Solutions to Assignment 1

1. Let G be a finite set with an associative law of composition, and e ∈ G an element with
xe = ex = x for all x ∈ G. If G has the property that

xa = ya implies x = y,

prove that G is a group.

Solution: Let |G| = n, and G = {g1 , . . . , gn }. Let a ∈ G be an arbitrary element; we need
to show it has an inverse. The elements g1 a, . . . , gn a must be distinct since gi a = gj a implies
gi = gj . Since g1 a, . . . , gn a are n distinct elements, and hence all of the group elements, we
must have gi a = e for some i. But then agi a = ae = ea, and so agi = e, and gi is an inverse
for a.

2. Let n be a positive integer, and consider the set G of positive integers less than or equal to n,
which are relatively prime to n. The number of elements of G is called the Euler phi-function,
denoted ϕ(n). For example, ϕ(1) = 1, ϕ(2) = 1, ϕ(3) = 2, ϕ(4) = 2, etc.

(a) Show that G is a group under multiplication mod n.

(b) If m and n are relatively prime positive integers, show that

mϕ(n) ≡ 1 mod n.

Solution: (a) Let x, y ∈ G. Since they are relatively prime to n, so is their product.
Consequently xy ≡ z mod n for some z ∈ G. The element 1 serves as the identity and since
G is finite, we may use Problem 1 above. Suppose x, y, a ∈ G with xa ≡ ya mod n. Then
n divides xa − ya = (x − y)a. Since a is relatively prime to n, we must have n|(x − y). But
then x and y are both positive integers less than or equal to n, so they must be equal.
(b) Since m is relatively prime to n, there exists x ∈ G with x ≡ m mod n. The order of x
divides |G| = ϕ(n), and so xϕ(n) ≡ 1 mod n. This implies mϕ(n) ≡ 1 mod n.

3. Let n be a positive integer. Show that

X
n= ϕ(d),
d|n

where the sum is taken over all positive integers d which divide n.
(Hint: Recall that a cyclic group of order n has a unique subgroup of order d for each integer
d which divides n.)
Solution: Let G = Z/hni. Note that if x ∈ G, then the order of x divides n.
X
n = |G| = number of elements of G which have order d
d|n
X
= number of generators for the unique subgroup of G of order d
d|n
X
= ϕ(d).
d|n

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Solutions to Assignment 1

4. Let G be a finite group with identity e, which has the property that for every positive integer
d, the number of elements x of G with xd = e is at most d. Show that G must be cyclic.
Solution: Let |G| = n, and a ∈ G be an element of order d. Then the cyclic group
hai = {a, a2 , . . . , ad } has d distinct elements satisfying xd = e, and so these must be the only
elements x with xd = e. Consequently if G has at least one element of order d, then it has
precisely ϕ(d) elements of order d. But
X X
n = number of elements of G which have order d ≤ ϕ(d) = n,
d|n d|n

and so for each d dividing n, the number of elements of G which have order d must be precisely
ϕ(d). In particular, G has ϕ(n) elements of order n, and therefore is a cyclic group.

5. Let G be a group, and H be a subgroup of finite index. Prove that the number of right cosets
of H is equal to the number of left cosets of H.
Solution: Consider the map

f : {left cosets of G} −→ {right cosets of G}, where f (xH) = Hx−1 .

Now xH = yH if and only if y −1 x ∈ H, and this occurs if and only if Hy −1 = Hx−1 .

Therefore f is a well defined injective map. It is clearly also surjective, and so gives a
bijection between the left cosets and right cosets.

6. If a subgroup of a group G has index 2, show that it must be a normal subgroup.

Solution: Let H be a subgroup of G of index 2. If g ∈ H then gH = Hg. Next let g ∈ G \ H.
Since there are only two cosets and gH 6= H, we must have gH = G \ H. By the previous
problem, H also has two right cosets, and so similarly Hg = G \ H. Hence gH = Hg for
every g ∈ G.

