.

MTTS 2004, May 17 - June 12, 2004

Level I - Group theory
B.Sury
I.S.I., Bangalore

This is an expanded version of the original notes. Some of the material is

taken from my book “Group theory - Selected Problems” published in 2004
by the Universities Press and distributed by Orient-Longman Pvt.Ltd.

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§0 Introduction

The presence of group theory can often be felt even when it is not seen. Let
me explain the meaning of the first statement. Generally, what one sees is
some geometrical object and the underlying group of symmetries is always
there, guiding the whole structure but, to be seen, it has to take an avataar.
Of course, the significance of a notion lies in the number of diverse situations
in which it plays a substantial role. In that respect, the notion of a group is
one of the most - if not the most - important in mathematics as well as in
other sciences. We digress for a moment to ask a question. We have been
saying that a group may be invisible but may be at work in a situation.
Here is a problem which exemplifies this.
Consider all integers of the form 14x + 20y and all integers of the form
24x + 34y where x, y run through integers. Then, look at the lattice points
(14x + 20y, 24x + 34y) in the plane. This set intersects the Y -axis and the
X-axis at times. What is the smallest natural number n for which both
(n, 0) and (0, n) are in this set? What is the proportion of lattice points on
the plane which lie on this set?
These questions can be easily and naturally answered by recognising the
groups figuring here. We shall do so later but you are urged to think about
it. I offer a small prize to whoever solves either question first.
Group as Latin square
All of you have probably already learnt a lot of group theory but let us
start with some easy and familiar things. Let us start with a visual way
of defining groups. This idea is due to Uri (not Sury !) Rimon of Hebrew
University, Jerusalem.
View multiplication table of group as a Latin square with a distinguished
vertex. That is to say, there is a square array of symbols (to represent the
elements of the group) with each symbol occurring exactly once in each row
and in each column. There is a distinguished symbol e such that whenever
one draws a rectangle eacb cornered at this symbol with ea along a column
and eb along a row, then the symbol c diagonally opposite to it can be taken
to represent the product ab. It is a nice exercise to deduce associativity and
verify that such a table is equivalent to a group structure on the symbols.
It is also a nice exercise to prove statements like (ab)−1 = b−1 a−1 .
A word of warning : Not every Latin square with a distinguished vertex
arises from a group. It is possibly true that if the operation using rectangles
as above is well-defined, then the Latin square indeed comes from a group.

As we remarked, it is fruitful to view the same notion in different guises

1
as it tells us some new aspect of its personality. One could think of a
group as a special equivalence relation which is often how it arises. By
the way, continuing in the same vein of visual viewing, one may think of
a general equivalence relation in the following nice way. Given a set of d
objects, a relation is just a d × d matrix R of 0’s and 1’s associated in an
obvious manner. Reflexivitiy and symmetricity correspond, respectively, to
the relation matrix having diagonal entries 1 and being symmetric.
Ex : What does transitivity correspond to?
Ans. Rij 6= 0 implies (R2 )ij 6= 0.

Without further ado, let us now start with the examples of groups that
one comes across. We shall then study them in different ways; study them
according to their size, study them according to the way they ‘act’, study
them according to their internal structure etc. Needless to say, there are
still many things to be properly understood. For instance, the finite simple
groups have been classified but it cannot be said that the classification of
finite simple groups (CFSG) is well-understood (indeed, no proof which is
5000 pages long can be so accepted).

Examples and basic notions.

• ZZ/n denotes the group of integers modulo n, under addition mod n.

• (ZZ/n)∗ = {a ≤ n : (a, n) = 1} under multiplication mod n.
The notations ZZ/n and (ZZ/n)∗ are perhaps less familiar compared to ZZn
etc.
• S 1 = {z ∈ C : |z| = 1} under multiplication.
All the above are abelian groups.
• Sn , the set of all permutations (i.e., bijections) of n symbols, under the
composition of permutations. In fact, for any set X, the set Sym(X) of all
bijections on X is a group under composition. The subset SF (X) consisting
of all those bijections which move only finitely many elements, is a subgroup.
We shall use the convention that the permutation στ is obtained by apply-
ing σ first and τ later.
• GL(n, Q), the set of all n × n rational matrices which have non-zero deter-
minants (the symbol GL stands for ‘general linear’) is a group under matrix
multiplication.
Similarly, one can define GL(n, R) and GL(n, C).
One has also the ‘special linear’ groups SL(n, Q), SL(n, R) etc. consisting
of the matrices which have determinant 1.

2
• SO(n) = {g ∈ SL(n, R) : gg t = In }, is known as the special orthogonal
group of degree n.
• SU (n) = {g ∈ SL(n, C) : ḡg t = In }, is known as the special unitary group
of degree n.    
t 0 In 0 In
• Sp(n) = {g ∈ SL(2n, R) : g g= }, is known as
−In 0 −In 0
the symplectic group of degree n; here 0 and In denote n × n block matrices.
These are nonabelian groups when n > 1.
• B(n, Q), the upper triangular rational matrices with nonzero diagonal
entries, U (n, Q), the subset consisting of those upper triangular matrices
which have all diagonal entries equal to 1 and T (n, Q), the diagonal matri-
ces with all diagonal entries nonzero rational numbers, are groups.
Note that T (n, Q), for any n and, U (2, Q) are abelian groups.
• GL(n, ZZ) = {g an integral matrix : det(g) = ±1}.
• GL(n, ZZ[1/p]) where p is a prime number.
Here, ZZ[1/p] denotes the set of rational numbers whose denominators are
divisible only by p and, the above group consists of all matrices whose de-
terminants are ± a power of p.
Note that
GL(n, ZZ) ≤ GL(n, ZZ[1/p]) ≤ GL(n, Q)
where we have used the notation H ≤ G to denote the fact that H is a
subgroup of G. We shall also write G ≥ H at times. Also, H < G for
groups would mean that H is a proper subgroup.
Let us note that the set of matrices
 
a a
G={ : a ∈ R∗ }
0 0

forms a group under matrix multiplication but it is not a subgroup of

GL(2, R).
We have
P erm(n) = {Pσ : σ ∈ Sn } ≤ GL(n, ZZ)
where Pσ is the ‘permutation matrix’ whose rows are the rows of the identity
matrix permuted according
T to σ i.e., the rows of Pσ are Rσ(1) , · · · , Rσ(n) .
Note that trivially I Hi ≤ G if Hi (i ∈ I) ≤ G while the union of two
subgroups may not be a subgroup. Further, if g is an element of a group
G, then its centraliser CG (g) := {x ∈ G T : xg = gx} ≤ G. Moreover, for any
subset S ⊂ G, the subgroup CG (S) := s∈S CG (s) is the centraliser of S.
For S = G, the centraliser CG (G) is usually denoted by Z(G), and is called

3
the center of G.
If µ(n) denotes the complex n-th roots of unity, we have a chain of subgroups
[
µ(n) ≤ µ(n) ≤ S 1 ≤ C∗ .
n

For any subset S ⊂ G, one denotes by < S >, the subgroup generated by
S. In concrete terms, it consists of all finite products of elements of S and
their inverses. For instance, for any group G and any positive integer n, the
subset Gn := {g n : g ∈ G} gives a subgroup < Gn >. Note that, in any
abelian group G, we have < Gn >= Gn .
Recall that a subgroup N is normal in G if gxg −1 ∈ N for all g ∈ G, x ∈ N .
This notion comes from Galois theory. Evidently, for any G and any n, the
subgroup < Gn > is normal in G; we write < Gn > EG.
Given a normal subgroup N , one can give a group structure to the set G/N
of all (left or right - both are same) cosets of the subgroup. There is a natural
onto homomorphism from G to G/N whose kernel is N and, conversely, the
kernel of any homomorphism from G to some group, is a normal subgroup
of G.

Examples of homomorphisms :
• i : H → G; inclusion of a subgroup H in G.
• For any n, Z → Z; a 7→ na.
• Z → Z/n; a 7→ a mod n.
• GL(n, C) → C∗ ; g 7→ det(g).
• Sn → P erm(n); σ 7→ Pσ .
• The composition of two homomorphisms.
Note that the composite of the last-mentioned homomorphism with the de-
terminant homomorphism is the ‘sign’ of the permutation.
• G → G; g 7→ g n if G is abelian.
• If 0 6= λ ∈ R, Z → R∗ ; n 7→ λn .
• R → R∗ ; t 7→ et .  
1 t
• R → U (2, R); t 7→ .
0 1  
1 t iθ Cosθ Sinθ
• S → SO(2) = {g ∈ SL(2, R) : gg = 1}; e 7→ .
−Sinθ Cosθ
• SO(2) → S 1 ; g 7→ last column.
• SU (2) → S 3 ; g 7→ last column. Pn
Here S n−1 = {(z1 , · · · , zn ) ∈ Rn : 2
i=1 zi = 1}, the unit sphere in n
dimensional space. The above two are isomorphisms. Some of the other
homomorphisms above are isomorphisms as well.

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One says that a group G is finitely generated if there exists a finite subset
S ⊂ G such that G =< S >.
Given a subset S ⊂ G, one also defines the normal subgroup generated by S
as the smallest normal subgroup < S >N of G which contains S. In concrete
terms, it consists of all finite products of conjugates of elements of S and
their inverses.
For any subset S ⊂ G, we define the normaliser

NG (S) = {g ∈ G : gSg −1 = S}.

Observe < S > ENG (S) ≤ G and NG (S) is the largest subgroup of G
containing S in which < S > is normal.
For any group G, and x, y ∈ G, we denote [x, y] = xyx−1 y −1 . Such elements
are called commutators in G. If S is the subset of all commutators of G,
one obtains the commutator subgroup of G, denoted by [G, G]. Sometimes,
one also denotes [G, G] by D(G) and refers to it as the derived group of G.
More generally, for H, K ≤ G, [H, K] denotes the subgroup of G generated
by {[h, k] : h ∈ H, k ∈ K}.
A group G is said to be solvable - the terminology comes from Galois theory
- if the chain

G ≥ D1 (G) := D(G) ≥ D2 (G) := D(D(G)) ≥ · · · · · ·

becomes the trivial group in a finite number of steps. Evidently, any abelian
group is solvable. A nonabelian example is the group B(n, Q), the upper
triangular rational matrices with nonzero diagonal entries for any n ≥ 2.
A complementary concept is that of a perfect group; a group G is said to be
perfect if G = [G, G].
A group G is said to be nilpotent if the lower central series

G ≥ C 1 (G) := D(G) ≥ C 2 (G) ≥ C 3 (G) · · ·

becomes trivial in a finite number of steps where C n+1 (G) = [G, C n (G)].
It is not only evident that abelian groups are nilpotent but also that nilpo-
tent groups are solvable because, it is seen by induction on n that Dn (G) ≤
C n (G) for any G. The group U (n, Q) is nilpotent but nonabelian when
n > 2.
In any group G, let xy denote yxy −1 and [x, y] denote the commutator
xyx−1 y −1 . It is also convenient to define x−y to be yx−1 y −1 . Also, one
defines inductively

[x1 , · · · , xn ] := [[x1 , · · · , xn−1 ], xn ].

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Free groups and presentations
Fn , the free group of rank n, is defined to be the set of all reduced (fi-
nite) words in n symbols x1 , · · · , xn and their ‘inverse symbols’ denoted by
x−1 −1
1 , · · · , xn . Here, a word is said to be reduced if it does not contain two
symbols xi and x−1 i in juxtaposition. The empty word is also counted and
is the identity element. The group multiplication is by concatenation of two
reduced words and cancelling off all consecutive symbols of the form xi x−1 i
or x−1
i x i .
Fn is nonabelian if n ≥ 2. It is evident that any group which can be gen-
erated by n elements can be identified upto isomorphism with a quotient
group of Fn .
A group G is said to be finitely presentable if G ∼ = Fn /K for some n where
N
K =< R > for some finite subset R of Fn . The choice of finite sets X ⊂ G
with |X| = n and R ⊂ Fn with Fn / < R >N ∼ = G is called a finite presenta-
tion of G.
If G =< X|R > and H =< Y |S >, then their free product is defined to
be the group < X t Y |R ∪ S > and is denoted by G ∗ H. Note that Fn is
a free product of ZZ with itself n times (taking free products is an associa-
tive operation). We shall see via some problems that familiar groups like
SL(2, ZZ)/{±I} are free products.

