an element z ∈ H n (G, M )(p) which lies in the kernel of resG
H . Composing this with the core-
striction map gives you multiplication by |G : H| by Proposition III.9.5[1], and so |G : H|z = 0. But the
order of z is |z| = pα and p does not divide |G : H|. Since pα and |G : H| both annihilate z, there is a
common factor between them which also annihilates z, so z = 0 (the only common factor is 1). Thus
resG
H is a monomorphism when restricted to the p-primary component.

18: Compute the homology group H1 (G) for any nonabelian simple group G.

The commutator group [G, G] is either 0 or G because the commutator group is a normal subgroup
of G and a normal subgroup of a simple group must be either the trivial group or G itself. Since G is
nonabelian, we cannot have [G, G] = 0, and so [G, G] = G (i.e. G is perfect). Thus H1 (G) ∼
= G/[G, G] =
G/G = 0.
In particular, H1 (An ) = 0 ∀ n ≥ 5 where An is the alternating group on n letters (subgroup of Sn
with index 2), by Theorem 4.6.24[2].

19: Show that I/I 2 ∼

= Gab where I is the augmentation ideal of ZG.

Apply Proposition III.6.1[1] to the short exact sequence 0 → I → ZG → Z → 0 to obtain the ex-
act sequence H1 (G, ZG) → H1 (G) → IG → (ZG)G → Z → 0 in low dimensions. The latter map is an
isomorphism (by the First Isomorphism Theorem) because the map is surjective and (ZG)G ∼ = Z. By
Proposition III.6.1[1], H1 (G, ZG) = 0 because ZG is free [hence projective]. Thus H1 (G) ∼
= IG = I/I 2
by exactness, and we know that H1 (G) ∼ = Gab , so the result follows.
An explicit isomorphism is given by g[G, G] 7→ (g − 1) + I 2 .

20: Find a module M with trivial action such that H 1 (D2n , M ) is nonzero, where D2n is the dihe-
dral group with respect to a regular 2n−1 -gon.
n−1
A presentation for this group is D2n = hα, β | α2 = β 2 = (αβ)2 = 1i. By the result of Exercise
III.1.2, H 1 (D2n , M ) = Hom(D2abn , M ). Now D2n quotiented by [D2n , D2n ] forces the trivial relation 1 =
n−1 n−1 n−1
(αβ)2 = α2 β 2 = 1 · 1 = 1, so D2abn = Z2 ⊕ Z2 . Thus H 1 (D2n , M ) = Hom(Z2 , M ) ⊕ Hom(Z2 , M ),
which is nonzero if M = Z2 , giving H 1 (D2n , Z2 ) ∼ = Z2 ⊕ Z2 .

21: Prove that a finitely generated abelian group G is cyclic if and only if H2 (G, Z) = 0.

If G is a cyclic group Zn , then H2 (Zn , Z) = 0 as explained on pg35[1].

If H2 (G, Z) = 0 then Hopf’s formula (Theorem II.5.3[1]) gives R ∩ [F, F ] ⊆ [F, R], where F/R is a presen-
tation for G. Since G is abelian we have 0 = [G, G] = [F/R, F/R] = [F, F ]R/R which implies [F, F ] ⊆ R.
Alternatively, since G = F/R is abelian, [F, F ] ⊆ R by Proposition 5.4.7[2]. Thus [F, F ] = [F, R], and
|F | < ∞ since G is finitely generated. If |F | = 1 then R = {xn } for some n ≥ 0, in which case G is
obviously cyclic.
It seems we’re stuck in terms of extracting any more information, but alternatively V2 we can refer to
Theorem V.6.3[1] which says H2 (G) is isomorphic to the second exterior power G for any abelian
group. Thus it suffices to show that G is cyclic if G ⊗ G ∼ = h{g ⊗ g}i ∼
= G. Now if rkZ (G) = n then
rkZ (G ⊗ G) = n2 , so n2 = n gives n = 1 or n = 0. Also, Zi ⊗ Zj ∼ = Zgcd(i,j) , so G must contain at most
one Zm -summand (for each m) because the tensor product contains at least the squared-amount of those
summands. Assume that G is not cyclic; then all m’s must be relatively prime (otherwise the tensor
product will contain additional summands not in G due to greatest common divisors). Thus G ∼ = Z ⊕ Zn
(for some n ≥ 2) since Zi ⊕ Zj ∼ = Zij iff gcd(i, j) = 1 [note that G must have the Z-summand in order to
not be cyclic]. But (Z ⊕ Zn ) ⊗ (Z ⊕ Zn ) ∼
= Z ⊕ Zn ⊕ Zn ⊕ Zn which is not isomorphic to G, a contradiction
(hence G is cyclic).
Aside: A finitely generated abelian group G = F/R is cyclic iff [F, F ] = [F, R].

22: Let Zm = hti, Zm2 = hsi, and define a Zm -action on Zm2 by t · s = sm+1 . Compute the re-
sulting Zm -module structure on Hj (Zm2 ), and then compute Hi (Zm , Hj (Zm2 )) where m is an odd prime.

65
First we must compute the change-of-rings map in integral homology for the map Za = hxi → Zb = hyi,
x 7→ y r , where b|ra. Mimicking the solution to Exercise AE.4, the induced map in odd-dimensional
homology is easily seen to be multiplication by r. Applying this result to the case where r = m + 1 and
a = b = m2 , the Zm -action on Hj (Zm2 ) is multiplication by p(m+1) for tp . If j is even, then the homolo-
gies in question are trivial. By a result on pg58[1], Hodd (Zm , Zm2 ) = Coker(N : (Zm2 )Zm → (Zm2 )Zm )
and Heven (Zm , Zm2 ) = Ker(N : (Zm2 )Zm → (Zm2 )Zm ). Since m is an odd prime, the only nontrivial
proper subgroup/quotient of Zm2 is Zm which is not fixed by the Zm -action, so the norm map is 0 → 0
and thus Hi (Zm , Hj (Zm2 )) = 0 ∀ nonzero i, j.

23: Prove that if Q is injective then E(G, Q) is trivial.

Any short exact sequence 0 → Q → E → G → 0 splits for Q injective, so there is only one equiva-
lence class of group extensions (namely, the split extension) and E(G, Q) = 0. Alternatively, Proposition
III.6.1[1] and Theorem IV.3.12[1] imply E(G, Q) ∼ = H 2 (G, Q) = 0 because Q is injective.

24: Prove that H 2 (A5 , Z2 ) 6= 0.

We have the central group extension 1 → Z2 → SL2 (F5 ) → A5 → 1 as in Exercise I.5.7(b). It suf-
fices to show that E = Z2 o A5 is not isomorphic to SL2 (F5 ), for then the above extension is nonsplit
and hence H 2 (A5 , Z2 ) ∼
= E(A5 , Z2 ) 6= 0 by Theorem IV.3.12[1]. First note that Z2 must have the trivial
A5 -action, so E = Z2 × A5 . Now it is a fact that SL2 (F5 ) is perfect, while [E, E] ⊆ A5 by Proposition
5.4.7[2] because E/A5 ∼ = Z2 is abelian (hence E is not perfect). Thus H1 (E) 6= 0 = H1 (SL2 (F5 )) and so
Z2 o A5  SL2 (F5 ).

