19 More Properties of Isomorphims

Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan

In this section we discuss more properties of group isomorphisms. We start with the following lemma.

Lemma 19.1 Let θ : G −→ H and γ : H −→ K be two group isomorphisms. Then

(a) θ−1 : H −→ G is an isomorphism. (b) γ ◦ θ : G −→ K is also an isomorphism.

Proof. −1 −1 −1 −1 (a) First we show that θ is one-to-one. If θ (h1) = θ (h2) then θ(θ (h1)) = −1 −1 θ(θ (h2)) or ιH (h1) = ιH (h2). Hence, h1 = h2. To see that θ is onto, let g ∈ G. Then θ(g) ∈ H and θ−1(θ(g)) = g. Finally, we show that θ−1 is a homomorphism. If a, b ∈ H then a = θ(u) and b = θ(v) for some u, v ∈ G, since θ is onto. Hence, if w = θ−1(ab) then θ(w) = ab = θ(u)θ(v) = θ(uv). Thus, θ−1(ab) = uv = θ−1(a)θ−1(b). Hence, θ−1 is an isomorphism. (b) To see that γ ◦ θ is one-to-one, suppose that γ ◦ θ(g1) = γ ◦ θ(g2). Then γ(θ(g1)) = γ(θ(g2)). Since γ is one-to-one then θ(g1) = θ(g2). Since θ is one- to-one then g1 = g2. To see that γ ◦ θ is onto, let k ∈ K. Since γ is onto then we can ﬁnd h ∈ H such that γ(h) = k. Since θ is onto then we can ﬁnd a g ∈ G such that θ(g) = h. Hence, γ(θ(g)) = γ(h) = k. To see that γ ◦ θ is a homomorphism, let a, b ∈ G. Since both θ and γ are homomorphisms then

(γ ◦ θ)(ab) = γ(θ(ab)) = γ(θ(a)θ(b)) = γ(θ(a))γ(θ(b)) = (γ ◦ θ)(a)(γ ◦ θ)(b).

Theorem 19.1 Isomorphism is an equivalence relation on the collection of all groups.

Proof. Reﬂexive: Let G be a group. Then the identity map ιG(a) = a for all a ∈ G is an isomorphism. Hence, G ≈ G. Symmetric: Suppose that G ≈ H for some groups G and H. Then there is an isomorphism θ : G −→ H. By Lemma 19.1(a), θ−1 : H −→ G is also an isomorphism. Hence, H ≈ G. Transitive: Suppose that G ≈ H and H ≈ K, where G, H, and K are groups. Then there are isomorphisms θ : G −→ H and γ : H −→ K. By Lemma 19.1(b), γ ◦ θ : G −→ K is also an isomorphism. Hence, G ≈ K.

The following theorem lists more properties shared by isomorphic groups. Thus, the simplest way to show that two groups are not isomorphic is to ﬁnd a property not shared by the two groups.

1 Theorem 19.2 Let G and H be groups such that θ : G −→ H is an isomorphism.

(a) |G| = |H|. (b) If G =< a > then H =< θ(a) > . That is, if G is cyclic, then H is also cyclic. (c) If K is a subgroup of G of order n then θ(K) is a subgroup of H of order n. (d) If a ∈ G and o(a) = n then o(θ(a)) = n.

Proof. (a) There’s a one-to-one onto map between the two sets. So counting the elements of one set simultaneously counts the elements of the other. (b) Suppose that G is cyclic with generator a. We will show that H =< θ(a) > . Clearly, since θ(a) ∈ H and H is a group then < θ(a) >⊆ H. Now suppose that h ∈ H then there is an g ∈ G such that θ(g) = h(since θ is onto). But then g = an for some n ∈ Z. Hence, by Theorem 18.2(iii), θ(g) = θ(an) = (θ(a))n. That is, h ∈< θ(a) > and so H ⊆< θ(a) > . (c) If K is a subgroup of G then by Theorem 18.2(iv), θ(K) is a subgroup of H. The mapping θ restricted to K is a one-to-one mapping from K onto θ(K). Thus, |K| = |θ(K)|. n n (d) If o(a) = n then a = eG. By Theorem 18.2 (iii), we have (θ(a)) = n θ(a ) = θ(eG) = eH . By Theorem 14.6(ii), o(θ(a))|n. On the other hand, since o(θ(a)) o(θ(a)) o(θ(a)) (θ(a)) = eH then θ(a ) = eH = θ(eG). Thus, a = eG (since θ is one-to-one) and by Theorem 14.6(ii), n|o(θ(a)). Hence, by Theorem 10.2(d), o(θ(a)) = n.