Group actions
Let us recall the concept of group actions which makes groups one of the
most powerful mathematical tools.
A group G is said to act on a set S if there is a homomorphism from G to
the permutation group Sym(S) of S. The action is said to be faithful if the
homomorphism is injective. It is customary to write g.s for the element of S
that s ∈ S is sent to by the permutation corresponding to g ∈ G. For s ∈ S,
the subset G.s := {g.s : g ∈ G} is called the orbit of s. For any g ∈ G, the
subset S g := {s ∈ S : g.s = s} is called the set of fixed points under g. For
s ∈ S, the subgroup Gs := {g ∈ G : g.s = s} is called the isotropy subgroup
at s. Note that, for s ∈ S, the map g 7→ g.s is a well-defined bijection from
the set G/Gs of left cosets of Gs to the orbit G.s of s. Thus, orbits under a
finite group action have cardinalities which are divisors of the order of the
group.
In the particular case of a group acting on itself by conjugation, we denote
the orbit of an element g (this is the conjugacy class of g) by G(g). One
also writes g ∼ h to mean that g and h are in the same orbit; that is, they
are conjugate.

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One calls an action of G on a set S transitive if each orbit is the whole set
S.
For instance, Sn acts transitively on {1, 2, · · · , n}.
The group GL(n, R) acts transitively on Rn − (0).
A group G is said to act r-transitively on a set S if G acts transitively on
the set
{(s1 , · · · , sr ) ∈ S r : gi distinct }.

Automorphism group
For an element g in a group G, one usually denotes the automorphism
x 7→ g −1 xg of G as Int (g). The group Aut (G) of all automorphisms of G is
defined by means of the composition in the following order: g(στ ) = (g(σ))τ ;
this is consistent with our convention of multiplying permutations. In this
convention, the set of automorphisms x 7→ Int(x) can be identified with a
subgroup of Aut G because

g(Int(xy)) = (xy)−1 g(xy) = g(Int(x))(Int(y)).

This is in fact, a normal subgroup.

Jordan-Hölder theorem
The last important notion we recall is that of composition series and the
Jordan-Holder theorem. Recall that a normal series of a group G is simply
a finite sequence of subgroups

{1} = G0 C G1 · · · C Gn = G.

Of course, we know that the Gi ’s need not be normal in the whole of G;

they are usually called subnormal subgroups of G. A normal series

{1} = H0 C H1 · · · C Hm = G

is said to be a refinement of a normal series

{1} = G0 C G1 · · · C Gn = G

if each Gi is some Hj ; it is called a proper refinement if m > n.

It is elementary to show that any two normal series have refinements which
are isomorphic; that is, refinements in which any successive factor Gi /Gi−1
of one is isomorphic to a successive factor of the other and vice versa.
One says that a normal series is a composition series if it has no proper

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refinements. Note that this is equivalent to the successive factors being
simple groups.
For instance, any normal series for ZZ will have a proper refinement because
the smallest nontrivial term in such a series is an infinite cyclic group and will
have proper nontrivial (normal) subgroups. Therefore, ZZ has no composition
series.
The theorem of Jordan-Holder asserts that for a finite group, each normal
series admits of a refinement which is a composition series and that any two
composition series are isomorphic.
Finally, we recall Zorn’s lemma which is equivalent to the axiom of choice.
The use of Zorn’s lemma becomes unvoidable when we need to prove facts
like every vector space has a basis. Let (S, ≤) be any non-empty partially
ordered set. This means the three properties : (i) s ≤ s for all s ∈ S; (ii)
s ≤ t, t ≤ s ⇒ s = t and, (iii) s ≤ t ≤ u ⇒ s ≤ u.
A chain is a subset T of S in which, for any two elements s, t ∈ T , either
s ≤ t or t ≤ s. Zorn’s lemma asserts that if every chain T in S has an upper
bound (that is, an element s0 ∈ S such that t ≤ s0 for all t ∈ T ), then S has
a maximal element (that is, an element m ∈ S which is not ≤ any element
of S other than itself).
Finally, we recall that a group is said to have exponent d if d is the smallest
natural number for which each element g satisfies g d = 1.
One word of convention - a subgroup M of a group G is maximal if M is a
proper subgroup which is not contained in any other proper subgroup of G.
Also, the identity element is always denoted by 1 unless there is an abelian
group written additively when it is denoted by 0. Also, we write O(g) for
the order of an element g of a group G while the order of the group itself
is written as |G|. The notation pr kn means that pr is the highest p-power
dividing n.

Now, we begin with the systematic study of groups. After studying cyclic
and abelian groups, we discuss group actions and permutation groups. Then,
we shall study nilpotent and solvable groups. Following that we study groups
of matrices and groups figuring in geometry. Finally, we discuss miscella-
neous results on groups which do not fit exactly into any of the above titles.
Each of these discussions will be interspersed with several related results
and miscellaneous comments. Throughout the notes, we use the defi-
nitions and notations introduced in section §0.

§1 Cyclic groups

8
A group generated by a single element is called cyclic.
The only infinite cyclic group, upto isomorphism, is Z and for any n, the
only finite cyclic group of order n is, upto isomorphism, the set Z/n of
integers modulo n considered under addition modulo n. Indeed, for finite
cyclic G, g 7→ 1̄ is an isomorphism onto ZZ/|G|.
Evidently, Subgroups and quotient groups of cyclic groups are cyclic as well
but it is not true that a subgroup of an n-generated group is n-generated
although a quotient of an n-generated group is n-generated.
Lemma 1.1
(i) The direct product G×H of nontrivial cyclic groups G, H is cyclic if, and
only if, both G, H are finite and their orders m, n are coprime. (ii) A finite
group is cyclic if, and only if, it has aPunique subgroup of each order dividing
the order of the group. Hence n = d|n φ(d) for each natural number n.
Lemma 1.2
Let G be a finite group in which, for every n/|G|, the set {g : g n = e} has
at most n elements. Then, G is cyclic and the set {g : g n = e} has exactly
n elements for each n||G|. In particular, any finite subgroup of K ∗ is cyclic
for any field K.
Note that for the application to field theory, we need only prove the result
for abelian G. Later, we will discuss this again and prove it as an application
of Sylow’s theorems.
Proof of 1.2
Let |G| = n and d|n.
Consider N (d) = #{g ∈ G : O(g) = d}.
If N (d) 6= 0, look at some element g with O(g) = d. As e, g, g 2 , · · · , g d−1
are distinct and are solutions of xd = e, these are all the solutions of the
equation xd = e. As elements of order d in G are among these and are
φ(d) in number, we have proved that N (d) = φ(d) if N (d) P 6= 0. As every
element
P of G has some
P order d dividing n, we have n = d|n N (d). Since
n = d|n φ(d) ≥ d|n N (d) = n, we must have the equality N (d) = φ(d)
for all d|n. In particular, N (n) = φ(n) 6= 0.
All of us are familiar with the division algebra H of Hamilton’s quaternions
H := {a0 + a1 i + a2 j + a3 k : ai ∈ R}
where the mulltiplication is defined by i2 = j 2 = k 2 = −1, and ij = k = −ji.
Note that H is a 4-dimensional vector space over R and is central over R.
In general, a finite-dimensional division algebra D over a field K is a finite-
dimensional vector space D over K such that D has a multiplication under

9
which the set D∗ of all non-zero elements forms a group; the mutliplication
is also required to satisfy x(y + z) = xy + xz and (x + y)z = xz + yz and,
λ(xy) = (λx)y for all λ ∈ K and all x, y, z ∈ D. Note that D is a field if,
and only if, D∗ is an abelian group.
Further, D is said to be central over K if the centre of D∗ is K ∗ .
A very similar construction to H above produces many examples of central
division algebras over Q.
Remark 1.3
Let D be any finite-dimensional central division algebra over a field F of
characteristic p > 0. Then, every finite subgroup of D∗ is again cyclic.
On the other hand, the above Hamilton quaternion algebra

H := {a0 + a1 i + a2 j + a3 k : ai ∈ R}

is such that H ∗ := H \ (0) is a group which contains a finite noncyclic group

- the quaternion group Q = {±1, ±i, ±j, ±k}.
To see how the first remark follows, let F = Z(D) be the centre of D. Note
that F ⊇ Fp , the field of p elements. If G ≤ D∗ is a finite subgroup, then
the set X
K := { αg g : α ∈ Fp }
g∈G

is a finite-dimensional Fp -vector space. So, it is a finite division algebra and,

therefore, it must be a field K and G ≤ K ∗ . Therefore, by lemma 1.2, G
must be cyclic.
Remark 1.4 (groups vs congruences)
A re-statement of Lagrange’s theorem for the groups (ZZ/p)∗ and (ZZ/n)∗ ,
one has Fermat’s little theorem asserting ap−1 ≡ 1 mod p for prime p and
(a, p) = 1 and Euler’s congruence asserting aφ(n) ≡ 1 mod n for (a, n) = 1.
In fact, we have the assertion n|φ(an − 1) for natural numbers a, n. This is
because, in the group (ZZ/(an − 1))∗ , the integer a has order n.
The Wilson congruence that for a prime p, one has (p − 1)! ≡ −1 mod p
follows by looking at the product of all elements in ZZ∗p . In the product, each
element cancels with its inverse except for those elements which are their
inverses. The only elements of (ZZ/p)∗ which are their own inverses are 1
and p − 1 because if i2 ≡ 1 mod p, then p|(i − 1) or p|(i + 1); so the resultant
product is p − 1.
We have a more general congruence :
Let a be a natural number and p be a prime. Suppose p|(an −1), p2 6 |(an −1)
for some n ≥ 1. Then, ap−1 6≡ 1 modulo p2 .

10
To prove this, let θ : (ZZ/p2 )∗ → (ZZ/p)∗ be the canonical homomorphism.
Considering a as an element of (ZZ/p2 )∗ , it follows that an ∈ ker θ. Clearly,
Ker θ has order p and an is a nontrivial element of (ZZ/p2 )∗ since p2 6 |(an −1)
by hypothesis. Therefore, an has order p in (ZZ/p2 )∗ . This means p divides
the order of a in (ZZ/p2 )∗ and, thus ap−1 6≡ 1 modulo p2 .
Theorem 1.5
(ZZ/n)∗ = {a ≤ n : (a, n) = 1} is cyclic if, and only if, n = 2, 4, pr or 2pr for
some odd prime p.
Any generator for the group for such n is called a primitive root modulo n
in number-theoretic parlance.
Exercise 1.6
Are ZZ[1/2] and ZZ[1/3] isomorphic?

§2 Abelian groups

Lemma 2.1
A finite abelian group A is isomorphic to the direct sum of cyclic p-groups
for various primes p dividing its order.
A free abelian group of rank n is defined to be a group isomorphic to Zn ,
the set of integral n-tuples under co-ordinatewise addition. Equivalently,
P it
is an abelian group G with a basis of n elements g1 , · · · , gn i.e., i ai gi = 0
implies ai = 0 for all i and G = {a1 g1 + · · · + an gn : ai ∈ Z}. The rank n is
uniquely determined since the number of homomorphisms from G to Z/2 is
2n .
One of the most important results on abelian groups is the next theorem
which, in turn, follows from the one following it :
Theorem 2.2 (Structure theorem for finitely generated abelian
groups)
A finitely generated abelian group is isomorphic to Zm × Zd1 × · · · × Zdr for
some m ≥ 0 and di dividing di+1 . The integer m as well as all the di ’s (upto
sign) are uniquely determined.
Theorem 2.3 (Invariant factor theorem)
If H is a subgroup of a free abelian group G of rank n, then H is free abelian
of rank r ≤ n. Further, there are bases {e1 , · · · , en } of G and {d1 e1 , · · · , dr er }
of H respectively where di divides di+1 for i < r. The integers di are uniquely
determined upto sign and are called the invariant factors of H.