25: Let A = Z2 ×Z2 and let Aut(A) ∼

= S3 act on A in the natural fashion. Prove that H 1 (S3 , Z2 ×Z2 ) = 0.

In the semi-direct product E = A o S3 = Hol(A) [called the holomorph of A] we have a Sylow 3-subgroup
P ∼= Z3 by Sylow’s Theorem, where |E| = 24 = 23 · 3. By Sylow’s Theorem, n3 | 8 and n3 ≡ 1 mod 3,
where n3 is the number of Sylow 3-subgroups of E. Thus n3 is either 1 or 4, and this implies |NE (P )|
is either 24 or 6 by the fact n3 = |E : NE (P )|. But we can exhibit at least two such subgroups [noting
that S3 ∼= Z3 o Z2 ], namely P1 = (0 × 0) o (Z3 o 0) and P2 = {(0, 0, 0, 0), (1, 0, 1, 0), (1, 0, 2, 0)}. Thus
|NE (P )| = 6 which corresponds to n3 = 4, and S3 is then the normalizer of the Sylow 3-subgroup
P = 0 o Z3 of E because P / S3 and |S3 | = 6. Given a complement G to A in E [a group G is a comple-
ment to A in E if E = A o G], one Sylow 3-subgroup is Q / G and hence P = eQe−1 for some e ∈ E by
Sylow’s Theorem. Now S3 = NE (P ) = NE (eQe−1 ) = eNE (Q)e−1 = eGe−1 , and so all complements are
conjugate. Noting that the conjugacy classes of complements to A in E are the A-conjugacy classes of
splittings of E, we have H 1 (S3 , Z2 ×Z2 ) = 0 by Proposition IV.2.3[1] because there is only one conjugacy
class.

26: In the proof of Theorem IV.3.12[1], all extensions were assumed to have normalized sections. Explain
why this simplification does not affect the result of the theorem.

Given a factor set (2-cocycle) f : G × G → A, we assert that this lies in the same cohomology
class as a normalized factor set (2-cocycle). Let δ1 c be the coboundary of the constant function
c : G → A defined by c(g) = f (1, 1). It suffices to show that F = f − δ1 c is a normalized factor
set, for then it belongs to the same cohomology class as the arbitrary f [it differs by a coboundary]. Now
F (1, 1) = f (1, 1)−[δ1 c](1, 1) = f (1, 1)−[c(1)+1·c(1)−c(1·1)] = f (1, 1)−f (1, 1) = 0 and so F : G×G → A
is normalized. It is indeed a 2-cocycle (factor set) because gF (h, k) − F (gh, k) + F (g, hk) − F (g, h) =
[gf (h, k) − f (gh, k) + f (g, hk) − f (g, h)] − {g[c(h) + hc(k) − c(hk)] − [c(gh) + ghc(k) − c(ghk)] + [c(g) +
gc(hk) − c(ghk)] − [c(g) + gc(h) − c(gh)]} = 0 − gc(h) − ghc(k) + gc(hk) + c(gh) + ghc(k) − c(ghk) − c(g) −
gc(hk) + c(ghk) + c(g) + gc(h) − c(gh) = [−gc(h) + gc(h)] + [−ghc(k) + ghc(k)] + [gc(hk) − gc(hk)] +
[c(gh) − c(gh)] + [−c(ghk) + c(ghk)] + [−c(g) + c(g)] = 0 + 0 + 0 + 0 + 0 + 0 = 0.
Since the bijection in the theorem is dependent on the cohomology class, and every class has a normalized
factor set (hence normalized section), we are able to restrict our attention to those sections satisfying
the normalization condition.

66
27: Show that H 2 (F, A) = 0 for F free by appealing to group extensions.
π
For any group extension 0 → A → E → F → 1 define the set map S → E by s 7→ s̃, where s̃ ∈ π −1 (s) is
a lifting of s ∈ F = F (S). Then by the universal mapping property of free groups, this set map extends
uniquely to a homomorphism ϕ : F → E which satisfies πϕ = idF by construction. Thus the extension
splits, and by Theorem IV.3.12[1] we have H 2 (F, A) ∼
= E(F, A) = 0.
Note that this also follows from direct computation using the free resolution 0 → I → ZF → Z → 0,
where the augmentation ideal I of ZF is free by Exercise IV.2.3(b). Indeed, the coboundary map is
δn : HomF (0, A) → HomF (0, A) for all n > 1 and hence H n (F, A) = 0 ∀ n > 1.

28: Compute H 2 (Q8 , Z2 ), and determine the number of group extensions of Z2 by Q8 . [These two
problems are independent of each other].

The quaternion group Q8 is a non-abelian group of order 8 with presentation hx, y | x4 = 1, x2 =

y 2 = (xy)2 i. Its center is C := Z(Q8 ) ∼ = Z2 which is also its commutator subgroup, and Inn(Q8 ) =
Q8 /Z(Q8 ) ∼ = Z2 × Z2 . Also, Aut(Q8 ) ∼= S4 and hence Out(Q8 ) = Aut(Q8 )/Inn(Q8 ) ∼ = S3 . As explained
on pg102+104[1], Q8 is an S4 -crossed module via the canonical map Q8 → Aut(Q8 ), and such an exten-
sion gives rise to a homomorphism ψ : Z2 → Out(Q8 ) ∼ = S3 . By Theorem IV.6.6[1] the set E(Z2 , Q8 , ψ)
of equivalence classes of extensions giving rise to ψ is either empty or in bijective correspondence with
H 2 (Z2 , C), where C is a Z2 -module via ψ [note: it is a fact that the center of a group is a characteristic
subgroup, so Aut(Q8 ) acts naturally on C and hence Out(Q8 ) acts on C because any inner automorphism
leaves C fixed]. Now the only possible action on C ∼ = Z2 is the trivial one (giving CZ2 = C = C Z2 ), so
2 ∼
H (Z2 , C) = CokerN = C/N C = C/0 = Z2 . There are two possible choices of ψ, namely, the trivial map
and the injection Z2 ,→ S3 = Z3 o Z2 . If ψ is the trivial map then it automatically lifts to the trivial ho-
momorphism Z2 → S4 and we have a direct product extension E ∼ = Q8 × Z2 . For the injective ψ we have
the semi-direct product Q8 o Z2 where Z2 switches the generators of Q8 . Thus E(Z2 , Q8 , ψ) is nonempty,
and Theorem IV.6.6[1] implies there are a total of 4 group extensions 1 → Q8 → E → Z2 → 1.
To compute the second cohomology group of the group Q8 with coefficients in Z2 , embed Z2 in the
H ∗ -acyclic module Hom(ZQ8 , Z2 ) by z 7→ (q 7→ qz), and note that Coker(Z2 → Hom(ZQ8 , Z2 )) = Z2
because the evaluation map Hom(ZQ8 , Z2 ) → Z2 given by f 7→ f (2) composes with the embedding
to give the trivial map z 7→ (q 7→ qz) 7→ 2z = 0. The dimension-shifting argument then implies
H 2 (Q8 , Z2 ) ∼
= H 1 (Q8 , Z2 ), and by Exercise III.1.2 we have H 1 (Q8 , Z2 ) ∼ = Hom(H1 (Q8 ), Z2 ). Now
H1 (Q8 ) = Q8 /[Q8 , Q8 ] = Q8 /C ∼ = Z2 × Z2 and so Hom(H1 (Q8 ), Z2 ) ∼ = Hom(Z2 , Z2 ) ⊕ Hom(Z2 , Z2 ) ∼
=
Z2 × Z2 . Therefore, H 2 (Q8 , Z2 ) ∼ = H1 (Q8 ) ∼
= H 1 (Q8 , Z2 ) ∼ = Z22 ≡ Z2 × Z2 .

