Solutions to Assignment 1

1. Determine all group structures on a set of n elements for 2 ≤ n ≤ 6. Determine the

subgroups, quotient groups of these groups and also their Jordan-Holder series.

Solution:

n = 1: The trivial group.

n = 2: The group Z2 .
n = 3: The group Z3 .
n = 4: Groups Z4 and Z2 × Z2 . Jordan Holder series are given by, Z4 ⊃ Z2 ⊃ {e}, Z2 ×
Z2 ⊃ Z2 ⊃ {e}.
n = 5: The group Z5 .
n = 6: The groups Z6 , S3 . Jordan Holder series, Z6 ⊃ Z3 ⊃ {e}, S3 ⊃ H ⊃ {e}, where
H is the subgroup of order 3 containing the identity and the 3-cycles.

2. Let A and B be subgroups of a group G.

(a) Prove that for AB to be a group it is necessary and sufficient that A and B are
permutable, that is AB = BA.

Solution: Suppose that AB is a group. Let a ∈ A, b ∈ B. Then, ab ∈ AB.

But, (ab)−1 = b−1 a−1 must also be in AB. Hence, b−1 a−1 = a1 b1 for some
a1 ∈ A, b1 ∈ B. Inverting again, ab = b−1 −1
1 a1 ∈ BA. Hence, AB ⊂ BA and by
symmetry BA ⊂ AB.
Now suppose, AB = BA. Let a1 , a2 ∈ A and b1 , b2 ∈ B. Now consider, a1 b1 a2 b2 .
Since, BA = AB, b1 a2 = a3 b3 for some a3 ∈ A and b3 ∈ B. Thus, AB is closed
under multiplication. Also, (a1 b1 )−1 = b−1 −1
1 a1 ∈ BA = AB. Thus, AB is also
closed under taking inverses.
(b) Prove that if A and B are permutable and C is a subgroup containing A, A is
permutable with B ∩ C and A(B ∩ C) = C ∩ (AB).

Solution: Let a ∈ A and b ∈ B ∩ C. Since, A, B are permutable, ab = b1 a1 for

some b1 ∈ B and a1 ∈ A. But since, A ⊂ C, a, b, ab, a1 are all in C, which implies
that b1 ∈ C. Hence, A is permutable with B ∩ C.
Clearly, A(B ∩ C) ⊂ C ∩ (AB). Now, let c = ab ∈ C ∩ (AB) for some a ∈ A, b ∈
B, c ∈ C. Now, b = a−1 c ∈ C because a ∈ C. Hence, ab ∈ A(B ∩ C).

3. If a subgroup of a group G has index 2, show that it is a normal subgroup of G.

Solution: Let H be a subgroup of G of index 2 and let g ∈ G \ H. Since, there are

only two cosets and gH 6= H, gH = G \ H. But the same is true for Hg. Hence,
gH = Hg for every g, proving that H is normal.

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Solutions to Assignment 1

4. Let G be a group such that all elements of G other than the identity has order 2. Show
that:

(a) G is commutative.
Solution: Let a, b ∈ G. Then, (ab)2 = abab = e. Hence, ab = b−1 a−1 . But,
b2 = e ⇒ b = b−1 and similarly a = a−1 . Hence, ab = ba.
(b) If G is finite, order of G is a power of 2.
Solution: We proved in class that for a finite Abelian group the order of G must
divide some power of any exponent of G. Clearly, 2 is an exponent and hence
order of G must be a power of 2.

5. 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) , G(n) are normal subgroups of G.

Solution: It is easy to see by induction that for any g1 , . . . , gk ∈ G, (g1 . . . gk )n =
g1n . . . gkn . It is also clear from the property that (xy)n = xn y n , that G(n) , G(n) are
subgroups.
Now, let g, x ∈ G and consider the element gxn g −1 . Clearly,

gxn g −1 = (gxg1 )(gxg −1 ) . . . (gxg −1 ) = (gxg −1 )n ∈ G(n) .

This shows that, G(n) is a normal subgroup.

Now let y ∈ G(n) . Hence, y n = e. Now, (gyg −1 )n = (g n y n g −n ) = e. Hence,
gyg −1 ∈ G(n) , which show that G(n) is a normal subgroup.
(b) If G is finite then the order of G(n) is equal to the index of G(n) .

Solution: We will establish a bijection between the elements of G(n) and the left
cosets of G(n) . Namely, let x = g n ∈ G(n) map to gG(n) . Clearly, the map is onto.
Now, suppose that g1 G(n) = g2 G(n) . Then, g1 y1 = g2 y2 for some y1 , y2 ∈ G(n) .
But, y1n = y2n = e. Hence, (g1 y1 )n = g1n y1n = g1n = (g2 y2 )n = g2n . Hence, g1n = g2n
proving that the map is 1-1.
(c) Show that for all x, y ∈ G, x1−n y 1−n = (xy)1−n and deduce that xn−1 y n = y n xn−1 .

Solution:
n
x1−n y 1−n = xx−n y −n y = xx−1 y −1 y
= xx−1 (y −1 x−1 )n−1 y −1 y = (y −1 x−1 )n−1 = (xy)−1 n − 1 = (xy)1−n .
Also,
xn−1 y n = (x−1 )1−n (y −1 )1 − ny = (x−1 y −1 )1−n y
= (yx)n−1 y = (yx)n−1 yxx−1 = (yx)n x−1 = y n xn−1 .