11
The proof is carried out by induction on n using the division algorithm as
follows. It is clear for n = 1. Assume n > 1 and that the theorem holds
for m < n. Corresponding to any basis of G, there is a positive integer
with the property that it is the smallest positive integer that occurs as a
coefficient in the expression of elements of H in terms of this basis. This
positive integer can depend on the basis and let l1 be the smallest such
with respect to all basesP of G. Let v1 , · · · , vn be a corresponding basis for
G such that v = l1 v1 + ni=2 ai vi ∈ H. Dividing all P the ai by l1 ,Pwe have
ai = qi l1 +Pri with 0 ≤ ri < l1 . Evidently, v = l1 (v1 + ni=2 qi vi ) + ni=2 ri vi
and v1 + ni=2 qi vi , v2 , · · · , vn is another basis of G. By the minimality Pn of
l1 , we must have ri = 0 for all i ≥ 2. Thus, writing w1 for v1 + i=2 qi vi ,
v = l1 w1 ∈ H. Look at the subset H0 of H which have coefficients of w1 to
be zero in terms of the basis w1 , v2 , · · · , vn of G. Clearly, H0 is aP subgroup
of H such that H0 ∩ Zv = {0}. Also, if h ∈ H, write h = b1 w1 + ni=2 bi vi .
Once again, dividing the bi ’s P by l1 , say, bi = mi l1 + si with 0 ≤ si < l1 ,
we have h − m1 v = s1 w1 + ni=2 bi vi ∈ H. Thus, by the minimality of
l1 we get s1 = 0 i.e., h − m1 v ∈ P H0 . Thus, H = H0 ⊕ Zv. Now, H0 is
n
contained in the subgroup G0 = i=2 Zvi . By induction hypothesis, G0
has a basis w2 , · · · , wn and there exists r ≤ n such that H0 has a basis of
the form d2 w2 , · · · , dr wr with d2 |d3 | · · · |dn . Clearly, therefore, H itself has
rank r ≤ n and l1 w1 , d2 w2 , · · · , dn wn is a basis for H. We have only to show
that l1 |d2 . Once again, writing d2 = cl1 + d with 0 ≤ d < l1 , we notice
l1 w1 + d2 w2 = l1 (w1 + cw2 ) + dw2 ∈ H where w1 + cw2 , w2 , · · · , wn is a basis
of G. Thus, minimality of l1 forces d = 0 i.e., l1 |d2 . The proof is complete.
Exercise
(i) Deduce 2.2 from 2.3.
(ii) Give an example of a group G and a subgroup H such that G can be
generated by some number n of elements while H cannot be n-generated.
(iii) Solve the prize problem posed in the beginning.
Lemma 2.4
The existence of bases as in the invariant factor theorem is equivalent to the
following statement about matrices :
Given any A ∈ Mm,n (ZZ) of maximum possible rank, there exist P ∈ GL(m, ZZ)
and Q ∈ GL(n, ZZ) such that P AQ is a matrix whose‘diagonal’ entries are
d1 , d2 , · · · where di |di+1 .
Further, GL(n, ZZ) is generated by elementary matrices I + Eij .
Proof
Suppose the matrix statement holds. Let H be a subgroup of a free abelian
group G of rank n. Then, H is also free abelian of rank m ≤ n (this we

12
are assuming known through other arguments). Let α : ZZm → H and
β : G → ZZn be isomorphisms. If i : H ≤ G denotes the inclusion map, we
have the composite map β ◦ i ◦ α corresponds to a matrix A ∈ Mn,m (ZZ) with
respect to the canonical ordered bases of ZZm and ZZn . The matrix statement
gives us P ∈ GL(n, ZZ) and Q ∈ GL(m, ZZ) such that

d1 ··· 0
 
 ... ..
.
.. 
.

AQ = P . 
 0 ··· dm 
 0 ··· 0 
0 ··· 0

where di |di+1 .
Hence, the bases
{v1 , · · · , vn } = {P e1 , · · · , P en }
of ZZn and
{w1 , · · · , wm } = {Qe1 , · · · , Qem }
of ZZm are so that
{β −1 (v1 ), · · · , β −1 (vn )}
is a basis for G and {α(w1 ), · · · , α(wm )} is a basis for H.
Now, note that the matrix identity above implies that AQ(ei ) = P (di ei )
where ei on the left side are in ZZm and those on the right side are in ZZn .
That is, βα(Qei ) = di P (ei ).
So, we have βα(wi ) = di vi , which means that the bases {β −1 (v1 ), · · · , β −1 (vn )}
of G and {α(w1 ), · · · , α(wm )} of H are as asserted in the invariant factor
theorem.
Conversely, let us assume that the invariant factor theorem holds. Consider
any A ∈ Mn,m (ZZ) of rank max(m, n). Without loss of generality, we shall
take m ≤ n for, otherwise, we could take the transpose. Now, A defines a
homomorphism
TA : ZZm → ZZn ; v 7→ Av.
Now the image of TA is a free abelian group generated by the n vectors
Ae1 , · · · , Aem .
Since the matrix A has rank m, the vectors Ae1 , · · · , Aem are linearly in-
dependent vectors over Q. Therefore, they are linearly independent over ZZ
also. In other words, Image TA is free abelian subgroup of ZZn of rank m.
By the invariant factor theorem, let us choose bases {v1 , · · · , vn } of ZZn and
{d1 v1 , · · · , dm vm } of Image TA such that di |d+1 . Call Awi = di vi for all

13
i ≤ m.
Let P ∈ GL(n, ZZ) denote the matrix effecting the change of basis from the
canonical basis to the vi ’s. Similarly, let Q ∈ GL(m, ZZ) be the matrix ef-
fecting the change of basis from the canonical basis to the wi ’s.
Then, P −1 AQ(ei ) = di vi for all i ≤ m. In other words, P −1 AQ has the
form asserted.
The above proof of the invariant factor theorem clearly shows the generation
of GL(n, ZZ) by the elementary matrices.
Exercise 2.5
Prove that SL(n, ZZ) is perfect for n ≥ 3.
Lemma 2.6
For any A ∈ Mm,n (ZZ) define hi (A) to be the GCD of all i × i minors of
A. If A has maximal rank, then for any P ∈ GL(m, ZZ) and Q ∈ GL(n, ZZ),
the numbers hi (A) = hi (P A) = hi (AQ) for all i. The invariant factors of a
matrix A ∈ Mm,n (ZZ) are h1 (A), hh12 (A)
(A) h3 (A)
, h2 (A) , · · · etc.
We know that GL(n, ZZ) is generated by the matrices of the form Xij =
I + Eij ; i 6= j and the matrices diag(±1, · · · , ±1). elsewhere. We shall check
for each r that
hr (AXij ) = hr (Xij A)
for all i 6= j ≤ n.
By the previous lemma, we need to consider only A of the ‘diagonal’ form
with non-zero entries d1 , · · · , dm with di |di+1 .
Therefore, it is clear that hr (AD) = hr (DA) for D = diag(±1, · · · , ±1).
Now, for such A, we have, if i > m that AXij = A and, if i ≤ m, AXij =
A + A0 where A0 is a matrix whose only nonzero entry is di at the (i, j)-th
place.
Clearly, hr (AXij ) = hr (A).
Similarly, we see also that hr (Xij A) = hi (A). Therefore, we have the first
assertion. For the second, we merely note that for ‘diagonal’ matrices A
as above, with di |di+1 , the numbers hi (A) = d1 · · · di . Thus, the invariant
factors are successive quotients of the hi ’s.
Exercise 2.7
If S is any set of generators of the additive group of Q, then S contains a
proper subset of generators. Further, show Q does not have proper maximal
subgroups.

§3 Group actions and permutation groups

14
One can use group actions to prove many of the basic results on groups as
follows.
Cauchy’s & Fermat’s little theorems 3.1
Let G be any group of order n and let p any prime number. Consider the
subset S of G × · · · × G (p times) defined by

S = {(g1 , · · · , gp ) : g1 g2 · · · gp = e}.

Evidently, |S| = np−1 . For each tuple (g1 , · · · , gp ) in S, there are exactly
p distinct tuples (g2 , · · · , gp , g1 ), (g3 , · · · , gp , g1 , g2 ) etc. in S unless g1 =
g2 = · · · = gp (here is where we use the fact that p is prime). Note that
g1 = g2 = · · · = gp if, and only if, g1p = e. Thus, we have |S| ≡ #{g : g p = e}
mod p.
If p|n, then p divides np−1 = |S| ≡ #{g : g p = e} mod p. In this case,
(since ep = e), there are at least p − 1 elements of order p in G. This proves
Cauchy’s theorem.
If p 6 |n, then one has g p = e for some g if, and only if, e = g (n,p) = g. Thus,
|S| ≡ 1 mod p. This proves Fermat’s little theorem.
We have the following strikingly novel proofs of Fermat’s famous theorem
asserting that a prime number of the form 4n + 1 is a sum of two squares.
The proofs are due to Zagier, HeathBrown and Mohan Nair and are variants
of the same argument.
Lemma 3.2
Let p any prime number of the form 4n + 1. Consider the finite set

S = {(x, y, z) ∈ N × N × N : x2 + 4yz = p}.

Define the Zagier map σ : S → S by mapping (x, y, z) to

(x + 2z, z, y − x − z) if x≤y−z ,

(2y − x, y, x + z − y) if y − z < x < 2y ,

(x − 2y, x + z − y, y) if x ≥ 2y .
Then σ is a permutation of order 2 and has a unique fixed point. Hence,
any prime number p of the form 4n + 1 is a sum of two squares.
First, note that in the definition of σ, one could have taken the < sign
wherever ≤ appears; the reason is that x = y − z and x = 2y are impossible
to hold in S.
Now, it is clear that σ has the unique fixed point (1, 1, n). Now, we note

15
the general fact that for any permutation τ of order 2 on a set, the non-
fixed points can be paired off and thus the number of fixed points is of
the same parity as |S|. Applying this to σ, we have that |S| must be odd.
Turning this around and applying the above observation to the permutation
τ : (x, y, z) 7→ (x, z, y), it follows that τ must have an odd number of fixed
points. Therefore, τ does have at least one fixed point. Any fixed point of
τ is a tuple (x, y, y) which means that p = x2 + 4y 2 .
Exercise 3.3
(i) Let p be as before but define
S1 = {(x, y, z) ∈ ZZ × N × N : x2 + 4yz = p}
and S2 = {(x, y, z) ∈ S1 : z > x + y}. Consider the Nair maps
α : S2 → S2 ; (x, y, z) 7→ (−x − 2y, y, z − x − y)
and β : S2 → S2 ; (x, y, z) 7→ (−x, y, z) or (x, z, y) according as to whether
z > y − x or z < y − x.
Then, prove α is an involution with a unique fixed point and draw the
conclusion about p using β.
(ii) Let p, S1 and S2 be as above. Consider the subset S3 = {(x, y, z) ∈ S1 :
z < x + y}. Prove that if no element of S1 is of the form (x, y, y), then all
the elements of S1 can be collected in groups of 4 where exactly 2 are in S2
and two in S3 . Consider the Heathbrown-Nair map
θ : S2 → S2 ; (x, y, z) 7→ (−x − 2y, y, z − x − y)
to conclude that |S2 | is odd and arrive at a contradiction.
Exercise 3.4
(i) For any n, prove that the permutations σ = (1 2) and τ = (1 2 · · · n)
generate the whole of Sn .
Further, if p is a prime, show that any transposition and any p-cycle gener-
ate Sp .
(ii) For general n, and for a transposition σ and any n-cycle τ , find a neces-
sary and sufficient condition for Sn to be generated by σ and τ .
Since any group acts faithfully on its underlying set by left multiplications,
one has Cayley’s theorem asserting the fact that any group is a group of
permutations. More generally, if H is a subgroup of G, then G acts by left
translations on the set of left cosets of H.
Lemma 3.5
If H has some finite index n in G, then H contains a normal subgroup N

16
whose index is a divisor of n!. In particular, if G is a finite group and p
is the smallest prime dividing O(G) and, if H ≤ G has index p, then it is
normal in G. Thus, subgroups of index 2 are normal.
Proof.
Take N to be the kernel of the action. Thus N ≤ H and G/N is isomorphic
to a subgroup of Sym(G/H) = Sn .
Now, ∃N ≤ H ≤ G with N normal in G and [G : N ]|p! Since [G : N ]|O(G)
as well, [G : N ] divides the GCD of O(G) and p! which is p since p is the
smallest prime dividing O(G). But, H contains N , which implies p|[G : N ].
Thus, [G : N ] = p = [G : H].