29: Let p be a prime, let A be an abelian normal p-subgroup of a finite group G, and let P be a
Sylow p-subgroup of G. Prove that G is a split extension of G/A by A iff P is a split extension of P/A
by A [Note: it is a fact that a normal p-subgroup is contained in every Sylow p-subgroup, so A ⊆ P ].
This result is known as Gaschütz’s Theorem.

Suppose G splits over A, so that G ∼ = A o G/A. Then the subgroup A o P/A is a Sylow p-subgroup of
G and hence A o P/A ∼ = P by Sylow’s Theorem, so P also splits over A. Conversely, suppose P splits
over A (i.e. P ∼ = A o P/A). Note that P/A = Sylp (G/A) and multiplication by |G/A : P/A| = |G : P | is
an automorphism of A [hence of H 2 (G/A, A)] since |A| is relatively prime to |G : P |. The composition
res cor
H 2 (G/A, A) → H 2 (P/A, A) → H 2 (G/A, A) is then an isomorphism by Proposition III.9.5[1] because it is
multiplication by |G : P |. In particular, the restriction homomorphism res : H 2 (G/A, A) → H 2 (P/A, A)
is injective, so the only element of H 2 (G/A, A) which corresponds to the trivial element [split extension]
of H 2 (P/A, A) is the trivial element [split extension]. Thus G splits over A, and the proof is complete.

30: The Schur multiplier of a finite group G is defined as H2 (G, Z) ∼

= H 2 (G, C∗ ) where the multiplica-
∗
tive group C = C−{0} is a trivial G-module. Prove that the Schur multiplier (of a finite group) is finite.

Since |G| < ∞ and Z is finitely generated (as an abelian group), H2 (G, Z) is a finite group by Ex-
ercise AE.16.
Alternatively, we shall show that every cohomology class contains a cocycle whose values lie in the nth

67
roots of unity hζi ∼= Zn (where n = |G| and ζ = e2πi/n ), for then there are only finitely many func-
2 n
tions/cocycles f : G2 → hζi and |H 2 (G, C∗ )| ≤ nn < ∞. From the exact sequence 0 → Zn → C∗ →
∗ 1 ∗ 1 ∗ 2 2 ∗
C → 0 we obtain a long exact sequence → H (G, C ) → H (G, C ) → H (G, Zn ) → H (G, C ) →
H 2 (G, C∗ ) → by Proposition III.6.1[1]. But the nth -power map on C∗ induces the n-multiplication map
on H 2 (G, C∗ ) which is the zero map since n annihilates H 2 (G, C∗ ) by Corollary III.10.2[1], so the above
long exact sequence gives us a surjection H 2 (G, Zn ) → H 2 (G, C∗ ) → 0. Now H 2 (G, Zn ) is finite by Exer-
cise AE.16, so H 2 (G, C∗ ) is necessarily finite (by the 1st Isomorphism Theorem) and the proof is complete.

31: Show that ZG ⊗ ZG0 ∼

= Z[G × G0 ] as (G × G0 )-modules.

The map ZG×ZG0 → Z[G×G0 ] defined by (zg, z 0 g 0 ) 7→ zz 0 (g, g 0 ) is obviously a Z-balanced map and hence
gives a rise to a unique group homomorphism ϕ : ZG ⊗ ZG0 → Z[G × G0 ] by Theorem 10.4.10[2]. The ob-
vious group homomorphism Z[G × G0 ] → ZG ⊗ ZG0 defined by z(g, g 0 ) 7→ z(g ⊗ g 0 ) = (zg ⊗ g 0 ) = (g ⊗ zg 0 )
is the inverse of ϕ because z(g, g 0 ) 7→ (zg ⊗ g 0 ) 7→ z1(g, g 0 ) = z(g, g 0 ) and (zg ⊗ z 0 g 0 ) 7→ zz 0 (g, g 0 ) 7→
zz 0 (g ⊗ g 0 ) = (zg ⊗ z 0 g 0 ). Thus ϕ is a group isomorphism, and it is a (G × G0 )-module isomorphism
because ϕ[(h, h0 ) · (zg ⊗ z 0 g 0 )] = ϕ(zhg ⊗ z 0 h0 g 0 ) = zz 0 (hg, h0 g 0 ) = zz 0 (h, h0 )(g, g 0 ) = (h, h0 ) · ϕ(zg ⊗ z 0 g 0 )
where (h, h0 ) ∈ G × G0 .

32: Suppose u1 ∈ H p G, u2 ∈ H q G, v1 ∈ H r G, and v2 ∈ H s G. Prove that (u1 × v1 ) ` (u2 × v2 ) =

(−1)qr (u1 ` u2 ) × (v1 ` v2 ) in H p+q+r+s (G, Z).

Note that u ` v := d∗ (u × v) where d : G → G × G is the diagonal map. Let D : G × G → G4 be the

analogous diagonal map, and let P : G4 → G4 be the permutation (g1 , g2 , g3 , g4 ) 7→ (g1 , g3 , g2 , g4 ). Then
D = P ◦(d×d), and we obtain (u1 ×v1 ) ` (u2 ×v2 ) = D∗ (u1 ×v1 ×u2 ×v2 ) = (d×d)∗ [P ∗ (u1 ×v1 ×u2 ×v2 )] =
(−1)qr (d × d)∗ [u1 × u2 × v1 × v2 ] = (−1)qr d∗ (u1 × u2 ) × d∗ (v1 × v2 ) = (−1)qr (u1 ` u2 ) × (v1 ` v2 ).
Another way is to perform the same calculation using u × v := p∗1 (u) ` p∗2 (v) and the fact that
p∗ (u ` v) = p∗ (u) ` p∗ (v), where p1 is the projection G × G → G × {1} = G and p2 is the pro-
jection G × G → {1} × G = G.

33: For the cap product, state the property of naturality with respect to group homomorphisms
α : G → H. Also, provide the existence of an identity element for the cap product.

