114 Groups and Rings 2008–09
Suggested solutions to exercise set 1
February 10, 2009
16.6 Let H = h(1 3)i. Since (1 3) has order 2, we have H = {e, (1 3)}.
We have to find all of the right cosets of H in S3 . By Lagrange’s theorem,
[S3 : H] = |S3 |/|H| = 6/2 = 3, so there are three cosets to find.
He = H is one coset, H(1 2) = {h(1 2) : h ∈ H} = {e(1 2), (1 3)(1 2)} =
{(1 2), (1 2 3)} is another coset, and all that remains in S3 is the set
{(2 3), (1 3 2)} which is the remaining coset. (In fact, {(2 3), (1 3 2)} =
H(2 3), and we know that this must be the case without doing any computations since the cosets partition S3 ).
So the answer is: {e, (1 3)}, {(1 2), (1 2 3)}, {(2 3), (1 3 2)}.
16.9 To see that (a) =⇒ (b), suppose that ab−1 ∈ H. Then ab−1 = h for some
h ∈ H, so that multiplying by b on the right we obtain a = hb for some
h ∈ H.
For (b) =⇒ (c), recall that Hb is defined as the set of all elements of G that
can be written as hb for some h ∈ H. So if a = hb for some h ∈ H then
a ∈ Hb.
For (c) =⇒ (d), suppose that a ∈ Hb, so that a = hb for some h ∈ H. Then
Ha = {h1 a : h1 ∈ H} = {h1 hb : h1 ∈ H}.
If x = h1 hb ∈ Ha where h1 ∈ H, then since h is a subgroup, we have h1 h ∈ H,
so x = (h1 h)b ∈ Hb. So Ha ⊆ Hb. On the other hand, if y = h2 b ∈ Hb for
some h2 ∈ H then y = h2 h−1 hb = h2 h−1 a and, since h2 h−1 ∈ H, we have
y = (h2 h−1 )a ∈ Ha. So Hb ⊆ Ha, and so Ha = Hb.
For (d) =⇒ (a), suppose that Ha = Hb. Since e ∈ H, we have a = ea ∈
Ha = Hb, so a = hb for some h ∈ H. Multiplying on the right by b−1 gives
ab−1 = h for some h ∈ H.
1
This last paragraph also proves that a ∈ Ha for each a ∈ G, which is the
last statement of Lemma 16.1.
17.1 By Lagrange’s theorem, [S3 : h(1 2)i] = |S3 |/|h(1 2)i|. Since |S3 | = 6 and
|h(1 2)i| = o((1 2)) = 2, we have [S3 : h(1 2)i] = 6/2 = 3.
18.9 We are given isomorphisms θ : G → H and φ : H → K and asked to prove
that φ ◦ θ : G → K is an isomorphism. To simplify the notation, we’ll write
ψ = φ ◦ θ.
Both θ and φ are bijections. The composition of two bijections is a bijection
by Theorem 2.1, so ψ = φ ◦ θ is a bijection from G onto K.
We must show that ψ(ab) = ψ(a)ψ(b) for all a, b ∈ G. If a, b ∈ G then
ψ(ab) = φ θ(ab) = φ θ(a)θ(b) = φ θ(a) φ θ(b) = ψ(a)ψ(b).
So ψ is an isomorphism.
19.5 The group Z4 × Z2 is the product of two cyclic groups of prime power order
but Z2 × Z2 × Z2 is the product three cyclic groups of prime power order.
By the fundamental theorem of abelian groups they are not isomorphic.
Alternatively, observe that each element of Z2 × Z2 × Z2 has order at most 2,
but the order of ([1], [0]) ∈ Z4 × Z2 is 4. By Theorem 19.1(e), Z2 × Z2 × Z2
cannot be isomorphic to Z4 × Z2 .
19.14 Since 27 = 33 is a prime power, the fundamental theorem of abelian groups
tells us that Z27 and Z9 × Z3 are two non-isomorphic groups of order 27.
(Another is Z3 × Z3 × Z3 ).
19.17 We are to prove that a finite cyclic group G = hai of order n is isomorphic
to Zn .
Recall from section 14 (specifically Theorem 14.3 and the following Corollary)
that the order of a is o(a) = |hai| = n, and that for t ∈ Z,
at = e if and only if n|t.
We have Zn = {[0], [1], . . . , [n − 1]} and G = {e, a, a2 , . . . , an−1 }. Let
θ : Zn → G,
[k] 7→ ak .
The map θ is well defined, since if [k] = [j] in Zn then n|k − j, so ak−j = e
and so aj = aj ak−j = ak .
2
To see that θ is one-to-one, suppose that [j], [k] ∈ Zn with θ([k]) = θ([j]).
Then ak = aj , so ak−j = e, so n|k − j, which implies that [j] = [k] in Zn . So
θ is one-to-one.
It is clear from the definition of θ that it is onto; so θ is a bijection.
Finally, for j, k ∈ Z we have
θ([j] ⊕ [k]) = θ([j + k]) = aj+k = aj ak = θ([j])θ([k]).
So θ is an isomorphism of Zn onto G; so G ≈ Zn .
3