Group actions are naturally used to prove Sylow’s theorems and their gen-
eralisations as follows.
Theorem 3.6 (Frobenius)
Let G be a finite group and pr be a prime power dividing the order of G.
Then, there exist subgroups of order pr in G and that these are ≡ 1 mod p
in number.
Theorem 3.7 (Snapper)
Let G be a finite group and p be a prime such that pn ||G|. Suppose H ≤ G
has order pm ≤ pn . Then, :
(i) there exist subgroups of order pn containing H and,
(ii) the number of subgroups of order pn which contain H, is ≡ 1 modulo p.
Proof of 3.6
Consider the action by left multiplication of G on the set Ω of all subsets
of G with pr elements. Let us write |G| = pn d with
n ≥ r and p 6 |d.
pn d
The cardinality of Ω is the binomial coefficient pr . It can be shown by
elementary number theory that |Ω| ≡ pn−r d mod pn−r+1 but as we observe
below this also follows from some group-theoretic counting. Break up Ω
into disjoint orbits, say, Ω = ti=1 G.Si where Si ∈ Ω. We claim that for
S
any S ∈ Ω, G = ∪g∈G gS. To see that this claim holds, take any S ∈ Ω
and any s ∈ S. Now, 1 = s−1 s ∈ s−1 S ⊂ G.S. Further, if g ∈ G, then
g = g.1 ∈ g.S. Thus, the claim is true. Hence, in any orbit, the number of
subsets is at least pn−r d and divides |G| = pn d. This means that exactly
one of the two things happen: either an orbit has exactly pn−r d elements
or it has a multiple of pn−r+1 number of elements. Note that |Ω| 6≡ 0 mod
pn−r+1 . Hence, not each orbit can have cardinality a multiple of pn−r+1 .
This already proves that there are orbits with exactly pn−r d elements. Note
that, obviously, the orbit of a set S in Ω has exactly pn−r d elements if, and

17
only if, the corresponding isotropy subgroup has order pr . But, we see that
each orbit with exactly pn−r d elements corresponds to exactly one subgroup
of order pr and conversely. This is because in each such orbit G.A, there is
exactly one group (that gA which contains the identity) since G = ∪g gA,
and so, the orbit is the set of left cosets of that group. Suppose the number of
such orbits with exactly pn−r d elements is t, then |S| ≡ tpn−r d mod n−r+1

p n−r .
pn d
Although, one can prove by elementary number theory that pr ≡ p d
mod pn−r+1 , one could obtain this already from the above sentence as it is
valid for any G of order pn d and applying this to the corresponding cyclic
pn d
group, one gets t = 1. Hence, pr ≡ tp d ≡ pn−r d mod pn−r+1 . Thus,
n−r

t ≡ 1 mod p. This proves Frobenius’s result.

Exercise 3.8
(i) Prove Sylow’s second theorem by using group actions.
(ii) Prove lemma 1.2 using Sylow theorems.
(iii) Consider the alternating group A(N) defined as the set of bijections of
N which move only finitely many natural numbers and move them as even
permutations. Prove that A(N) is simple.
(iv) Prove that any finite group is isomorphic to a subgroup of a finite simple
group.
(v) Show that An is (n − 2)-transitive on {1, 2, · · · , n}.
(vi) Prove that in the infinite group SL(2, F ) for any infinite field F , each
element is expressible as a finite product of elements of finite order.
Exercise 3.9
(i) Prove that any element in An is a commutator xyx−1 y −1 in Sn where x
is an n-cycle.
(ii) In S2n+1 , prove that the cycle (1 2 · · · 2n+1) is expressible as xyx−1 y −1
where x is a n + 1-cycle.
(iii) In any Sn , show that every element is a product of at the most two
cycles.
(iv) Let F be a finite field. Prove that Sym(F ) is generated by the per-
mutations σ : x 7→ x−1 for x 6= 0; σ(0) = 0 and τa,b : x 7→ ax + b for
a, b ∈ F .

Examples 3.10
1. The group of rotations of a cube which leave it invariant are
(I) 90 degree (clockwise or anti-clockwise) rotations about the axes joining
the centres of the opposite faces - there are 6 such;
(II) 180 degree rotations about each of the above axes - there are 3 such;

18
(III) 120 degree (clockwise or anti-clockwise) rotations about the axes join-
ing the opposite vertices - there are 8 such;
(IV) 180 degree rotations about the axes joining the midpoints of the oppo-
site edges - there are 6 such and;
(V) the identity.
2. The cyclic group Cn can be regarded as the group of permutations of the
vertices of a regular n-gon. That is, it is the subgroup of Sn generated by
an n-cycle (1, 2, · · · , n).
3. For n > 2, the dihedral group Dn is defined as the group of rotations of
the regular n-gon given by n rotations about an n-fold axis perpendicular to
the plane of the n-gon and reflections about the n two-fold axes in the plane
of the n-gon like the spokes of a wheel, where the angle between consecutive
spokes is 2π π
n or n according as n is odd or even. It has order 2n.
It can be regarded as a subgroup of Sn as follows. The n rotations corre-
sponding to the powers of σ = (1, 2, · · · , n) and the group Dn is the subgroup

{Id, σ, · · · , σ n−1 , τ, τ σ, · · · , τ σ n−1 }

where τ = (2, n)(3, n − 1) · · · So, the dihedral group D6 is the symmetry

group of the hexagon. One can represent it as the subgroup of S6 generated
by (16)(25)(34) and (123456).

Group actions can also be used to get information on the number an of

subgroups of a given index n in a group G in the following manner.
Proposition 3.11
Let G be any group.
(i) Denote by tn , the number of transitive actions of G on {1, 2, · · · , n}. Then
an = tn /(n − 1)!.
(ii) If hn = |Hom (G, Sn )|, then one has the relation
n−1
X hn−k
an = hn /(n − 1)! − ak .
(n − k)!
k=1

(iii) For the free group Fr , one has Hall’s formula

n−1
X
r−1
an (Fr ) = n(n!) − (n − k)!r−1 ak (Fr ).
k=1

(iv) A d-generated group G has at most n(n!)d−1 subgroups of any index n.

(v) The number of subgroups of index n in ZZ2 is σ(n), the sum of the divisors

19
of n.
(vi) An application of (ii) to ZZ2 yields the well-known partition identity
n−1
X
np(n) = σ(i)p(n − i) + σ(n).
i=1

Proof
(i) If ρ : G → Sn is any transitive representation, then the subset H := {g ∈
G : ρ(g)(1) = 1} is a subgroup of G, of index n. Conversely, if H ≤ G is a
subgroup of index n in G, the set G/H of left cosets has n elements and G
acts transitively on it. There are (n − 1)! ways to identify G/H with the set
{1, 2, · · · , n} where the coset H is identified
 with 1. Thus, an = tn /(n − 1)!.
n−1
(ii) For each 1 ≤ k ≤ n, there are ways to choose the orbit of 1,
k−1
tk transitive actions on it, and hn−k actions on its complement. This proves
the relation
n−1
Xn − 1
hn = t h + tn .
k − 1 k n−k
k=1

Rewriting the relation in terms of the an , one has

n−1
X hn−k
an = hn /(n − 1)! − ak .
(n − k)!
k=1

(iii) is immediate from (ii).

hn
(iv) If G is d-generated, then hn ≤ (n!)d . Hence an ≤ (n−1)! ≤ n(n!)d−1 .
(v) Now, any subgroup H of ZZ2 is of the form gZZ2 for some g ∈ M (2, ZZ).
It is of finite index if g ∈ GL(2, Q). Note that H = gZZ2 = gxZZ2 for
 x ∈GL(2, ZZ). We claim that x can be chosen
any  sothat gx is of the form
a 0 a b
where a > 0, 0 ≤ c < d. Now, if g = , then ∃ u, v ∈ ZZ such
c d  c d
u −b/(a, b)
that au + bv = (a, b). Then, x = ∈ GL(2, ZZ) and gx is of
  v a/(a, b)  
a1 0 ±1 0
the form . By multiplying by matrices like , we may
c1 d1   0 ±1
a1 0
assume that gx is of the form with a1 , d1 > 0. Now, multiplying
c1 d1   
1 0 a1 0
on the right by a matrix of the form , we get where
l 1 c1 + d1 l d1
we may assume that 0 ≤ c1 + d1 l < d1 . Therefore, we have shown that any

20
  
2 a 0
subgroup H of finite index is of the form gZZ where g = ∈ M (2, ZZ),
c d
with a, d > 0 and 0 ≤ c < d. Clearly, the index  of His ad. We claim
2 2 a1 0
finally that gZZ = hZZ for another matrix h = ∈ M (2, ZZ), with
c1 d1
a1 , d1 > 0 and 0 ≤ c1 < d1 if, and only if, g = h. To see this, suppose
g −1 h ∈ GL(2, ZZ). This gives a1 = a, d1 = d and d|(c1 − c). As 0 ≤ c1 , c < d,
we get c1 = c as well. Therefore, the  numberof subgroups of index n is
n/d 0
the number of matrices of the form , with d|n and 0 ≤ c < d.
c d
Thus,P for each divisor d|n, there are exactly d subgroups of index. Therefore,
an = d|n d = σ(n).
(vi) We claim that hn (ZZ2 ) = n!p(n). The reason is as follows. x can be
arbitrarily
P chosen in Sn , P and y chosen in its centraliser CSn (x), so that
hn = x |CSn (x)| = |Sn | 1/|[x]| = |Sn ||[x]|1/|[x]| = |Sn |p(n) = n!p(n).
This yields us the identity
n−1
X
np(n) = σ(i)p(n − i) + σ(n).
i=1

Exercise 3.12
Let G be a group such that the set S of all torsion elements of G is a finite
set. Then, prove that S is a group.

§4 Nilpotent groups

Lemma 4.1
(i) A group G is nilpotent if, and only if, there exists a series

G = G0 ⊇ G1 ⊇ · · · ⊇ Gn = {1}

for some n where each Gi is normal in G and Gi−1 /Gi is contained in the
center of G/Gi .
(ii) The center of any nontrivial nilpotent group is nontrivial.
(iii) All p-groups are nilpotent.
(iv) Subgroups and quotient groups of nilpotent groups are nilpotent.
(v) B(n, Q) is not nilpotent for n ≥ 2 whereas its normal subgroup U (n, Q)
and the quotient group B(n, Q)/U (n, Q) are.
(vi) A group G is nilpotent if, and only if, G/Z(G) is nilpotent where Z(G)
is the center of G.

21
Proof
(i) If G is nilpotent, then by our definition, the lower central series

G ≥ C 1 (G) ≥ C 2 (G) ≥ · · · C n (G) = {1}

for some n. Evidently, C i+1 (G) C C i (G) and C i (G)/C i+1 (G) is contained
in the center of G/C i (G) by the very definition.
Conversely, let
G = G0 ⊇ G1 ⊇ · · · ⊇ Gn = {1}
for some n where each Gi is normal in G and Gi−1 /Gi is contained in the
center of G/Gi .
Note that C 0 (G) = G = G0 . We prove by induction that for any r ≤ n,
C r (G) ≤ Gr . If we assume C m (G) ≤ Gm for m < n, then C m+1 (G) =
[G, C m (G)] ≤ [G, Gm ] ≤ Gm+1 as Gm /Gm+1 is contained in the centre of
G/Gm+1 . Therefore, the inductive proof follows, and shows C n (G) = {1}.
(ii) The penultimate term C n−1 (G) in the lower central series

G ≥ C 1 (G) ≥ C 2 (G) ≥ · · · C n (G) = {1}

is contained in the center.

(iii) Note first that any p-group G has a nontrivial centre Z (this follows
by using the conjugation action of the group on itself). Applying induction,
we may assume that G̃ = G/Z is nilpotent. Evidently, C i (G) ≤ C̃ i for all
i where C i (G/Z) = C̃i /Z. If C n (G̃) = 1, then we have C n (G) ≤ Z; so
C n+1 (G) = 1. Thus, G is nilpotent.
(iv) It is clear that for any H ≤ G, we have C r (H) ≤ C r (G) for every r, by
induction. Thus, subgroups of nilpotent groups are also nilpotent.
Let N C G and let C n (G) = {1}. Consider the series

G/N ≥ C 1 (G)N/N ≥ · · · C n (G)N/N = {1}.