Checking definitions, we have u a α∗ z = α∗ (α∗ u a z) which is associated to the “commutative” di-

agram
H p (G, M ) ⊗ Hq (G, N )
a
/ Hq−p (G, M ⊗ N )
D
α∗ α∗ α∗
 
H p (H, M ) ⊗ Hq (H, N )
a
/ Hq−p (H, M ⊗ N )
There is a left-identity element 1 ∈ H 0 (G, Z) = Z which is represented by the augmentation map ε,
regarded as a 0-cocycle in HomG (F, Z). Let F be the standard resolution, let z ∈ H∗ (G, M ) with repre-
sentation z = (g0 , . . . , gi )⊗m, and take the diagonal approximation ∆ to be the Alexander-Whitney map.
Pi
Then ε ⊗ z maps under a to p=0 (−1)degε·p (g0 , . . . , gp ) ⊗ ε(gp , . . . , gi ) ⊗ n = 0 + (−1)0·i (g0 , . . . , gi ) ⊗
P
1 ⊗ n = z, as explained on pg113[1]. Thus the element satisfies 1 a z = z for all z ∈ H∗ (G, M ), where
we make the obvious identification Z ⊗ M = M of coefficient modules.

34: Prove that a finitely generated projective Z-module M is a finitely generated free Z-module; thus
the two are actually equivalent (since free modules are projective).

By the Fundamental Theorem (Theorem 12.1.5[2]) M has the decomposition M = ∼ Zr ⊕ Zn ⊕ · · · ⊕ Zn ,

1 i
where we note that Z is a Principal Ideal Domain. Since M is projective, all of its direct summands
must be projective. Now Zr is free, hence projective. But Zn is not Z-projective because if it were then
it would be a direct summand of a free Z-module F , and F would then have elements of finite order (a
contradiction); alternatively we could note that applying the functor Hom(Zn , −) to the exact sequence
n
0 → Z → Z → Zn → 0 yields the sequence 0 → 0 → 0 → Zn → 0 which is not exact. Thus M ∼ = Zr for

68
some r ∈ N ∪ {0} and hence M is Z-free of finite rank.

35: Let G = P SL2 (Z) and let A be a G-module. Show that for every q ≥ 2 and for every x ∈ H q (G, A),
we have 6x = 0.

The modular group is the group of Möbius transformations T (z) = az+b cz+d in the complex plane such


that a, b, c, d ∈ Z with ad − bc = 1. It is isomorphic to G via the map T 7→ ac db , and G is called the
projective special linear group [the quotient of SL2 (Z) by Z(SL2 (Z)) ∼ = Z2 ]. Now it is a fact that G
contains a free subgroup H of index |G : H| = 6 [this fact can be found in the paper The Number of
Subgroups of Given Index in the Modular Group by W. Stothers]. By Exercise AE.27, H q (H, A) = 0 for
all q ≥ 2. Thus we can apply Proposition III.10.1[1] which states H q (G, A) is annihilated by |G : H|, i.e.
6x = 0 for all x ∈ H q (G, A) with q ≥ 2.

36: Let G be a finite cyclic group and let M be a G-module. The Herbrand quotient is defined to
be h(M ) = |H 2 (G, M )|/|H 1 (G, M )|, assuming both cohomology groups are finite. Show that h(Z) = |G|
where Z has the trivial G-action, and show that h(M ) = 1 for M finite.

We know that H 2 (G, M ) = ∼ M G /N M and H 1 (G, M ) = NM/IM , where N is the norm element and
I ≡ hσ − 1i is the augmentation ideal of G = hσi and NM := Ker(N : M → M ). For M = Z with trivial
action we have H 2 (G, M ) ∼
= Z|G| and H 1 (G, M ) = 0, so h(Z) = |ZG |/|{0}| = |G|/1 = |G|. If M is finite
then h(M ) = (|M | · |IM |)/(|N M | · |NM |) = |M G | · |(σ − 1)M |/|M |, where we note that M/NM ∼
G
= NM.
σ−1 G G
But the kernel K of the surjective map M → (σ − 1)M is equal to M because m ∈ M maps to
σm − m = m − m = 0 and if m ∈ K then σ |G|−1 m = σ |G|−2 m = · · · = σm = m, i.e. m ∈ M G . Thus
|M |/|M G | = |(σ − 1)M | and so h(M ) = |M |/|M | = 1.

37: Given the short exact sequence of G-modules 0 → A → B → C → 0 with G finite cyclic, show that
h(B) = h(A)h(C) where we assume the cohomology groups for A and C [hence B] are finite. Here h is
the Herbrand quotient (defined in the previous exercise).
δ ϕ φ
Consider the long exact cohomology sequence H 0 (G, C) →
0
H 1 (G, A) → H 1 (G, B) → · · · → H 2 (G, B) →
δ
H 2 (G, C) → 2
H 3 (G, A). It is a fact (Exercise V.3.3(b)) that cupping this sequence with the generator
2
of H (G, Z) gives an isomorphism of that sequence onto itself (raising dimensions by 2), and hence
gives the equality δ0 = δ2 which implies Kerϕ ∼ = Cokerφ. We can break the above sequence and ob-
ϕ φ
tain an exact sequence 0 → Kerϕ → H 1 (G, A) → · · · → H 2 (G, C) → Cokerφ → 0. I claim that
Qm i
if 0 → H1 → · · · → Hm → 0 is an exact sequence of finite abelian groups, then i=1 |Hi |(−1) =
1. Assuming this claim holds, and noting that the cohomology groups are finite abelian, we have
|H 1 (G,A)| |H 1 (G,C)| |H 2 (G,B)|
1 = |Cokerφ| 1 1
|Kerϕ| |H 1 (G,B)| |H 2 (G,A)| |H 2 (G,C)| = 1 · h(A) h(C) h(B) which implies h(B) = h(A)h(C) as desired.
It suffices to prove the claim. The case m = 2 is trivial since the sequence yields the isomorphism
H1 ∼ = H2 and hence |H1 |/|H2 | = 1, so we proceed by induction on m > 1. From the exact sequence
φ ϕ
0 → H1 → · · · → Hm−1 → Hm → Hm+1 → 0 we obtain the exact sequence 0 → H1 → · · · Hm−2 →
φ Qm−1 i m
Hm−1 → Imφ → 0. Applying the inductive hypothesis we have i=1 |Hi |(−1) · |Imφ|(−1) = 1, and by
exactness of the original sequence we see that |Imφ| = |Kerϕ|. But |Hm |/|Kerϕ| = |Imϕ| = |Hm+1 |, so
Qm−1 i m Qm+1 i
|Kerϕ| = |Hm |/|Hm+1 | and hence 1 = i=1 |Hi |(−1) ·(|Hm |/|Hm+1 |)(−1) = i=1 |Hi |(−1) as desired.

38: If G and H are abelian groups with isomorphic group rings ZG ∼

= ZH, show that G ∼
= H.