Example 19.1 Z2 × Z2 and Z4 are not isomorphic. Both groups have 4 elements, however, every element of Z2 × Z2 has order 1 or 2, while Z4 has two elements of order 4(namely, [1] and [3].)

The following result classiﬁes all groups of prime order.

Theorem 19.3 If G is a cyclic group of order n then G ≈ Zn. Proof. k Suppose that G =< a > . Deﬁne θ : G −→ Zn by θ(a ) = [k] where 1 ≤ k <

2 k m k−m n. We show that θ is well-deﬁned. Indeed, if a = a then a = eG so that by Theorem 14.6 (ii), n|(k − m). Thus, k ≡ m(mod n) and by Theorem 9.2, [k] = [m]. Next, we show that θ is one-to-one. Indeed, if θ(ak) = θ(am) then [k] = [m] and this implies that n|(k − m). Thus, k − m = nq for some q ∈ Z. k−m nq n q k m Therefore, a = a = (a ) = eG. Hence, a = a . Next, we show that θ is onto. Let b ∈ Z. Then by the Division Algorithm, b = nq + r, 0 ≤ r < n. Hence, ab = (an)qar = ar ∈ G and θ(ab) = θ(ar) = [r]. Since b − r = nq then n|(b − r so that b ≡ r(mod n). By Theorem 9.2, [b] = [r]. Finally, it remains to show that θ is a group homomorphism. Indeed,

θ(akam) = θ(ak+m) = [k + m] = [k] ⊕ [m] = θ(ak)θ(am).

Example 19.2 If p is prime then Zp × Zp is not cyclic. For if it is, then by the previous theorem, Zp × Zp ≈ Zp2 . But every nonidentity element of Zp × Zp has order 2 2 p whereas Zp2 has an element of order p , namely, [p − 1].

Corollary 19.1 If |G| = p, where p is a prime number, then G ≈ Zp.

Proof. Since G has a prime order then by Corollary 17.3, G is cyclic. By the previous theorem, G ≈ Zp.

Theorem 19.4 If G is an inﬁnite cyclic group then G ≈ Z.

Proof. Suppose that G =< a > with o(a) = ∞. Deﬁne θ : Z −→< a > by θ(n) = an. Then θ is a well-deﬁned mapping. Indeed, if n = m then an = am. We will show next that θ is one-to-one. Suppose that θ(n) = θ(m). Without loss of generality we can assume that n < m. Then

e = a0 = an−n = ana−n = ama−n = am−n.

This contradicts the fact that o(a) = ∞ Since m − n ∈ N. Thus, we must have n = m. To show that θ is onto, pick an x ∈ G. Then x = an for some n ∈ Z and θ(n) = an = x. It remains to show that θ is a homomorphism: θ(n + m) = an+m = anam = θ(n)θ(m). Thus, Z ≈< a > or < a >≈ Z since

3 ≈ is symmetric. This ends a proof of the theorem.

A principal problem in ﬁnite group theory is the problem of classifying groups of ﬁnite orders. For example, the problem of determining all isomorphism classes is settled by the following theorem whose proof is omitted.

Theorem 19.5 (Fundamental Theorem of Finite Abelian Groups) Every ﬁnite abelian group is isomorphic to a direct product of cyclic groups in the form

r r r Cp1 1 × Cp2 2 × · · · × Cpt t , where the pi are (not necessarily distinct) primes; the product is unique up to reordering of the factors.

Example 19.3 There are six isomorphism classes of Abelian groups of order 360. If G is Abelian with |G| = 360 = 23.32.5, then the possible sets of prime powers are as follows:

{23, 32, 5}, {23, 3, 3, 5}, {2, 22, 32, 5}, {2, 22, 3, 3, 5}, {2, 2, 2, 32, 5}, {2, 2, 2, 3, 3, 5}.

Hence there are six mutually non-isomorphic abelian groups of order 360:

Z8 × Z9 × Z5, Z8 × Z3 × Z3 × Z5 Z2 × Z4 × Z9 × Z5 Z2 × Z4 × Z3 × Z3 × Z5 Z2 × Z2 × Z2 × Z2 × Z9 × Z5 Z2 × Z2 × Z2 × Z2 × Z3 × Z3 × Z5