Clearly, each C i (G)N/N C G/N and let us check that any successive factor
C i (G)N/N
C i+1 (G)N/N
is contained in the centre of C i+1G/N
(G)N/N
. Now, [G/N, C i (G)N/N ] =
[G, C i (G)]N/N ≤ C i+1 (G)N/N . Thus, the above series shows that G/N is
nilpotent by (i).
(v) Call B = B(n, Q) and U = U (n, Q) for simplicity.
An easy matrix computation shows that [U, U ] ≤ U and that [U, U ] consists
of matrices which have the entries (1, 2), (2, 3), · · · , (n − 1, n) to be zero.
By induction, one can easily show that C r (U ) consists of matrices whose
(i, j)-th entries for i < j ≤ i + r − 1 are all zero. Hence, C n (U ) = {1}.

22
Clearly, B = T U with T ∩ U = {1} where T ≤ B is the subgroup of diagonal
matrices. Also, U = Ker(det : B → Q∗ ) is a normal subgroup. Since T is
abelian, it is nilpotent.
However, B is not nilpotent since [B, B] = U and [B, U ] = U by the same
computation as above.
(vi) We already know that quotient groups of nilpotent groups are nilpotent.
So, we assume that G/Z(G) is nilpotent and show that G must be nilpotent.
Using (i) for G/Z(G), we have normal subgroups Gi of G containing Z(G)
such that

G/Z(G) = G0 /Z(G) ⊇ G1 /Z(G) · · · ⊇ Gn /Z(G) = {1}

where Gi−1 /Gi is contained in the center of G/Gi . Consider the series

G = G0 ⊇ G1 · · · ⊇ Gn = Z(G) ⊇ Gn+1 = {1}.

Evidently, Gn /Gn+1 is contained in (in fact, equal to) the center of G/Gn+1 .
Thus, G is nilpotent.
Here is a general result interesting in its own right and to be used in the
next proof.
Prime factorisation in groups 4.2
Let G be any finite group and let g ∈ G. Write O(g) = pk11 · · · pkr r where
pi ’s are distinct primes. Then, prove that every element g ∈ G can be
uniquely expressed as a product g = g1 · · · gr where gi ’s commute pairwise
and O(gi ) = pki i for i = 1, · · · , r.
Proof.
Let n1 = O(g) k1
k1 . Then, since p1 6 |n1 , we may write ap1 + bn1 = 1 for some
p1
integers a, b.
k1
Take g1 = g bn1 and g2 = g ap1 .
Of course, g1 g2 = g2 g1 = g. We shall check that g1 , g2 have orders pk11 and
n1 respectively.
k1 k1
p
Now, g1 1 = g bn1 p1 = g bO(g) = 1; so O(g1 )|pk11 .
Moreover, if g1d = 1, then we get g bn1 d = 1; so O(g)|bn1 d. This gives pk11 |bd;
however (p1 , b) = 1 as seen from apk11 + bn1 = 1. Thus, pk11 |d which proves
that O(g1 ) = pk11 .
k1 k1
Similarly, g2n1 = g ap1 n1 = 1 so that n1 |O(g2 ). If g2l = 1, then g ap1 l = 1 so
that O(g) = pk11 n1 |apk11 l.
Hence n1 |l as (n1 , a) = 1. This implies that O(g2 ) = n1 .
Now, we may proceed by induction with g replaced by g2 . Ultimately, we

23
will have elements xi , i ≤ r which are powers of g and have orders pki i such
that g = x1 · · · xr .
Finally, we show that such an expression is unique. If g = y1 · · · yr is another
expression where yi has order pki i and commute pairwise, then they commute
with g. As xi ’s are powers of g, the xi ’s and the yj ’s all commute pairrwise.
So,
x−1 −1 −1
1 y1 = x2 · · · xr y2 · · · yr

has order dividing pk11 as seen from the left side and at the same time has
order dividing pk22 · · · pkr r as seen from the right side. Thus, x1 = y1 . Pro-
ceeding by induction, we get xi = yi for all i.
Lemma 4.3
For a finite group G, the following are equivalent:
(i) G is nilpotent,
(ii) every subgroup is subnormal,
(iii) every proper subgroup H is properly contained in NG (H),
(iv) all maximal subgroups are normal,
(v) all p-Sylow subgroups are normal,
(vi) elements of coprime order commute and,
(vii) G is the direct product of its Sylow subgroups.
Proof.
Let G be nilpotent and H ≤ G. Suppose

G ≥ C 1 ≥ C 2 ≥ · · · C n = {1}

be its lower central series. Since, we have that C i /C i+1 is contained in the
centre of G/C i+1 , it follows that

G = GH ≥ C 1 H ≥ · · · C n H = H

is a chain of subgroups where each term is normal in the previous one. Thus,
H is subnormal in G; that is, (i) implies (ii).
Now, suppose (ii) holds. Let H < G. We have a series

H = H0 C H1 C · · · Hr = G.

If i is the smallest number for which Hi > H, it follows that Hi ≤ NG (H)

and Hi 6≤ H. This proves (ii) implies (iii).
For a maximal subgroup, M 6= NG (M ) shows that NG (M ) = G i.e., M is
normal. Thus (iii) implies (iv).
Assume (iv). If P is a p-Sylow subgroup which is not normal, then NG (P ) ≤

24
M for some maximal subgroup M . Since M is normal, we get for any g ∈ G,
gP g −1 is again a p-Sylow subgroup of gM g −1 = M . Thus, by Sylow’s second
theorem applied to M , there is some m ∈ M with gP g −1 = mP m−1 i.e.,
m−1 g ∈ NG (P ) ≤ M . Hence g ∈ M , a contradiction, since a maximal
subgroup, by definition, is a proper subgroup. Hence (iv) implies (v).
Assume (v) holds; that is, the Sylow subgroups are normal. We shall show
first that elements of orders powers of different primes commute. Let O(x) =
pr and O(y) = q s where p 6= q are primes. If P, Q are the unique p-Sylow
subgroup and the unique q-Sylow subgroup, then x ∈ P and y ∈ Q. As
xyx−1 y −1 ∈ P ∩ Q = {1}, we have xy = yx.
Now, we deal with the general case. We saw in 4.2 that in any finite group,
every element can be written as commuting elements of prime power orders
dividing the order of the element. Let g, h ∈ G be elements in our group
which have coprime orders m, n. Then, g = x1 · · · xr , h = y1 · · · , ys where
xi ’s commute among themselves, yj ’s commute among themselves and each
of them has prime power order. Also O(xi )|O(g) = m and O(yj )|O(h) = n
which are coprime. Thus, each xi and each yj commute. Hence gh = hg.
Thus, (v) implies (vi).
Suppose now that elements of coprime orders commute. We write

|G| = pk11 · · · pkr r .

Let Pi be any pi -Sylow subgroup of G ; 1 ≤ i ≤ r. Then, since [Pi , Pj ] = {1}

for i 6= j, the product P1 · · · Pr is a group. As the orders of Pi ’s are pairwise
coprime to each other,

|P1 · · · Pr | = |P1 | · · · |Pr | = |G|.

Hence G = P1 · · · Pr and Pi ∩ j6=i Pj = {1}, which means that G ∼

Q
= P1 ×
· · · × Pr .
So, we have proved that (vi) implies (vii).
Finally, since p-groups are nilpotent, (vii) clearly implies that P1 × · · · Pr is
nilpotent; that is, (i) follows.
The Frattini subgroup Φ(G) of a finite group G is defined to be the intersec-
tion of all (proper) maximal subgroups.
Lemma 4.4
(i) Φ(G) is the set of ‘nongenerators’ of G i.e., those elements which can be
dropped from any generating set for G.
(ii) Φ(G) is a characteristic subgroup.
(iii) Φ(G) is nilpotent.

25
(iv) For any p-group G, Φ(G) = [G, G] < Gp >.
(v) For a finite abelian p-group A, A/Φ(A) is an elementary abelian group
of order pd(A) where d(A) is the minimal number of generators needed to
generate A.
Proof
(i) Let G =< S > and s ∈ S ∩ Φ(G). It suffices to show that s ∈< S \ {s} >.
Now, if H =< S \ {s} >< G, then there is some maximal subgroup M of G
containing H. But then S \ {s} is contained in M as well as s ∈ M . This
means that G =< S >≤ M , which is a contradiction. Thus, we have shown
that elements of Φ(G) can be dropped from any generating set for G.
Conversely, suppose g ∈ G be such that whenever < S ∪ {g} >= G, we have
< S >= G. Let, if possible, M be a maximal subgroup not containing g.
Then, < M ∪ {g} >= G. By the hypothesis, this gives < M >= M = G, a
contradiction. Hence all maximal subgroups contain g and (i) is proved.
(ii) Since Φ(G) is the intersection of all maximal subgroups of G, and since
any automorphism permutes maximal subgroups of G, it follows that Φ(G)
is a characteristic subgroup of G.
(iii) By the previous problem, it suffices to prove that the p-Sylow subgroups
of Φ(G), for any p, are normal in it. By the Frattini argument (problem 15
(ii)), we have G = Φ(G)NG (P ).
In particular, NG (P ) ∪ Φ(G) generate G which implies that NG (P ) =<
NG (P ) >= G; that is, P C G.
In particular, P E Φ(G).
(iv) Let M1 · · · , Mr be the (proper) maximal subgroups of G. Consider
their images in the elementary abelian group G/[G, G] < Gp >. They are
maximal subgroups and every maximal subgroup of G/[G, G] < Gp > is
the image of one of these. Of course, two different Mi ’s may have the same
image. Note that the intersection of all maximal subgroups in an elementary
abelian p-group is trivial. This is so because an elementary abelian p-group
is isomorphic to ZZ/p × · · · ZZ/p, and in this group, the subproducts with one
factor being the trivial groups are maximal subgroups and intersect in the
identity. Therefore,
r
\
Φ(G) := Mi ≤ [G, G] < Gp > .
i=1

Conversely, note that in p-groups, all maximal subgroups are of index p.

This means that in our p-group G, < Gp >≤ Φ(G). Also, since maximal
subgroups M in G are necessarily normal, G/M is an abelian group (of order

26
p) and so, [G, G] ≤ M . Thus, [G, G] ≤ Φ(G) as well. This proves that

Φ(G) = [G, G] < Gp > .

(v) By (iv), it follows that the Frattini subgroup of an abelian p-group A

is simply Ap . Therefore, clearly A/Φ(A) = A/Ap is an elementary abelian
p-group. If A/Ap has order pr , then it can be generated by r elements
x1 Ap , · · · , xr Ap say. Therefore, A =< Ap , x1 , · · · , xr >.
However, Ap is the set of nongenerators and, therefore, A =< x1 , · · · , xr >
which implies that r ≥ d(A). However, it is obvious that for any abelian
p-group A, the elementary abelian p-group A/Ap is a direct sum of at the
most d(A) copies of the group of order p; so |A/Ap | = pr ≤ pd(A) . Therefore,
we have d(A) = r.
Proposition 4.5
Any subgroup of a finitely generated nilpotent group is also finitely gener-
ated. Therefore, a finitely generated nilpotent torsion group is finite.
Proof
Let S be a finite set of generators of a nilpotent group G. Consider the
sets S0 = S, Si+1 = {aba−1 b−1 : a ∈ S, b ∈ Si } for all i ≥ 0. Of course,
each Si is a finite set. Since G is nilpotent, there is some n so that Sr = 1
for all r ≥ n. For each i ≥ 0, consider the subgroup Gi of G generated
by the set ∪r≥i Sr . As Sr = 1 for r ≥ n, each Gi is finitely generated. It
is clear that [G, Gi ] = Gi+1 since S, Si , Si+1 generate G, Gi and Gi+1 and
Si+1 = {aba−1 b−1 : a ∈ S, b ∈ Si }. In particular, each Gi is normal in
G and the subgroup Gi /Gi+1 is central in G/Gi+1 . Now, if H is any sub-
group of G, then (H ∩ Gi )/(H ∩ Gi+1 ), being a subgroup of the finitely
generated abelian group Gi /Gi+1 , is finitely generated. Since H ∩ Gn = 1,
this gives inductively that each H ∩ Gi is finitely generated. In particular,
H = H ∩ G0 = H ∩ G is finitely generated.
To prove the second assertion, look at the lower central series of G. As
C n (G)/C n+1 (G) is a subgroup of the finitely generated, nilpotent group
G/C n+1 (G), this subgroup is finitely generated as well. But, this is an
abelian, torsion group and is, hence, finite. Thus, G itself is finite.