The homology groups of G and H are independent of the choice of resolution up to canonical iso-
morphism, and the groups are defined by Hi G = Hi (FG ) where F is a projective resolution of Z over ZG
(similary for H). Since ZG ∼ = ZF , the G-modules Fi can be regarded as H-modules via restriction of
scalars and hence the projective resolution F for the homology of G can also be used for the homology
of H. We have FG ∼ = FH since Z ⊗ZG F ∼ = Z ⊗ZH F by the obvious map 1 ⊗ f 7→ 1 ⊗ f [using the
isomorphism ϕ : ZG → ZH we have 1 ⊗ f = 1 ⊗ gf 7→ 1 ⊗ ϕ(g)f = 1 ⊗ f ], and so Hi (G) ∼ = Hi (H)
for all i. In particular, G/[G, G] ∼
= H1 (G) ∼
= H1 (H) ∼= H/[H, H]. Since G and H are abelian groups,

69
[G, G] = 0 = [H, H] and hence G ∼
= H.

39: Let F be a field and G a finite group of order |G| > 1. Show that the group algebra F G has
zero divisors (hence is not a field) and show that the augmentation ideal I is a maximal ideal of F G.
ε
Noting that F G/I ∼ = F from the exact sequence 0 → I → F G → F → 0, I is a maximal ideal by
Proposition 7.4.12[2] since F is a field [in a commutative ring R, an ideal M is maximal iff R/M is a
field]. Now take g ∈ G such that g m = 1 with m > 1. Then (1−g)(1+g+· · ·+g m−1 ) = 1−g m = 1−1 = 0
and hence 1 − g is a zero divisor.

40: Regarding Z2 as a module over the ring Z4 , construct a resolution of Z2 by free modules over
Z4 and use this to show that ExtnZ4 (Z2 , Z2 ) is nonzero for all n.
2
The 2-multiplication map Z4 → Z4 has kernel {0, 2} and image {0, 2}, and the projection Z4  Z2
2 2
has kernel {0, 2} and image {0, 1}. Thus we have a free resolution · · · → Z4 → Z4 → Z4  Z2 → 0.
Since Hom(Z4 , Z2 ) = Z2 , the 2-multiplication maps become trivial maps, and we obtain the sequence
0 0
· · · ← Z2 ← Z2 ← Z2 after applying Hom(−, Z2 ) to our free resolution. Thus ExtnZ4 (Z2 , Z2 ) ∼
= Z2 6= 0
for all n.
i j
41: Let H be a subgroup of G of finite index, and let 0 → A0 → A → A00 → 0 be an S.E.S. of G-
modules. Show that the connecting homomorphisms are consistent with corestriction, i.e. the following
diagram commutes.
δ /
H n (G, A00 ) H n+1 (G, A0 )
O O
cor cor

H n (H, A00 )
δ / H n+1 (H, A0 )

Let F be the resolution associated to G and let F 0 be the resolution associated to H, and let ∂ ∗
denote the [cochain complex] coboundary map. Consider the exact sequence of cochain complexes
i jH
0 → HomH (Fn0 , A0 ) → H
HomH (Fn0 , A) → HomH (Fn0 , A00 ) → 0, with similar notation for the sequence
associated to G. The connecting map δ in cohomology is defined as follows. For the cohomology-class
representative fb ∈ HomH (Fn0 , A00 ), there exists an f ∈ HomH (Fn0 , A) which maps onto fb via jH (by
surjectivity of jH ). Then it is realized that ∂ ∗ f = iH (f ) for some f ∈ HomH (Fn0 , A0 ) which is unique
by injectivity of iH , and f is the class representative of δ[fb] where [fb] is the cohomology class of fb.
Thus in order for δ to commute with corestriction, it suffices to show that corestriction commutes with
{∂ ∗ , i∗ , j ∗ } where i∗ and j ∗ refer to both maps associated to H, G. Let’s show commutativity of the
diagram
HomG (Fn , A)
jG
/ HomG (Fn , A00 )
O O
cor cor

HomH (Fn0 , A) / HomH (Fn0 , A00 )

jH

Now cor[jH (f )] = cor[fb = j ◦ f ] = g∈G/H g fb(g −1 ) = g∈G/H gj[f (g −1 )], and in the other di-
P P

rection we have jG [cor(f )] = jG [ g∈G/H gf (g −1 )] = g∈G/H gf\ (g −1 ) = g∈G/H j[gf (g −1 )] =

P P P
−1
P
g∈G/H gj[f (g )] where we note that j is a G-module homomorphism and hence commutes with the
G-action. Thus the diagram is commutative, and similar calculations give commutativity with i∗ and ∂ ∗
since both are G-module homomorphisms.

42: Let p be a prime and let Sp be the symmetric group of degree p. Then each Sylow p-subgroup
P is cyclic of order p, one such being the subgroup generated by the cycle (1 2 · · · p). Thus, H ∗ (P, Z) ∼
=
Z[ν]/(pν) where degν = 2. Show that NSp (P )/P ∼ = Z∗p and that it acts on P ∼= Zp in the obvious way.
Conclude that H ∗ (Sp , Z)(p) ∼
= Z[ν p−1 ]/(pν).

70
Since P is abelian, P ⊆ CSp (P ). Now |CSp (P )| = p(p − p)! = p as explained on pg127[2] where
we note that P and its generator have the same centralizer, so we must have CSp (P ) ∼ = P . Corol-
lary 4.4.15[2] states that NSp (P )/CSp (P ) is isomorphic to a subgroup of Aut(P ), and by Proposition
4.4.16[2] we know that Aut(P ) ∼ = Z∗p is cyclic of order ϕ(p) = p − 1 [ϕ is the Euler function]. Thus
∼ ∗
NSp (P )/P = H ⊆ Zp . The number of p-cycles in Sp is (p − 1)! as explained on pg127[2], and every
conjugate of P contains exactly p − 1 p-cycles, so there are (p − 1)!/(p − 1) = (p − 2)! conjugates of P
which is equal to the index |Sp : NSp (P )| by Proposition 4.3.6[2]; thus |NSp (P )| = p!/(p − 2)! = p(p − 1).
This means |H| = |NSp (P )/P | = p − 1 and hence H ∼ = Z∗p ⇒ NSp (P )/P ∼ = Z∗p . This group acts
by conjugation on P because it is a quotient of the normalizer, and this action is well-defined because
conjugation by elements of P is trivial, noting that P is abelian.
Theorem III.10.3[1] along with a theorem of Swan (see Exercise III.10.1) states that H ∗ (Sp , Z)(p) ∼ =
H ∗ (P, Z)NSp (P ) . By Proposition II.6.2[1] the conjugation action by P induces the identity on H ∗ (P, Z), so
the resulting action (Corollary II.6.3[1]) is the NSp (P )/P -action induced on H ∗ (P, Z). Thus H ∗ (Sp , Z)(p) ∼
=
Z∗
(Z[ν]/(pν)) . For a particular dimension j, the elements of the jth-cohomology group belong to Zp and
p

hence the action on an element ν would be ν 7→ z∗ · ν = zν with 0 < z < p and z∗ ∈ Z∗p . Then in the
cohomology ring, the action is given by ν i 7→ z∗ · ν i = (z∗ · ν) ` · · · ` (z∗ · ν) = (zν)i = z i ν i . So for
an element to belong to the group of Z∗p -invariants, we must have ν i = z i ν i ⇒ z i ≡ 1 mod p for all
z < p. This is only satisfied when i = P p − 1 by Fermat’s Little Theorem (z is relatively prime to p), so
the invariant elements are of the form i zi ν i(p−1) . Therefore, H ∗ (Sp , Z)(p) ∼
= Z[ν p−1 ]/(pν).

