114 Groups and Rings 2008–09
Suggested solutions to exercise set 3
February 23, 2009
22.9 (a) Suppose that [G : N ] is prime. Since |G/N | = [G : N ], Theorem 19.3
shows that G/N ≈ Zp . Since Zp is cyclic, so is G/N (by Theorem 19.1).
(b) The converse is false. For example, Z4 is a cyclic group of non-prime order,
so if N = {[0]4 } is the trivial subgroup of Z4 then N C Z4 and [Z4 : N ] = 4
is not prime, but and Z4 /N ≈ Z4 is cyclic.
22.14
G/N is Abelian ⇐⇒
⇐⇒
⇐⇒
⇐⇒
(N a)(N b) = (N b)(N a) for all a, b ∈ G
N ab = N ba for all a, b ∈ G
ab(ba)−1 ∈ N for all a, b ∈ G (by Lemma 19.1)
aba−1 b−1 ∈ N for all a, b ∈ G.
[Here we use the fact that (ba)−1 = a−1 b−1 , which is easy to check].
23.4 The only normal subgroups of S3 are hei, h(123)i and S3 itself. By the
fundamental homomorphism theorem, every homomorphic image of S3 is
(isomorphic to) the quotient of S3 by one of these normal subgroups; that is,
to either
S3 /hei ≈ S3 ,
S3 /h(123)i ≈ Z2
or S3 /S3 ≈ {e}.
23.6 Observe that the map [k]8 7→ [k]2 is a well-defined homomorphism of Z8 onto
Z2 with kernel h[2]8 i; so Z8 /h[2]8 i ≈ Z2 by the fundamental homomorphism
theorem. Since all subgroups of an Abelian group are normal, we have
Z4 ≈ h[2]8 i C Z8
and Z8 /h[2]8 i ≈ Z2 .
So Z8 is an extension of Z4 by Z2 .
Also, we have seen that G ≈ G × {e} and (G × H)/(G × {eH }) ≈ H. Since
all subgroups of an Abelian group are normal, we have
Z4 ≈ Z4 × {[0]2 } C Z4 × Z2
and (Z4 × Z2 )/(Z4 × {[0]2 }) ≈ Z2 .
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So Z4 × Z2 is an extension of Z4 by Z2 .
Since [1]8 is an element of order 8 in Z8 , but Z4 × Z2 contains no element
of order 8 (it’s easy to see that the largest order of an element of this group
is 4), Z8 and Z4 × Z2 are not isomorphic.
23.11 Let G be a simple Abelian group with |G| > 1 and let a ∈ G with a 6= e.
Since G is Abelian, the cyclic subgroup hai of G is normal in G, and hai =
6 {e}
since a 6= e. So hai = G.
If G = hai is infinite then a has infinite order. We claim that ha2 i is a proper
normal subgroup of G; for if a ∈ ha2 i = {a2k : k ∈ Z} then a = a2k for some
k ∈ Z, so a2k−1 = e; since a has infinite order, this implies that 2k − 1 = 0
which is impossible for k ∈ Z.
So G = hai is finite; so o(a) = n < ∞ and G ≈ Zn by Exercise 19.17. Since G
is simple, Zn is also simple. If n is not prime then n = st for some s, t ∈ Z
with 1 < t < n. But then h[s]n i = {[0]n , [s]n , . . . , [s(t − 1)]n } is a normal
subgroup of Zn of order t, so it is neither {[0]n } nor Zn . This contradicts the
fact that Zn is simple. So n must be prime.
23.18 Any homomorphic image of G is isomorphic to a quotient of G by a normal
subgroup, by the fundamental homomorphism theorem. If G is simple then
the only normal subgroups of G are {e} and G; so the only quotients of G are
G/{e}, which has order |G|/1 = |G|, and G/G which has order |G|/|G| = 1.
23.20* (a) We have K C G and H a subgroup of G. To check that HK is a subgroup
of G, observe that e ∈ H ∩ K, so e = e · e ∈ HK, and that if h1 , h2 ∈ H and
k1 , k2 ∈ K then h−1
2 k1 h2 ∈ K by normality, so
h1 k1 h2 k2 = (h1 h2 )(h−1
2 k1 h2 k2 ) ∈ HK
since h1 h2 ∈ H and h−1
2 k1 h2 k2 ∈ K. Similarly, for h ∈ H and k ∈ H we have
hk −1 h−1 ∈ K, so
(hk)−1 = k −1 h−1 = h−1 (hk −1 h−1 ) ∈ HK.
(b) We have θ : H → HK/K, θ(h) = Kh for h ∈ H. This map is clearly a
well defined map, and for a, b ∈ H,
θ(ab) = K(ab) = KaKb = θ(a)θ(b).
To see that θ is onto, consider Kb ∈ HK/K, the right coset of K by b ∈ HK.
Then b = hk for some h ∈ H and some k ∈ K, so Kb = Khk = Khkh−1 h =
2
Kh = θ(h) (here we have used the fact that Kxy = Ky if x ∈ K, and that
hkh−1 ∈ K since K C HK). So θ maps onto HK/K.
(c) For h ∈ H we have h ∈ ker θ ⇐⇒ θ(h) = Kh = K ⇐⇒ h ∈ K. So
ker θ = {h ∈ H : h ∈ K} = H ∩ K. By the fundamental homomorphism
theorem,
H/ ker θ = H/(H ∩ K) ≈ HK/K.
23.21* We have subgroups H, K of a group G and K C H, K C G and H C G.
Consider θ : G/K → G/H, Ka 7→ Ha for a ∈ G.
This is well defined, since Ka = Kb ⇐⇒ ab−1 ∈ K, and since K ⊆ H this
implies that ab−1 ∈ H ⇐⇒ Ha = Hb. It is clearly surjective, and it is a
homomorphism since
θ(KaKb) = θ(Kab) = Hab = HaHb = θ(Ka)θ(Kb).
If a ∈ G then Ka ∈ ker θ ⇐⇒ θ(a) = eG/H ⇐⇒ Ha = H ⇐⇒ a ∈ H, so
ker θ = {Ka : a ∈ H} = H/K.
By the fundamental homomorphism theorem,
(G/K)/ ker θ = (G/K)/(H/K) ≈ G/H.
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