The existence of fixed-point free automorphisms on finite groups have very

interesting implications on the abelianness or nipotency of a group as the
following proposition shows.
Proposition 4.6
Let T be an automorphism of prime order p of a finite group G such that
T (x) = x if, and only if, x = 1. Then, we have :

27
(i) The function F (g) := g −1 T (g) is a bijection on G.
(ii) For any g in G, the product g T (g) T 2 (g) . . . T p−1 (g) is the identity.
(iii) |G| is congruent to 1 modulo p.
(iv) For any prime q dividing the order of G, there is a q-Sylow subgroup Q
of G such that T (Q) = Q.
(v) For any prime q, prove there is a unique q-Sylow subgroup Q of G such
that T (Q) = Q.
(vi) Let q be a prime. Then, the q-Sylow subgroup Q fixed by T contains
any q-subgroup of G fixed by T .
(vii) For p = 2, G is abelian.
(viii) For p = 3, G is nilpotent.
The analogue of (viii) was proved for any prime p by the Fields medalist
J.Thompson in his thesis.
Proof
(i) Now, x−1 T (x) = y −1 T (y) for some x, y if, and only if T fixes yx−1 ; that
is, when x = y. Thus, the map x 7→ x−1 T (x) is 1-1. Being a map of finite
sets, this is onto as well.
(ii) From (i), an arbitrary element of g can be written as g = x−1 T (x) so
that we get
gT (g) · · · T (g)p−1 = 1.
(vii) Taking p = 2, this gives T (g) = g −1 for all g. As T is an automorphism,
G is abelian.
(iii) Consider the subgroup Z of Aut(G) generated by T . This is a cyclic
group of order p acting on G. Each orbit has cardinality dividing p and,
as T fixes only the identity element, each orbit other than that of 1 has p
elements. So, |G| has 1 mod p number of elements.
(iv) Look at the action of Z on the set S of all q-Sylow subgroups of G.
Again, the orbits which are not fixed points under Z, have order p. So, if
there is no fixed point in S, then the number of elements in S would be
a multiple of p. But, we know that the number of elements in S (i.e. the
number of q-Sylow subgroups) is a divisor of |G|. Thus, p would divide |G|,
a contradiction of (iii).
(v) Note that if T (Q) = Q, then T (N ) = N where N is the normaliser of
Q in G. Now, if T fixes another q-Sylow subgroup gQg −1 , then rewriting
T (gQg −1 = gQg −1 , we have g −1 T (g) belongs to N . But, applying (i) to
N itself, any element of N is of the form x−1 T (x) for some x in N . So,
g −1 T (g) = x−1 T (x), which gives g = x belongs to N . So, gQg −1 = Q.
(vi) Let R be any q-group in G which is fixed by T . Let S be a q-group which
contains R and is maximal with respect to this property (i.e., S is fixed by

28
T , contains R and is not contained in a strictly larger q-group which is T -
fixed). We claim that R = Q. For this, first look at the normaliser N (S)
of S in G. Since T (S) = S, we have also T (N (S)) = N (S) clearly. by
(v) applied to the subgroup M of N (S) which is T -fixed. Now, S is a q-
subgroup of N (S), say, yM y −1 where y is in N (S). But then S = y −1 Sy
is contained in M . By maximality property of S, we get S = M i.e., S is
a q-Sylow subgroup of N (S). But, any Sylow subgroup of a subgroup in
any group is the intersection of the subgroup with a Sylow subgroup of the
big group. So, we have a q-Sylow subgroup L of G such that S = N (S)
intersection L. In other words, S = NL (S), the normaliser of S in L. But,
in a q-group (indeed, in any nilpotent group), the normaliser of a proper
subgroup contains the subgroup properly. So, we have S = L. In other
words, S is a q-Sylow subgroup of G itself. By uniqueness of the T -fixed
q-Sylow subgroup of G itself. Thus, we have shown that R is contained in
S = P.
(viii) Finally, we prove the result for p = 3.
Now x T (x) T 2 (x) = e for any x in G. So, T (x−1 x−1 = (x T (x))−1 = T 2 (x)
from the above. Putting y = x−1 , we get T (y)y = T 2 (y −1 ) = (T 2 (y))−1 .
Thus, both yT (y) and T (y)y are equal to (T 2 (y))−1 . In other words, every
element x in G commutes with the element T (x). Similarly, x commutes
with T 2 (x) also.
Now, to prove the result, let us consider any prime q and the unique q-Sylow
subgroup Q of G which is T -fixed. Let g be any element of G which has
q-power order. Consider the subgroup Q0 is generated by the three elements
g, T (g) and T 2 (g). Clearly, Q0 is T -fixed. Also, since g, T (g), T 2 (g) commute
among themselves, Q0 is a q-group. Therefore, by (vi), Q0 is contained in Q.
Thus, Q contains all elements of G of q-power order. Hence, Q is the unique
q-Sylow subgroup of G. So, it is normal. This proves the result for p = 3.

§5 Solvable groups

Lemma 5.1
(i) A group G is solvable if, and only if, there exists a series

G = G0 ⊇ G1 ⊇ · · · ⊇ Gn = {1}

for some n where Gi are normal subgroups of G and Gi /Gi+1 is abelian.

(ii) Subgroups and quotient groups of solvable groups are solvable.
(iii) If N is a solvable, normal subgroup of a group G such that G/N is
solvable, then prove that G is solvable as well.

29
Proof
(i) Evidently, the derived series is a series as asserted and, so any solvable
group has such a series.
Conversely, suppose G admits a finite series as above. We prove by induction
that Dr (G) ≤ Gr for all r. That would prove Dn (G) = {1}. The assertion
holds for r = 0. Supposing Dr (G) ≤ Gr , we have

Dr+1 (G) = [Dr (G), Dr (G)] ≤ [Gr , Gr ] ≤ Gr+1

as Gr /Gr+1 is abelian. This proves (i).

(ii) & (iii) Suppose G is solvable. Start with a series for G as in (i). The
subgroups Hr = Gr N/N form a series for G/N as in (i), and since Hn = {1},
it follows that G/N is solvable. Of course, any subgroup K of G must be
solvable as the terms of its derived series are contained in the corresponding
term of the derived series for G.
Conversely, suppose N E G and G/N are solvable. Let

N = N0 ⊇ N1 ⊇ · · · ⊇ Nr = {1}

and
G/N = G0 /N ⊇ G1 /N ⊇ · · · ⊇ Gs /N = {1}
be series for N and G/N as in (i). But then

G = G0 ⊇ · · · ⊇ Gs = N ⊇ N1 · · · Nr = {1}

is a series for G as in (i). Thus, G is solvable.

Remark (Odd order and Burnside theorems) 5.2 :
One of the most famous, beautiful theorems in finite group theory is the
statement that every group of odd order is solvable. The proof by Walter
Feit and John Thompson occupies a whole volume of the Pacific Journal of
Mathematics apart from using many results proved by others earlier.
A theorem of Burnside asserts that any group of order pr q s for primes p, q
is solvable.
Exercise 5.3
(i) Let G be a finite group such that all its proper subgroups are abelian.
Then, G must be solvable.
(ii) Let G be a finite group such that all its proper subgroups are nilpotent.
Then, again G must be solvable.

§6 Matrix groups

30
As we pointed out earlier, the usefulness of a group in a situation depends
on its avataar. A group is especially useful when it appears as a subgroup
of GL(n, CI ).
Lemma 6.1
SU (2) ∼
= H 1 , the group of unit quaternions; that is, the quaternions a + bi +
cj + dk which satisfy a2 + b2 + c2 + d2 = 1.
Further, there exists a homomorphism from SU (2) to SO(3) whose kernel
is {±I2 }.
Proof.  
a + ib c + id
Any element of SU (2) looks like with a, b, c, d ∈ R such
−c + id a − ib
that a2 + b2 + c2 + d2 = 1.
This reminds us of Hamilton’s quaternions. Therefore, let us define the map
 
a + ib c + id
θ : SU (2) → H ; 7→ a + ib + cj + dk.
−c + id a − ib

Note that the image is contained in the group H 1 of ‘unit’ quaternions; that
is, a + ib + cj + dk for which a2 + b2 + c2 + d2 = 1 or, equivalently,

(a + ib + cj + dk)−1 = a − bi − cj − dk.

It is trivial to check that θ is a homomorphism. Also, clearly it is 1-1 as well

as onto H 1 . Therefore, SU (2) ∼ = H 1.
Now, as the group of nonzero elements H ∗ acts on the 3-dimensional real
vector space V generated by i, j, k by conjugation, we may think of SU (2)
as acting on this space. Thus, we have a homomorphism

ρ : SU (2) → GL(V ).

To explicitly evaluate
 it, we usethe isomorphism θ to view SU (2) as H 1 .
a + ib c + id
For any g = ∈ SU (2), we may compute qiq −1 , qjq −1
−c + id a − ib
and qkq −1 where q = θ(g)a + bi + cj + dk. We arrive at the following matrix
with respect to the ordered basis {i, j, k} :
 2
a + b2 − c2 − d2

2(bc − ad) 2(ac + bd)
ρ(g) =  2(ad + bc) a2 − b2 + c2 − d2 2(cd − ab) .
2(bd − ac) 2(ab + cd) 2 2
a −b −c +d 2 2

Now, since g −1 = g t is obtained by changing b, c, d to theire negatives, it is

clear that ρ(g)−1 = ρ(g)t ; that is, ρ(g) ∈ O(3, R).

31
Note that if g ∈ Kerρ, then the corresponding quaternion q commutes with
i, j, k and, hence, with the whole of H.
In particular, this gives g ∈ Z(SU (2)) = {±I}.
Of course, it is clear that −I does indeed belong to Ker ρ.
The final assertion left is to show that the determinant of ρ(g) is always 1
is proved by direct (but messy) calculation. A better way would be to use a
little bit of topology which shows that SU (2) is connected, ρ on SU (2) and
the determinant function on GL(3, R) are continuous, that the image of det
◦ρ is a connected subset of R contained in {±1} and containing I.
Bruhat lemma 6.2
For any field K, GL(n, K) is the disjoint union of double cosets BwB over
permutation matrices w. Here B := B(n, K), the group of invertible upper
triangular matrices with entries from K.
Minkowski lemma and more 6.3
(i) For any odd prime number p, an n×n integral matrix A 6= Id with entries
congruent to the corresponding entries of the identity matrix modulo p is
of infinite order. In other words, the ‘principal congruence subgroup’ Ker
(GL(n, ZZ) → GL(n, ZZ/p) is torsion-free for an odd prime p.
(ii) For p = 2, any matrix A in Ker (GL(n, ZZ) → GL(n, ZZ/p)) of finite
order, has order 1 or 2 and is expressible as

M.diag(1, 1, · · · , 1, −1, −1, · · · , −1).M −1

for some M ∈ GL(n, ZZ).