43: Prove that the Pontryagin product on H∗ (G, Z) for G finite cyclic is the trivial map in positive
dimensions (Z has trivial G-action).

The map is given by Hi G ⊗ Hj G → Hi+j G for each i, j. Since Hn G ∼ = G for n odd and Hn G = 0
for n even, the domain of the map [hence the map] is trivial for i or j even. But if both i := 2c + 1
and j := 2c0 +1 are odd, then i+j = 2(c+c0 +1) is even, so the image of the map [hence the map] is trivial.

44: Compute H 2r (Zp ⊕ Zq ) where p and q are not necessarily relatively prime.

The cohomology Künneth formula of Exercise V.2.2 gives H 2r (Zp ⊕Zq ) ∼

 L2r i 2r−i

= i=0 H (Zp )⊗H (Zq ) ⊕
(Zq ) ∼ = (Z ⊗ Zq ) ⊕ (Zp ⊗ Zq )r−1 ⊕ (Zp ⊗ Z) ⊕ [0] ∼
 L2r+1
= Zp ⊕ Zq ⊕ Zr−1
Z i 2r−i+1

  
i=0 Tor1 H (Zp ), H gcd(p,q)
for r ≥ 1.
The Tor-parts were trivial because each summand became Tor(H even , H odd ) = 0 and Tor(H odd , H even ) =
0. Note that if p and q are relatively prime, then the result becomes Zp ⊕ Zq which agrees with the fact
that Zp ⊕ Zq ∼= Zpq is cyclic.
n


45:
Prove that H i (Zn , Z) ∼
= Z i for i, n ∈ N, where we interpret n<i = 0. Deduce that H i (Zn , k) ∼


i =
n
k i for any commutative ring k.

For n = 1 we have H 0 (Z, Z) = H 1 (Z, Z) ∼ = Z and H i (Z, Z) = 0 for all i > 1 as proven on pg58[1].
This agrees with the proposed formula, so we argue by induction on n (assume the result holds up to and
including n). The Künneth formula of Exercise V.2.2 gives H i (Zn+1 ) = H i (Zn ⊕ Z) ∼
Li
= [ p=0 H p (Zn ) ⊗
TorZ H p (Zn ), H i−p+1 (Z) ] ∼
Li+1
Li
H i−p (Z)] ⊕ [ p=0 1 = H p (Zn ) ⊗ H i−p (Z), where the latter isomor-
p=0
phism follows from the fact that TorZ1 (−, Z) = 0 and H i−p+1 (Z) is either Z or 0 as stated
above.
Now
n n
(Z) ∼
= (H i (Zn ) ⊗ Z) ⊕ (H i−1 (Zn ) ⊗ Z) ∼ = H i (Zn ) ⊕ H i−1 (Zn ) ∼
Li p n i−p
p=0 H (Z ) ⊗ H = Z i ⊕ Z i−1 , so
n+1


H i (Zn+1 , Z) ∼= Z i since ni + i−1 n n! n! n!(n−i+1) n!i (n+1)!



= i!(n−i)! + (i−1)!(n−i+1)! = i!(n−i+1)! + i!(n−i+1)! = i!(n+1−k)! =
n+1


i , and the inductive process is complete.
The cohomological analog of the universal
coefficient sequence of Exercise III.1.3 gives H i (Zn , k) ∼=
n n


i n Z i+1 n ∼ ∼
(H (Z ) ⊗ k) ⊕ Tor1 (H (Z ), k) = (Z i ⊗Z k) ⊕ 0 = k i , where we note that Tor1 (Z , −) = 0. Z r

Note: These integral cohomology groups are dual to the associated
integral
homology groups; see pg38[1].
This observation of Poincaré duality is reflected in the relation ni = n−i n
.

71
46: Let p be a prime, G a finite p-group, k a field of characteristic p, and n ∈ N. If H n (G, k) = 0, prove
that H n+1 (G, Z) = 0.

The cohomological analog of the universal coefficient sequence of Exercise III.1.3 gives H n (G, k) ∼ =
(H n (G) ⊗ k) ⊕ TorZ1 (H n+1 (G), k) = 0, which in particular implies TorZ1 (H n+1 (G), k) = 0. Then since
k contains the subring Zp , we must have TorZ1 (H n+1 (G), Zp ) = 0 and hence no element in H n+1 (G)
has order p. But |G| = pα and so pα H n+1 (G) = 0 by Corollary III.10.2[1], which means any element
must have order a power of p. The only way this is satisfied with no element having order p is for every
element to have order 1 (if an element h had order pa with a ≥ 2, then pa−1 h would be an element of
order p). Only the identity has order 1, so all elements are the same; thus H n+1 (G, Z) = 0.

47: Prove that the shuffle product is strictly anti-commutative.

The shuffle product on the bar P resolution is given by

[g1 | · · · |gn ] · [g1 | · · · |gn ] = σ (−1)sgnσ [gσ−1 (1) | · · · |gσ−1 (n) |gσ−1 (n+1) | · · · |gσ−1 (2n) ],
where σ ranges over the (n, n)-shuffles. But given any shuffled tuple (−1)s [gi1 | · · · |g| · · · |g| · · · |gi2n ] we
have a corresponding shuffle which simply swaps the two g’s, leaving the 2n-tuple fixed and altering the
sign to (−1)s+1 ; the sign change arises from moving the left g an l amount of times and then moving the
right g an l + 1 amount of times (the extra +1 is due to moving the right g around the left g) and this
gives a total of 2l + 1 amount of moves with (−1)s+2l+1 = (−1)s+1 . So if there are an odd number of
shuffled tuples (due to n being even, i.e. the tuple is of even degree) then the sum will consist of paired
tuples with different signs plus one extra tuple, giving a nontrivial sum. But if the degree is odd (i.e.
an even number of shuffled tuples) then the sum will consist of only paired tuples with different signs,
giving a sum of 0. Thus x2 = 0 if degx is odd, where x = [g1 | · · · |gn ], and this is precisely the definition
of strict anti-commutativity.

48: Let p be a prime and let Cp be the cyclic group of order p with trivial Fp -action. Explain how
the fact H 2 (Cp , Fp ) ∼
= Fp and the classification of extensions of Cp by Fp matches up with the classifica-
tion theorem for groups of order p2 .