7. Let G be a finite group in which x2 = e for all elements x ∈ G. Prove that the order of G is
a power of 2.
Solution: Let a, b ∈ G. Then (ab)2 = abab = e, and so ab = b−1 a−1 . But b2 = e implies
b = b−1 , and similarly a = a−1 . Hence ab = ba, and so G is abelian.
We proceed by induction on |G|. If |G| = 1 = 20 there is nothing to be proved, so assume
|G| > 1. Pick x ∈ G with x 6= e. Then H = {e, x} is a normal subgroup, and G/H is also a
group which satisfies the hypothesis that the square of every element is the identity. Since

|G/H| = |G|/2 < |G|,

the induction hypothesis implies that |G/H| is a power of 2. But then |G| = |H||G/H| =
2|G/H| is a power of 2 as well.

8. Let G be a group such that for a fixed integer n > 1, (xy)n = xn y n for all x, y ∈ G. Let
G(n) = {xn |x ∈ G} and G(n) = {x ∈ G|xn = e}.

(a) Prove that G(n) and G(n) are normal subgroups of G.

(b) If G is finite, show that the order of G(n) is equal to the index of G(n) .

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Solutions to Assignment 1

(c) Show that for all x, y ∈ G, we have x1−n y 1−n = (xy)1−n . Use this to deduce that
xn−1 y n = y n xn−1 .
(d) Conclude from the above that the set of elements of G of the form xn(n−1) generates a
commutative subgroup of G.

Solution: (a) Consider the map f : G → G with f (x) = xn for all x ∈ G. The condition
(xy)n = xn y n tells us that f is a homomorphism. The kernel of f is precisely G(n) , and it
follows that G(n) is a normal subgroup. Since G(n) is the image of f , it must be a subgroup.
Let g ∈ G and xn ∈ G(n) . Then

gxn g −1 = (gxg −1 )n ∈ G(n) ,

and so G(n) is a normal subgroup.

(b) The homomorphism f induces an isomorphism G/G(n) ≈ G(n) and so, if G is finite, we
get
(G : G(n) ) = |G|/|G(n) | = |G(n) |.

(c)

x1−n y 1−n = xx−n y −n y = x(x−1 y −1 )n y

= x(x−1 y −1 ) · · · (x−1 y −1 )y
= (y −1 x−1 ) · · · (y −1 x−1 ) = (y −1 x−1 )n−1 = (xy)1−n .

Next, taking inverses, we get

(xy)n−1 = y n−1 xn−1 .
Consequently
xn y n = (xy)n = xy(xy)n−1 = xyy n−1 xn−1 = xy n xn−1 ,
and so xn−1 y n = y n xn−1 .
(d) For a, b ∈ G,

an(n−1) bn(n−1) = (an )n−1 (bn−1 )n

= (bn−1 )n (an )n−1 = bn(n−1) an(n−1) .

Consequently elements of G of the form an(n−1) commute with each other, and hence generate
a commutative subgroup.

9. Let G be a group such that

(a) the map x 7→ x3 permutes elements of G, and

(b) (xy)3 = x3 y 3 for all x, y ∈ G.

Prove that G is abelian.

Solution: Using the previous problem, elements of the form x6 commute with each other,
i.e.,
x6 y 6 = y 6 x6 for all x, y ∈ G. (∗)

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Solutions to Assignment 1

For arbitrary elements a, b ∈ G, condition (a) implies that there exist x, y ∈ G with a = x3
and b = y 3 , and (∗) then implies that a2 b2 = b2 a2 . Now

(ab)(ab)(ab) = a3 b3 = a(a2 b2 )b = a(b2 a2 )b = ab(ba)ab,

using which we get ab = ba.

10. Let G be a group. If S is a simple group, S is said to occur in G if there exist two subgroups
H and H 0 of G, with H C H 0 , such that H 0 /H ≈ S. Let In(G) be the set of isomorphism
classes of simple groups occurring in G.

(a) Prove that

In(G) = ∅ ⇔ G = {e}.
(b) If H is a subgroup of G, show that In(H) ⊂ In(G); if H is normal, show that

In(G) = In(H) ∪ In(G/H).

(c) Let G1 , G2 be two groups. Show that the following two properties are equivalent:
(i)
In(G1 ) ∩ In(G2 ) = ∅,
(ii) Every subgroup of G1 × G2 is of the form H1 × H2 with H1 ⊂ G1 and H2 ⊂ G2 .