Proof of (i).
Write A = I + pB, with B ∈ M (n, ZZ). The characteristic polynomial of A
is χA (t) = det (tI − A) = det (t − 1)I − pB). Therefore, χA (α) = 0 if, and
only if, χB ( α−1
p ) = 0. Now, if A has finite order, its eigenvalues are all roots
of unity. Since |α| = 1, we have

α−1 |α| + 1 2
| |≤ ≤ < 1.
p p p

Thus, all the roots of χB (t) have absolute values < 1. Since χB (t) is an
integral polynomial whose top coefficient is 1, this implies that χB (t) = tn ;
that is, all the eigenvalues of B are zero. In other words, all eigenvalues of
A are 1. As A has finite order, it must be the identity matrix.
Proposition 6.4    
0 −1 1 1
(i) The matrices S = and X = generate SL(2, ZZ).
1 0 0 1

32
(ii) P SL(2, ZZ) is isomorphic to the free product ZZ/2 ∗ ZZ/3.
(iii) SL(2, ZZ) =< x, y|x2 y −3 , x4 >.
(iv) The abelianisation of SL(2, ZZ) is the cyclic group of order 12.
Proof
(i) We shall prove, equivalently, −1 −1
  that S and A := S X generate SL(2, ZZ).
1 0
Note that AS = Y := .
1 1 
a b
Now, start with any g = ∈ SL(2, ZZ). We shall show that left
c d
and right mutliplications by powers of X and Y lead to ±I by the usual
Euclidean division algorithm.  
l a + lc b + ld
For any integer l, we have X g = . This shows us that one
c d
can divide a by c and replace a by its residue mod c.
Similarly, one can see that by left multiplication by some Y l , one can reduce
c mod a. Repeating these finitely many times, the division algorithm implies
that one of a and c becomes zero; the other has to be ±1 as the determinant
is always 1.    
0 ±1 ±1 b
So, g becomes g1 = or g2 = .
∓1 d 0 ±1
Now,  
±d 0 ±1
g1 X = = S ∓1
∓1 0
and g2 = X b or −X −b .
Since −I = S 2 , the assertion follows.
(ii) Since S 2 = −I also represents the identity element in P SL(2, ZZ), the
image s of S has order 2 in P SL(2,ZZ). 
0 −1
Also, the image b of B := SX = in P SL(2, ZZ) has order 3 as
1 1
(SX)3 = −I.
We know that the elements s, b generate the whole group; so, we need only
show that no matrix
SB a1 SB a2 · · · SB ar
with each ai either 1 or 2, can be the matrices I, −I.
Since SB = −X and SB 2 = Y , it follows  that
 any word in the positive
2 a b
powers of SB and SB is a matrix g = in which a, b, c, d are of the
c d
same sign. Therefore, if b 6= 0, then the corresponding entry −b − d of SBg
and b of SB 2 g are non-zero as well. Similarly, if c 6= 0, the corresponding
entries of SBg and SB 2 g are nonzero. Since SB and SB 2 have the property

33
that either the (1, 2)-th entry or the (2, 1)-th entry is non-zero, any word g
in their positive powers has this property; hence g can not be the identity
matrix.
(iii) follows from (ii) by sending x to S and y to SX.
(iv) Now,

SL(2, ZZ)ab =< x, y|x2 y −3 , x4 , xyx−1 y −1 >=< e, f |2e − 3f, ef − f e, 4e >

∼
= (ZZe ⊗ ZZf )/ < 2e − 3f, 4e > .
The invariant factors of thesubgroup
 < 2e−3f, 4e > above are the invariant
2 −3
factors of the matrix A = . The latter is computed by computing
4 0
h1 (A) = 1 = d1 and h2 (A) = 12 = d1 d2 . Clearly, d1 = 1 and d2 = 12.
Therefore, the abelianisation of SL(2, ZZ) is the cyclic group of order 12.

§7 Miscellaneous results

Lemma 7.1
The groups C I ∗ and S 1 are isomorphic.
Proof.
First of all, the polar co-ordinates provide an isomorphism C∗ ∼
= R>0 ×S 1 ∼
=
1 1
R × S . So, it suffices to show that R × S = S .∼ 1

Start with any Q-vector space basis B of R which contains 1. Already, we

have used Zorn’s lemma here when trying to extend the linearly independent
set {1} to a basis. Then, the set
[
C := (B × {0}) ({0} × B)

is a basis of the Q-vector space R × R. Once again, it is a consequence of

Zorn’s lemma that B and C are in bijection. Let θ : C → B be a bijection
which maps (0, 1) to 1. Then, there is a unique extension of θ to a Q-vector
space isomorphism from R × R onto R. Since θ(0, 1) = 1 and is a Q-vector
space isomorphism, it gives an isomorphism of {0} × ZZ onto ZZ. Therefore,
θ gives
(R × R)/({0} × ZZ) ∼= R/ZZ.
Evidently, the left hand side above is isomorphic to R/{0} × R/ZZ. This
completes the proof, since t 7→ e2iπt gives an isomorphism from R/ZZ onto
S1.

34
Ore’s covering lemma 7.2
Let H, K ≤ G be subgroups of the same finite index. Then, there exist
g1 , · · · , gn such that
G = tni=1 gi H = tni=1 Kgi .
Even the case H = K is interesting.
Proof.
Writing G as a union of the double cosets KgH, it is necessary and sufficient
to write each double coset in the form asserted since each double coset KgH
is a union of right cosets of K in G and also a union of left cosets of H in
G.
We note the two bijections :

KgH/H → K/(K ∩ gHg −1 );

kgH 7→ k(K ∩ gHg −1 )

K\KgH → (H ∩ g −1 Kg)\H;
Kgh 7→ (H ∩ g −1 Kg)h.
Here, the notation is explained as follows. If B is a subgroup of a group
A and if S is a subset of A which is a union of left cosets of B in A, then
one has written S/B to denote the set of left cosets of B contained in S.
Similarly, if T is a subset of A which is a union of right cosets of B, then
one has written B\T to denote the set of right cosets of B contained in T .
It is easy to verify that the above are bijections. Note that the right sides of
these bijections count cosets of subgroups in groups (and not merely sets).
If there is a bijection,

K/(K ∩ gHg −1 ) → (H ∩ g −1 Kg)\H

say
ki (K ∩ gHg −1 ) 7→ (H ∩ g −1 Kg)hi
where i = 1, · · · , r, then we see that

KgH = tri=1 ki ghi H = tri=1 Kki ghi .

Therefore, we only have to establish a bijection between K/(K ∩ gHg −1 )

and (H ∩ g −1 Kg)\H.
Now, we note that K/(K ∩gHg −1 ) is in bijection with g −1 Kg/(g −1 Kg ∩H).
Now, since H, K have the same finite index in G, so do the subgroups H

35
and g −1 Kg of G. Thus, it suffices to show that the subgroup g −1 Kg ∩ H
has finite index. But, this is true since both H and g −1 Kg do.
Frobenius’s theorem 7.3
Let G be a finite group and, for any natural number n, denote by fn (G) the
cardinality of the set {x ∈ G : xn = 1}. Then, :
(i) The number of elements of order n is a multiple of φ(n). In particular,
for a prime p such that pk k|G|, one has fpk (G) − fpr (G) ≡ 0 mod pr for all
r < k. More particularly, if fpk (G) ≡ 0 mod pk , then fpr (G) ≡ 0 mod pr for
all r < k.
(ii) Let pk kn and let Q be a set of representatives of conjugacy classes of ele-
k
ments y such that y |G|/p = 1. Then, fn (G) = y∈Q [G : CG (y)]fpk (CG (y)).
P

(iii) If pk k|G|, then fpk ≡ 0 mod pk .

(iv) If n/|G|, then fn (G) ≡ 0 mod n.
Proof
(i) The relation x ∼ y ⇔ hxi = hyi is an equivalence relation. The equiv-
alence class of x looks like {xi /(i, O(x)) = 1}; it has φ(O(x)) elements.
Writing the set of elements of order n as a union of equivalence classes (of
elements of order n), the assertion follows. Further, if pk /|G| and r < k,
then fpk (G) − fpr (G) is the number of elements of orders pr+1 , . . . , pk in
G. As φ(pr+1 ) = pr (p − 1), φ(pr+2 ) = pr+1 (p − 1) etc., we have that the
number of elements of orders pr+1 , . . . , pk is certainly a multiple of pr . The
last assertion is obvious.
(ii) By 4.2, we have each g = xy uniquely where xy = yx and O(x) is a
power of p and (O(y), p) = 1. Thus, g 7→ (x, y) gives a bijection between G
and ordered pairs (x, y) for which (O(y), p) = 1 and x ∈ CG (y) has p-power
n
k
order. So, g n = 1P⇔ xp = 1 = y pk .
Hence fn (G) = fpk (CG (y)) is constant where y varies through its conju-
y∈G
k
this conjugacy class has [G : CG (y)] elements, y n/p = 1
gacy class, and as P
we have fn (G) = [G : CG (y)]fpk (CG (y)).
y∈Q
(iii) The proof is by induction on |G|. It is vacuously true for G = {1}.
Assume for all proper subgroups H that fpr (H) ≡ 0 mod pr where pr k|H|.
Then, by (i), fpl (H) ≡ 0 mod pl if pl /|H|.
Now, apply (ii) with n = |G|. We get
X
|G| = fn (G) = [G : CG (y)]fpk (CG (y))
y∈Q
X X
= fpk (CG (y)) + [G : CG (y)]fpk (CG (y))
y∈Q∩Z(G) y∈Q\Z(G)

36
 X
= |Q ∩ Z(G)|fpk (G) + [G : CG (y)]fpk (CG (y)).
y∈Q\Z(G)

Look at any of the proper subgroups CG (y) for y ∈ Q\Z(G). If p` k|CG (y)|,
with ` ≤ k, then [G : CG (y)] ≡ 0 mod pk−` and, p` |fp` (CG (y)) by induction
hypothesis.
Note fp` (CG (y)) = fpk (CG (y)). Therefore each term in the sum above is a
multiple of pk . As pk ||G| we have pk ||Q ∩ Z(G)|fpk (G). But p 6 ||Q ∩ Z(G)|
for, if it did then it would have an element y of order p which would also
n
satisfy y pk = 1. This is a contradiction, as ( pnk , p) = 1. Therefore pk |fpk (G).
(iv) Let n||G|. We write = pk11 . . . pkr r for distinct primes pi . Then, it suffices
to show that pki i /fn (G). Now, by (ii), fn (G) is expressible as a sum of terms
of the form [G : H]fpki (H) for subgroups H. By (iii), each such H (including
i
G itself) satisfies pri /fpri (H) where pri k|H|. Therefore pki i = pki i −r . pri divides
[G : H]fpri (H) ∀ H. Thus pki i divides the sum which is fn (G).

Examples of presentations 7.4

1. ZZ = hx | φi; ZZn = hx | xn i.
2. ZZ ⊕ ZZ = hx, y | [x, y]i.
3. ZZn = hx1 , . . . , xn | {[xi , xj ] : 1 ≤ i < j ≤ n}i
Note that n is the minimal number of elements needed to generate ZZn .
Moreover, there does not exist a presentation with less than n(n − 1)/2
relations.
4. Every group has a presentation. For example,

G = hG | x.y.(xy)−1 : x, y ∈ Gi.

5. hx, y | x2 y 3 , x3 y 4 i is a presentation for the trivial group.

6. The symmetric group S3 of degree 3 has presentation

S3 = hr, s | r3 , s2 , srsri.

7. Let Dn , n > 1 be the symmetry group of the regular n-gon Pn . This

group is generated by the rotation r with angle 2π/n and a reflection s in
the line through the centre and one of the vertices.

Dn = hr, s | rn , s2 , (sr)2 i.
8. Let D∞ be the infinite dihedral group consisting of the
motions of IR which map the integers to integers, i.e. the transformations
IR → IR, x 7→ ±x + k, with k ∈ ZZ.

37
 D∞ = hs, t | s2 , ststi.

I = h{xn : n ≥ 1} | {xn = xknk : n, k ≥ 1}i

9. Q
10. Let n ≥ 2, X = {x1 , . . . , xn−1 } and R = {x2i , (xi xi+1 )3 f or 1 ≤ i ≤
n − 2, [xi , xj ] f or |j − i| > 1}. Then, the symmetric group Sn = hX | Ri.
Proof: Consider ϕ : X −→ Sn , xi 7→ σi = (i, i + 1), 1 ≤ i ≤ n − 1
and observe that Sn is a homomorphic image of G = hX | Ri. Consider
H = hx2 , . . . , xn i and its cosets

K1 = H, K2 = K1 x1 , K3 = K2 x2 , . . . , Kn = Kn−1 xn−1

and verify that for each i (1 ≤ i ≤ n), j (1 ≤ j ≤ n − 1) we have Ki xj = Kl

for some l (1 ≤ l ≤ n). Thus H has index at most n and by induction H
has order at most (n − 1)!.
11. The group
G = hx, y | x2 , y 3 , (xy)5 i
is the alternating group A5 .
Proof: The elements σ = (12)(34), τ = (135) of A5 satisfy σ 2 = 1, τ 3 =
1, (στ )5 = 1, and hence generate a subgroup of order 30 or 60; since A5 has
no subgroup of order 30 (being simple), therefore σ, τ generate A5 .
12. Consider the functions α and β on the set CI ∪ {∞} defined by the rules
x
α(x) = x + 2, β(x) = .
2x + 1
1 ∞
here the symbol ∞ is subject to such formulae as 0 = ∞ and ∞ = 1. Then
α and β are bijections since they have inverses
x
α−1 (x) = x − 2, β −1 (x) = .
1 − 2x
Thus α and β generate a group F of permutations of C I ∪ {∞}.
F is free on the set {α, β}.
Proof: Note that every non-zero power of α maps the interior of the circle
|z| = 1 to the exterior of the unit circle and a non-zero power of β maps the
exterior of the unit circle to the interior with 0 removed.
13. The group with generators a1 , a2 , a3 and relations

a−1 2
1 a2 a1 = a2 ;

38
 a−1 2
2 a3 a2 = a3 ,
a−1 2
3 a1 a3 = a1
is the identity group.
14. The group G generated by a1 , a2 , a3 , a4 subject to the defining relations
a−1 2
1 a2 a1 = a2 ;

a−1 2
2 a3 a2 = a3 ,
a−1 2
3 a1 a3 = a4 ,
a−1 2
4 a1 a4 = a1
has no proper normal subgroup of finite index. Since every finitely generated
group has at least one maximal normal subgroup, it follows that there exists
a finitely generated infinite simple group, namely the quotient group G/N
where N is any maximal normal subgroup of G.
15. Let p be an odd prime. The subgroup Gp of the symmetric group on R
generated by x 7→ x + 1 and x 7→ xp is free of rank 2. This was proved by
S.White in J.Algebra 118 (1988) but a more transparent proof by S.D.Cohen
& A.M.W.Glass appears in Journal of the London Math. Society 55 (1997).