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Solutions to Assignment 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:
Let x, y ∈ G. Now,

xn(n−1) y n(n−1) = (xn−1 )n (y n−1 )n = (xn−1 y n−1 )n =

2
(xn−1 y n y −1 )n = (y n xn−1 y −1 )n = y n x(n−1)n y −n =
2 2
y n (xn )n−1 (y −1 )n = y n (y −1 )n (xn )n−1 = y n(n−1) xn(n−1) .
Thus, he set of elements of G of the form xn(n−1) commute with each other and
hence, they generate a commutative subgroup.

6. Let G be a group. If S is a simple group, S is said to occur in G is there exist

two subgroups H, H 0 of G, with H C H 0 such that H 0 /H ∼
= S. Let in(G) be the
isomorphism classes of simple groups occuring in G.

(a) Prove that,

in(G) = ∅ ⇔ G = {e}.
Solution: Clearly if G = {e} then in(G) = ∅. Now, suppose that in(G) = ∅. If
G 6= {e} then there exists an element g ∈ G, g 6= e. Consider the cyclic subgroup
hgi generated by g. Now, if hgi is infinite then hgi/hg 2 i ∼ = Z2 , which must belong
to in(G). If hgi is finite, then it is simple if |hgi| is prime. Otherwise, let H be a
maximal proper subgroup of hgi, and hgi/H must be simple and belong to in(G).
(b) If H is a subgroup of G, show that in(H) ⊂ in(G); if moreover, H is normal, then

in(G) = in(H) ∪ in(G/H).

Solution: If H is normal it follows easily from the third isomorphism theorem,

that in(H) ∪ in(G/H) ⊂ in(G).
Now, let S be a simple group, such that there exist two subgroups H1 , H2 of G,
with H1 C H2 , and H2 /H1 ∼ = S.
Now, H2 ∩ H C H2 , and hence H1 (H2 ∩ H) C H2 . Since, H2 /H1 is simple, there
is no proper normal subgroup of H2 properly containing H1 .
Hence, either H1 (H2 ∩ H) = H2 or H1 (H2 ∩ H) = H1 . In the first case, S ∼ =
∼ ∼
H2 /H1 = H1 (H2 ∩ H)/H1 = (H2 ∩ H)/(H1 ∩ H2 ∩ H) ∈ in(H), where we used
the second isomorphism theorem.
In the second case, H 0 = H2 ∩H ⊂ H1 and H2 H/H ∼ = H2 /H 0 contains a subgroup
isomorphic to S. Hence, S ∈ in(G/H).
(c) Let G1 , G2 be two groups. Show that the following two properties are equivalent:
i.
in(G1 ) ∩ in(G2 ) = ∅.

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Solutions to Assignment 1

ii. Every subgroup of G1 ×G2 is of the form H1 ×H2 with H1 ⊂ G1 and H2 ⊂ G2 .

Solution: Let in(G1 ) ∩ in(G2 ) = ∅. Let H be a subgroup of G1 × G2 and
H1 = p1 (H) and H2 = p2 (H2 ) where p1 , p2 are the projection homomorphisms.
We claim that, H ∼= H1 × H2 .
Let H1 = p1 (p2 (e2 ) ∩ H) and H20 = p2 (p−1
0 −1 0
1 (e1 ) ∩ H). It is easy to verify that, H1
is a normal subgroup of H1 and similarly, H20 is a normal subgroup of H2 .
Consider, the homomorphism

φ1 : H1 → H/(H10 × H20 )

defined by, φ1 (h1 ) = (h1 , h2 ), where (h1 , h2 ) is any element of p−1

1 (h1 ) ∩ H. This is
well-defined since for any other element (h1 , h02 ) of p−11 (h1 )∩H, (h 0 −1
1 , h2 )(h1 , h2 ) =
0−1
(e1 , h2 h2 ) ∈ H1 × H2 . Also, h1 ∈ ker(φ) iff h1 ∈ H1 . Hence, ker(φ) = H10 .
0 0 0

Moreover, let (h1 , h2 ) ∈ H/(H10 × H20 ). Clearly, φ1 (h1 ) = (h1 , h2 ) and hence φ1 is
surjective. Hence, H1 /H10 ∼ = H/(H10 × H20 ). By symmetry,

H1 /H10 ∼
= H/(H10 × H20 ) ∼
= H2 /H20 .

Now, in(H1 /H10 ) ⊂ in(H1 ) ⊂ in(G1 ) and in(H2 /H20 ) ⊂ in(H2 ) ⊂ in(G2 ). Since,
in(G1 )∩in(G2 ) = ∅, and H1 /H10 ∼= H2 /H20 , it follows that in(H1 /H10 ) = in(H2 /H20 ) =
∅, and using part (a) this implies that H10 = H1 and H20 = H2 . It is easy now to
deduce that H = H1 × H2 .
In order to prove the converse, suppose there exists S ∈ in(G1 ) ∩ in(G2 ). This
implies that there exists subgroups H10 C H1 of G1 and H20 C H2 of G2 , such that
S∼ = H1 /H10 ∼
= H2 /H20 . Chosse an isomorphism φ : H1 /H10 → H2 /H20 .
Let H = {(h1 , h2 )|h1 ∈ H1 , h2 ∈ H2 , h¯2 = φ(h¯1 )}.
Then, it is easy to check that H is a subgroup of H1 × H2 , p1 (H) = H1 and
p2 (H) = H2 . However for any h1 ∈ H1 , there exists h2 ∈ H2 such that h¯2 6= φ(h¯1 )
since, H1 /H10 is not trivial. Hence, H is not of the form P1 × P2 for subgroups P1
of G1 and P2 of G2 .