Solution: (a) Clearly if G = {e} then In(G) = ∅. If G 6= {e}, pick g ∈ G with g 6= e. If

hgi ≈ Z, then we have hg 2 i C hgi, and so

hgi/hg 2 i ≈ Z/h2i

is a simple group which occurs in G. If hgi is finite, then it is simple if |hgi| is prime.
Otherwise, let H be a maximal proper subgroup of hgi. Then hgi/H must be simple and
occurs in G.
(b) If S occurs in H, then S ≈ H1 /H2 where Hi are subgroups of H with H2 C H1 . But then
Hi are subgroups of G as well, so S occurs in G.
If H C G and S occurs in G/H, then S ≈ (H1 /H)/(H2 /H) where Hi /H are subgroups of
G/H with H2 /H C H1 /H. But then H2 C H1 , and so

H1 /H
S≈ ≈ H1 /H2
H2 /H

occurs in G. Consequently we have In(H) ∪ In(G/H) ⊆ In(G).

Now suppose S occurs in G. Then there exist two subgroups H1 , H2 of G where H2 C H1
and H1 /H2 ∼
= S is a simple group. Then H1 ∩ H C H1 , and hence H2 C H2 (H1 ∩ H) C H1 .
Since H1 /H2 is simple, either H2 (H1 ∩ H) = H1 or H2 (H1 ∩ H) = H2 .
In the first case,

S ≈ H1 /H2 = H2 (H1 ∩ H)/H2 ≈ (H1 ∩ H)/(H1 ∩ H ∩ H2 ) = (H1 ∩ H)/(H2 ∩ H),

and so S occurs in H.

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Solutions to Assignment 1

In the second case, H1 ∩ H ⊂ H2 , and it is easily checked that H1 ∩ H C H2 . Note that

H1 /(H1 ∩ H)
H1 /H2 ≈ ,
H2 /(H1 ∩ H)

which implies that S occurs in H1 /(H1 ∩ H) ≈ H1 H/H. But H1 H/H is a subgroup of G/H,
and so S occurs in G/H.
(c) (i) ⇒ (ii) Let H be a subgroup of G1 × G2 , and set H1 = p1 (H) and H2 = p2 (H), where
p1 , p2 are the projection homomorphisms. Let

H10 = {x ∈ H1 : (x, e2 ) ∈ H} and H20 = {y ∈ H2 : (e1 , y) ∈ H}.

It is easily verified that H10 C H1 and H20 C H2 . Consider the homomorphism

φ : H −→ H1 /H10 where φ((x, y)) = xH10 .

Then

Ker φ = {(x, y) ∈ H : x ∈ H10 }

= {(x, y) ∈ H : (x, e2 ) ∈ H}
= {(x, y) ∈ H1 × H2 : (x, e2 ) ∈ H, (e1 , y) ∈ H} = H10 × H20 .

Consequently H/(H10 × H20 ) ≈ H1 /H10 and so, by symmetry,

H1 /H10 ≈ H/(H10 × H20 ) ≈ H2 /H20 .

If H is not a direct product of a subgroup of G1 with a subgroup of G2 , then H/(H10 × H20 ) 6=

{e}, and so we have a nonempty set

In(H1 /H10 ) = In(H/(H10 × H20 )) = In(H2 /H20 )

contained in the intersection In(G1 ) ∩ In(G2 ).

(ii) ⇒ (i) Suppose there exists S ∈ In(G1 ) ∩ In(G2 ), then there exist subgroups H10 C H1 of
G1 and H20 C H2 of G2 , such that S ≈ H1 /H10 ≈ H2 /H20 . Let φi be the composition
≈
Hi −→ Hi /Hi0 −→ S, for i = 1, 2.

Then (φ1 , φ2 ) : H1 × H2 → S × S. Consider the subgroup ∆ = {(s, s) ∈ S × S} and its

inverse image H = (φ1 , φ2 )−1 (∆). Then H is a subgroup of H1 × H2 , and we claim it is not
the direct product of a subgroup of G1 with a subgroup of G2 .
It is easy to check that p1 (H) = H1 and p2 (H) = H2 , and so it suffices to show that
H 6= H1 × H2 . Let h1 ∈ H1 \ H10 and h2 ∈ H20 . Then (φ1 , φ2 )((h1 , h2 )) = (φ1 (h1 ), e) where
φ1 (h1 ) 6= e, and so (h1 , h2 ) ∈ (H1 × H2 ) \ H.