If P is a group with |P | = p2 then it has nontrivial center, |Z(P )| =

6 1, by Theorem 4.3.8[2]. If |Z(P )| = p2
then P = Z(P ) and P is abelian. The only other scenario is |Z(P )| = p, in which case |P/Z(P )| = p; this
implies P/Z(P ) is cyclic and hence it is a fact that P is abelian. By the Fundamental Theorem of Finitely
Generated Abelian Groups, P must then be either the cyclic group Cp2 or the elementary abelian group
Cp × Cp ; another proof of this is given in Corollary 4.3.9[2]. This means there are only two extension
groups of Cp by Fp , but this does not necessarily mean that there are only two classes of extensions (pos-
sible short exact sequences). Theorem IV.3.12[1] states that E(Cp , Fp ) ∼ = Fp and hence there are a total
of p classes of group extensions. There is the canonical split extension 0 → Fp → Cp × Cp → Cp → 1,
and so the other extensions must fit into p − 1 classes and arise from projections Cp2  Cp (this is
because the only injection Cp ,→ Cp2 is the canonical inclusion, i.e. the pth -power map). Switching to
βi
multiplicative notation, these extensions are 1 → Cp ≡ hai → Cp2 ≡ hbi → Cp → 1 with a 7→ bp and
βi (b) = ai for 1 ≤ i ≤ p − 1.

49: Let G be a finite group, let H ⊆ G, and let K be any group. Let F be a field which acts triv-
∼
=
ially on G and K, and consider the Künneth isomorphism κ : H ∗ (G, F) ⊗F H ∗ (K, F) → H ∗ (G × K, F).
G×K G G×K G
Show that resH×K ◦ κ = κ ◦ (resH ⊗ id) and trH×K ◦ κ = κ ◦ (trH ⊗ id).

Both equations are straightforward, so we will only prove the latter (concerning the transfer map). For
uH ∈ H ∗ (H, F) and v ∈ H ∗ (K, F), κ is defined on H ∗ (H, F)⊗F H ∗ (K, F) by κ(uH ⊗v) = huH ×v, x⊗yi =
huH , xi · hv, yi, whereP we hide the factor (−1)degv·degx for convenience. Then P (tr ◦ k)(uH ⊗ v) =
−1
trhuH × v, x ⊗ yi = (g,k)∈(G×K)/(H×K) h(g, k) · uH × v, (g, k) x ⊗ yi = (g,1)∈(G/H)×{1} hguH ×
1v, g −1 x⊗1yi = g∈G/H hguH , g −1 xi·hv, yi. But [κ◦(tr⊗id)](uH ⊗v) = κ( g∈G/H guH (g −1 x)⊗v(y)) =
P P
−1
P
g∈G/H hguH , g xi · hv, yi, so the two compositions are in fact equal, as desired.

50: For an abelian group G whose order is divisible by the prime p, Theorem V.6.6[1] states that

72
 V ∼
=
the isomorphism ρ : Zp (Gp ) ⊗Zp ΓZp (pG) → H∗ (G, Zp ) is natural if p 6= 2. Why?

Referring to pg126[1], where pG = Tor(G, Zp ) and Gp = G ⊗ Zp = G/pG, we have a split-exact uni-

V2
V coefficient sequence 0 →
versal (Gp ) → H2 (G, Zp ) → pG → 0. Now ρ(x ⊗ y) = ψ(x)ϕ(y), where
ψ : (Gp ) → H∗ (G, Zp ) is the natural map of Theorem V.6.4[1] and ϕ : Γ(pG) → H∗ (G, Zp ) is the
Zp -algebra homomorphism extended from a splitting φ : pG → H2 (G, Zp ) of the above sequence. Since ψ
is natural and ϕ is an extension of the splitting φ, the question of naturality of ρ reduces to the question
of naturality of φ (in dimension 2). The splitting is made by choice, and if p is odd, we may use the
V2
canonical splitting H2 (G, Zp ) → (Gp ) given in Exercise V.6.4(b) since 2 is invertible in Zp for p 6= 2
(i.e. prime p odd); remember that a splitting can be made on either side of the sequence. Thus the
isomorphism ρ is natural if p 6= 2. But if p = 2 then we do not have a known canonical splitting, and we
cannot prove naturality in this case since we do not have an explicit map.

51: Referring to the proof of Proposition VI.2.6[1], if η : M → Q0 is an admissible injection (i.e. a

split injection of H-modules) and M is projective as a ZH-module, then why is Cokerη projective as a
ZH-module?
η
Since η is H-split, we have a split-exact sequence 0 → M → Q0 → Cokerη → 0, and so Q0 ∼= M ⊕Cokerη.
But Q0 is ZG-projective by Corollary VI.2.2.[1], hence ZH-projective by Exercise I.8.2. Now a direct
sum is projective iff each direct summand is projective, so Cokerη must be ZH-projective.

b q (G, C)
52: Let G be a goup with order r. Show that for each q there exists G-modules C with H
cyclic of order r.

For q = 0 we can take C = Z since H b 0 (G, Z) = Z/|G|Z = Zr . Then using the dimension-shifting tech-
nique, we can find G-modules C1 and C2 such that H b 0 (G, Z) ∼
=H b 0 (G, Z) ∼
b 1 (G, C1 ) and H =Hb −1 (G, C2 );
see property 5.4 on pg136[1]. Therefore, through repeatable applications of dimension-shifting, we can
b q (G, C) ∼
range over all q to obtain H = Zr .

53: Fill in the details to the proof of Proposition VI.7.1[1] which states that the evaluation pairing
ρ : H i (G, M 0 ) ⊗ Hi (G, M ) → Q/Z is a duality pairing, where M 0 = Hom(M, Q/Z).

Let F be a projective resolution of Z over ZG. The evaluation pairing is obtained by composing the
pairing h·, ·i : HomG (F, M 0 ) ⊗ (F ⊗G M ) → M 0 ⊗G M with the evaluation map M 0 ⊗G M → Q/Z.
The pairing hu, z = x ⊗ mi is given by u ⊗ (x ⊗ m) = u(x) ⊗ m, and composing this with the evalu-
ation map gives ρ(u ⊗ z) = [u(x)](m). Note that ρ gives rise to the map ρ̄ : H i (G, M 0 ) → Hi (G, M )0
defined by [ρ̄(u)](z) = ρ(u ⊗ z) = [u(x)](m). We have HomG (F, M 0 ) = HomG (F, Hom(M, Q/Z)) ∼ =
(∗)
HomG (F ⊗M, Q/Z) = Hom(F ⊗G M, Q/Z) = (F ⊗G M )0 . The equality (∗) arises from Exercise III.1.3, as
Q/Z has trivial G-action. Since ( )0 is exact, we can pass to homology to obtain H ∗ (G, M 0 ) ∼ = H∗ (G, M )0 .
∼
This isomorphism resulted from HomG (F, Hom(M, Q/Z)) = HomG (F ⊗M, Q/Z) which is given by Theo-
rem 10.5.43[2], and this is precisely the map ρ̄. Therefore, since ρ̄ is an isomorphism, ρ is a duality pairing.

54: Let G = Z2 = hgi and let A = Z8 , written additively. Make G act on A by x 7→ 3x, and let
G act trivially on B = Z2 . Show that A is cohomologically trivial, but A ⊗ B is not.