Exercise (Josephus permutation) 7.4

The story goes that Flavius Josephus and 39 of his comrades were sur-
rounded when holding a revolt against the Romans during the 1st century
A.D. Rather than become slaves, they decided to kill themselves. They ar-
ranged themselves along a circle. Starting somewhere, they went around the
circle and the 7th person was killed. This continued with each 7th among
the surviving ones being killed at each step. Apparently, Josephus was a
clever mathematician and arranged himself in such a position that he would
be the last survivor. The story goes that he did not kill himself but came
and joined the Romans. The problem is to find out Josephus’s position. Try
also to solve the problem in general (i.e., with 7 and 39 replaced by other
numbers). I offer a prize for the first solution.

Exercise (Sam Lloyd generalized) 7.5

The famous puzzle of Sam Lloyd, for a solution of which he offered a thou-
sand dollars in 1879 goes as follows. Look at the picture here of a 4×4 square
on which 15 coins have been placed leaving out the last square empty.
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15

39
 The idea is to slide the coins utilising the empty square and to find out
what kind of arrangements are possible. Sam Lloyd offered 1000 dollars to
anyone who could get the arrangement with 14 & 15 switched above. He
knew his money would be safe with him. Analyse how to decide whether a
given position of coins in 15 of the 16 places is ‘good’ (i.e., can be brought
back to the beginning position by the sliding process. Do this for a square
of general size.

Exercise (Polya’s enumeration of isomers) 7.6

Suppose D is a set of m objects to be coloured using a range R of k colours.
Let G be the group of symmetries of D. Then, prove that the number of
1
colour patterns = |G| z(G; k, k, . . . , k) where z(G; s1 , s2 , . . . , sn ), the ‘cycle
index’ of G is the polynomial expression given by
1 X λ1 (g) λ2 (g)
z(G; s1 , s2 , . . . , sn ) = s1 s2 . . . snλn (g) .
|G|
g∈G

Here λi (g) denotes the number of i-cycles in G.

§8 Combinatorial group theory

Connections with topology 8.1

The notions of free groups, free products, and of free products with amalga-
mation come naturally from topology. For instance, the fundamental group
of the union of two path-connected topological spaces joined at a single point
is isomorphic to the free product of the individual fundamental groups.
The Seifert-van Kampen theorem asserts that if X = V ∪ W is a union of
path-connected spaces with V ∩ W non-empty and path-connected, and if
the homomorphisms π1 (V ∩ W ) → π1 (V ) and π1 (V ∩ W ) → π1 (W ) induced
by inclusions, are injective, then π1 (X) is isomorphic to the free product of
π1 (V ) and π1 (W ) amalgamated along π1 (V ∩ W ).
All these notions have found many group-theoretical applications. Recall
that:
If Gi , i ∈ I are groups, then a group G along with injective homomorphisms
φi : Gi → G is said to be their free product if, for every group H and ho-
momorphisms θi : Gi → H, there is a unique homomorphism φ : G → H so
that φ ◦ φi = θi for all i ∈ I.
In other words, G has the universal repelling property with respect to ho-
momorphisms from Gi ’s to groups.

40
To construct G, one starts with a presentation < Xi |Ri > of each Gi and
takes < X|R > as a presentation of G where X is the disjoint union of the
Xi ’s and R is the union of the Ri ’s. The homomorphisms φ : Gi → G are,
therefore, simply inclusions. The uniqueness of such a free product G up to
isomorphism follows from the uniqueness property of φ above.
One writes G = ∗i∈I Gi . If I is a finite set, say, I = {1, 2, · · · , n}, then it is
customary to write G = G1 ∗ G2 ∗ · · · ∗ Gn .
For example,
The free group of rank r is the free product ZZ ∗ · · · ∗ ZZ of r copies of ZZ.
More generally, a free group F (X) on a set X is the free product ∗x∈X < x >.
The group P SL(2, ZZ) is the free product of a cyclic group of order 2 and a
cyclic group of order 3.
Recall that if A is a group, Gi , i ∈ I is a family of groups and αi : A → Gi
(i ∈ I) are injective homomorphisms, then a group G is said to be the
free product of Gi ’s amalgamated along A, if there are homomorphisms
φi : Gi → G satisfying φi ◦ αi = φj ◦ αj for all i, j ∈ I such that the fol-
lowing universal property holds: for every group H and homomorphisms
θi : Gi → H with θi ◦ αi = θj ◦ αj for all i, j ∈ I, there is a unique homo-
morphism θ : G → H with φ ◦ φi = θi .
One denotes G by ∗A Gi if there is no confusion as to what the maps αi are.
Sometimes, the maps αi are taken to be not necessarily injective and still
the above definition can be carried out. Note that, if αi are trivial, then
∗A Gi = ∗Gi , the free product.
The construction is as follows. If Gi =< Xi |Ri >, i ∈ I, then
G :=< ti∈I Xi | ∪ Ri ∪ ∪i,j Rij >
where Rij = {αi (a)αj (a)−1 ; a ∈ A} > .
The uniqueness of G up to isomorphism is clear once again by the unique-
ness of θ.
For instance, SL(2, ZZ) = ZZ/4 ∗ZZ/2 ZZ/6.
The fundamental group of the Möbius strip is isomorphic to ZZ ∗2ZZ ZZ. A
free product with amalgamation could be the trivial group even if the groups
αi (A) are not.
For example, let α1 : ZZ → P SL(2, Q I ) be an injective homomorphism and
let α2 : ZZ → ZZ/2 be the natural homomorphism. Then, G1 ∗ZZ G2 = {1}.
Finally, we recall the notion of HNN extensions named after G.Higman,
B.H.Neumann & H.Neumann. The construction is akin to adjoining ele-
ments to fields to get field extensions.
Let G =< X|R > be a group and let A be a subgroup. For an injective
homomorphism φ : A → G, the HNN extension of G with respect to φ is the

41
group G∗ =< X ∪ {t}|R ∪ {tat−1 φ(a)−1 } >.
It is a fact that G∗ is independent of the presentation of G chosen and that
G embeds naturally into G∗ . It can be shown that, given two elements a, b of
equal order in a group G, this construction enables one to embed the group
G into a bigger group in which a, b are conjugate. The HNN construction
also finds a natural topological interpretation.
For, suppose V and W are open, path-connected subspaces of a path-
connected space X and suppose that there is a homeomorphism between
V and W inducing isomorphic embeddings of π1 (V ) and π1 (W ) in π1 (X).
One constructs a space Y by attaching the handle V ×[0, 1] to X, identifying
V × {0} with V and V × {1} with W . Then, the fundamental group π1 (Y )
of Y is the HNN extension of π1 (V ) relative to the isomorphism between its
subgroups π1 (V ) and π1 (W ).

Nielsen-Schreier Theorem 8.2

If W is a subgroup of a free group F , then W is a free group. Moreover, if
W has finite index m in F , the rank of W is precisely nm + 1 − m, where
n is the rank of F (which may be infinite).
Proof. Let F be a free group on a set X; let W be an arbitrary subgroup of
F . Label the right cosets of W in F by means of an index set I containing
the symbol 1, with the convention that W1 = W . Choose a right transversal
to W in F , the representative of the coset Wi being written as W i , with
the stipulation that W = 1.
If u ∈ F , the elements W i u and Wi u belong to the same right coset of
W , so that
W i u(Wi u)−1 ∈ W.
For each i ∈ I and x ∈ X, choose a symbol yix , and let F̂ be the free
group on the set of all yix . Consider the homomorphism

τ : F̂ −→ W, yix 7−→ W i x(Wi x)−1 .

Claim: τ is surjective.
Coset maps: For i ∈ I and u ∈ F associate an element uWi ∈ F −1 as
follows:

F −→ F̂ u 7−→ uWi
as follows;
Wi −1
1Wi = 1, xWi = yix , and (x−1 ) = (xWi x )−1 (x ∈ X).

42
Generally, if u = vy in reduced form with y ∈ X ∪ X −1 , define uWi by
induction on the length of u by means of the equation

uWi = v Wi y Wi v .

If u and v belong to F , then

−1
(uv)Wi = uWi v Wi u and (u−1 )Wi = (uWi u )−1 .

If u ∈ F and i ∈ I, then τ (uWi ) = W i u(Wi u)−1 .

Let
ψ : W −→ F̂
be the restriction of the coset map u 7−→ uW to W . Observe that ψ is a
homomorphism and
ψτ = 1.
Hence ψ is injective and τ is surjective. Write

χ = τ ψ,

an endomorphism of F̂ .
The group W has a presentation τ : F̂ −→ W with generators yix and
−1 χ
defining relators yix yix (i ∈ I, x ∈ X).
The elements uW , where u runs over the set of nontrivial elements in the
transversal, form a set of defining relators for the presentation τ : F̂ −→ W .
A transversal for W is called a Schreier transversal if it has the following
property:

Schreier property: Every initial subword of a representative is a repre-

sentative.

There exists a right Schreier transversal to W in F .

The Nielsen-Schreier theorem can now be proved.
Reidemeister-Schreier Theorem 8.3
Let G be a group. Suppose that ϕ : F −→ G is a presentation of G with
generators X and relators R. Let W be the preimage of H under ϕ. Then
with the above notation:

(i) τ ϕ : F̂ −→ H is a presentation of H with generators yix and defin-

ing relators rWi , uW , i ∈ I, r ∈ R, and u is a non-trivial element of a

43
transversal to W in F .

(ii) If [G : H] is finite and G is n-generator group, then H can be generated

by mn + 1 − m elements.
Corollary 8.4
If G is finitely presented and H is a subgroup of finite index in G, then H
is finitely presented.

Example 8.5
Consider the free group F = hx, y | φi. The homomorphism ϕ : F −→
ZZn = {0, 1, 2, . . . , n − 1}, n ≥ 2 defined by x, y 7→ 1 and let U be the kernel
of ϕ. As coset representatives we take 1, y, y 2 , . . . , y n−1 ; clearly this system
satisfies the Schreier property. Then the Reidemeister-Schreier generators
for U are the non-trivial elements of the following set:
{y i .x.(y i x)−1 , y i .y(y i+1 )−1 | i = 0, 1, 2, . . . , n − 1}
We obtain the non-trivial elements
y n , xi = y i .x.y −(i+1)
for i = 0, . . . , n − 2 and xn−1 = y n−1 .x. The rank of this group is n + 1.
Note that this shows that: The free group of rank 2 contains a free group of
rank n as a subgroup of index n − 1.

Exercise 8.6

1. Show that the modular group P SL(2, ZZ) contains a free subgroup of
rank 2 and index 6.
2. If F is a free group of rank two, then prove that the commutator
subgroup of F is an infinitely generated free group.
3. Prove that the group
 
a b p
Γ={ ∈ GL(2, IR) | a = 2n and b = q f or some n, p, q ∈ ZZ}
0 1 2
   
2 0 1 1
is generated by the two matrices and , while
0 1 0 1
 
1 b p
Γ0 = { ∈ GL(2, IR) | b = q f or some p, q ∈ ZZ}
0 1 2
is not finitely generated.

44