First note that H b ∗ ({1}, M ) = 0 for any M ; it is clearly trivial in all dimensions not equal to −1 and 0,
and in those two dimensions it is the kernel and cokernel of the norm map M → M which is the identity
(so the kernel and cokernel are trivial). Thus for A to be cohomologically trivial it suffices to show that
Hb ∗ (G, A) = 0. The complete resolution from Exercise VI.3.1 implies that H b n (G, A) = CokerN for n even
and H b n (G, A) = KerN for n odd (see pg58[1]). The action on A gives AG = Z8 /Z4 = {0+Z4 , 1+Z4 } ∼ = Z2
since 1 7→ 3 ≡ 1 mod 2, and AG = Z2 = {0, 4} since 4 7→ 12 ≡ 4 mod 8. Thus the norm map N : Z2 → Z2
is given by 1 7→ N 1 = 1 · 1 + g · 1 = 1 + 3 = 4, which is the identity. Thus KerN = CokerN = 0
and H b ∗ (G, A) = 0. However, repeating the above with coefficient module A ⊗ B we see that the norm
map is the trivial map, 1 ⊗ 1 7→ 1 ⊗ 1 + g · (1 ⊗ 1) = 1 ⊗ 1 + 3 ⊗ 1 = 4 ⊗ 1 = 2 ⊗ 2 = 0. Thus

73
b 2i+1 (G, A) = KerN = Z2 ⊗ Z2 ∼
H = Z2 and A ⊗ B is not cohomologically trivial.

55: Let H / G and let M be a G-module. Then M H is naturally a G/H-module, and the pair
(ρ : G → G/H, α : M H → M ) is compatible in the sense that α[ρ(g) · m] = g · α(m). Thus we
have a homomorphism inf : H n (G/H, M H ) → H n (G, M ) called the inflation map.
Using this, show that for the semi-direct product G = H o K and module M with trivial G-action, the
group H n (K, M ) is isomorphic to a subgroup H n (G, M ).

From the inclusion i : K ,→ G and the surjection ρ : G → G/H = K we can pass to cohomology to ob-
tain the restriction res : H n (G, M ) → H n (K, M ) and the inflation inf : H n (K, M ) → H n (G, M ). Since
ρ ◦ i = idK , the composite inf ◦ res is also the identity. Thus inf is injective, so H n (K, M ) ⊆ H n (G, M )
up to isomorphism.

55: In Exercise AE.34 it was shown that a module is finitely generated Z-projective iff it is finitely
generated Z-free. Weakening the hypothesis, show that Z-projective = Z-free.

Free modules are projective, so it suffices to show that Z-projective implies Z-free. If P is Z-projective
then it is a submodule of a Z-free module. But any submodule of a Z-free module is free by Theorem
I.7.3[5], so P is Z-free.

57: Prove that the symmetric group S3 has periodic cohomology, and find its period.

Since S3 ∼ = Z3 o Z2 and |S3 | = 3! = 6, its Sylow subgroups are Z2 and Z3 which are both cyclic.
Then by Theorem VI.9.5[1], S3 has periodic cohomology.
Alternatively, any proper subgroup must have order 1 or 2 or 3 and hence must be cyclic, so Theorem
VI.9.5[1] implies that S3 has periodic cohomology.
Alternatively, H 4 (S3 , Z) ∼
= Z6 by Exercise III.10.1 and so S3 has periodic cohomology by Theorem
VI.9.1[1].
We can see this periodicity via Exercise III.10.1, because H n (S3 ) ∼
= H n+4 (S3 ) for all n > 0. Thus the
period is 4.

58: Let N : Z → ZG denote the G-module homomorphism z 7→ N z, where N ∈ ZG is the norm

element, and let ε : ZG → Z be the augmentation map. Prove that if α : Z → ZG is a G-module
homomorphism then α = aN for some a ∈ Z, and if β : ZG → Z is a G-module homomorphism then
β = bε for some b ∈ Z.

First note that α is determined by where it sends the identity, α(1) = x. Then for α to be com-
patible with the G-action we must have g · x = g · α(1) = α(g · 1) = α(1) = x, so x ∈ (ZG)G = Z · N
where the equality is shown in Exercise AE.1; thus α = aN for some a ∈ Z. Now β is also determined
by β(1) = z, so we must have z = g · z = g · β(1) = β(g) = z 0 and hence β maps G onto a single integer
b; thus β = bε for some b ∈ Z (where we note that ε(G) = 1).

59: If G1 and G2 are perfect groups with universal central extensions E1 and E2 , respectively, prove
that E1 × E2 is a universal central extension of G1 × G2 .

We have universal central extensions 0 → H2 (Gi ) → Ei → Gi → 1 by hypothesis (see Exercise

IV.3.7). Since the direct sum of two extensions is an extension, we have a central extension 0 →
H2 (G1 )⊕H2 (G2 ) → E1 ×E2 → G1 ×G2 → 1. I claim that this extension is a universal central extension.
Indeed, H2 (G1 )⊕H2 (G2 ) is isomorphic to H2 (G1 ×G2 ) by the Künneth formula, since H1 (Gi ) = 0 by per-
fectness of Gi . Now, the universal central extension of G1 × G2 is 0 → H2 (G1 × G2 ) → E → G1 × G2 → 1
by definition, so E1 × E2 ∼= E by the Five-Lemma (applied to the two sequences of G1 × G2 ) and hence
E1 × E2 is the universal central extension of G1 × G2 .

60: In the proof of Proposition VIII.2.4[1], with Γ0 ⊂ Γ, where did we use the hypothesis that
|Γ : Γ0 | < ∞?

74
We considered a free Γ-module F , and noted that if F 0 is a free Γ0 -module of the same rank then
= IndΓΓ0 F 0 . We then applied Shapiro’s lemma to yield H n (Γ0 , F 0 ) ∼
F ∼ = H n (Γ, F ). But Shapiro’s
n 0 0 ∼ n Γ 0
Lemma is actually given by H (Γ , F ) = H (Γ, CoindΓ0 F ). We therefore used the isomorphism
CoindΓΓ0 F 0 ∼
= IndΓΓ0 F 0 of Proposition III.5.9[1], which holds if |Γ : Γ0 | < ∞.

75
11 References
[1] Kenneth Brown, Cohomology of Groups, Springer GTM 87 (1982).

[2] David Dummit and Richard Foote, Abstract Algebra (3rd Ed.), John Wiley & Sons (2004).

[3] Allen Hatcher, Algebraic Topology, Cambridge University Press (2006).

[4] James Munkres, Elements of Algebraic Topology, Perseus Books Publishing (1984).

[5] Serge Lang, Algebra (Revised 3rd Ed.), Springer GTM 211 (2002).

[6] James Munkres, Topology (2nd Ed.), Prentice Hall (2000).

Other Useful Literature

Alejandro Adem & James Milgram, Cohomology of Finite Groups (2nd Ed.), Springer (2004).
David Benson, Representations and Cohomology I+II, Cambridge University Press (1995).
Jon Carlson & others, Cohomology Rings of Finite Groups (with Appendix), Springer (2003).
Henri Cartan & Samuel Eilenberg, Homological Algebra, Princeton University Press (1956).
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