Arkansas Tech UniversityMATH 4033: Elementary Modern Algebra

Dr. Marcel B. Finan

19 More Properties of Isomorphims

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

Lemma 19.1Let 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.(a) First we show that θ−1 is one-to-one. If θ−1(h1) = θ−1(h2) then θ(θ−1(h1)) =θ(θ−1(h2)) or ιH(h1) = ιH(h2). Hence, h1 = h2. To see that θ−1 is onto, letg ∈ G. Then θ(g) ∈ H and θ−1(θ(g)) = g. Finally, we show that θ−1 is ahomomorphism. 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). Thenγ(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 thenwe can find h ∈ H such that γ(h) = k. Since θ is onto then we can find ag ∈ G such that θ(g) = h. Hence, γ(θ(g)) = γ(h) = k. To see that γ ◦ θ is ahomomorphism, 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 19.1Isomorphism 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 ∈ Gis an isomorphism. Hence, G ≈ G.Symmetric: Suppose that G ≈ H for some groups G and H. Then there isan isomorphism θ : G −→ H. By Lemma 19.1(a), θ−1 : H −→ G is also anisomorphism. Hence, H ≈ G.Transitive: Suppose that G ≈ H and H ≈ K, where G, H, and K aregroups. Then there are isomorphisms θ : G −→ H and γ : H −→ K. ByLemma 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 finda property not shared by the two groups.

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Theorem 19.2Let 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 alsocyclic.(c) If K is a subgroup of G of order n then θ(K) is a subgroup of H of ordern.(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 theelements 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(sinceθ 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 subgroupof H. The mapping θ restricted to K is a one-to-one mapping from K ontoθ(K). Thus, |K| = |θ(K)|.(d) If o(a) = n then an = eG. By Theorem 18.2 (iii), we have (θ(a))n =θ(an) = θ(eG) = eH . By Theorem 14.6(ii), o(θ(a))|n. On the other hand, since(θ(a))o(θ(a)) = eH then θ(ao(θ(a))) = eH = θ(eG). Thus, ao(θ(a)) = eG (since θ isone-to-one) and by Theorem 14.6(ii), n|o(θ(a)). Hence, by Theorem 10.2(d),o(θ(a)) = n.

Example 19.1Z2 × 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 order4(namely, [1] and [3].)

The following result classifies all groups of prime order.

Theorem 19.3If G is a cyclic group of order n then G ≈ Zn.

Proof.Suppose that G =< a > . Define θ : G −→ Zn by θ(ak) = [k] where 1 ≤ k <

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n. We show that θ is well-defined. Indeed, if ak = am then ak−m = eG so thatby 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.Therefore, ak−m = anq = (an)q = eG. Hence, ak = am. Next, we show that θis 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 thenn|(b− r so that b ≡ r(mod n). By Theorem 9.2, [b] = [r]. Finally, it remainsto show that θ is a group homomorphism. Indeed,

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

Example 19.2If p is prime then Zp × Zp is not cyclic. For if it is, then by the previoustheorem, Zp×Zp ≈ Zp2 . But every nonidentity element of Zp×Zp has orderp whereas Zp2 has an element of order p2, namely, [p2 − 1].

Corollary 19.1If |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 previoustheorem, G ≈ Zp.

Theorem 19.4If 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 willshow next that θ is one-to-one. Suppose that θ(n) = θ(m). Without loss ofgenerality 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 musthave n = m. To show that θ is onto, pick an x ∈ G. Then x = an for somen ∈ 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

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

A principal problem in finite group theory is the problem of classifying groupsof finite orders. For example, the problem of determining all isomorphismclasses 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 groupsin the form

Cp1r1 × Cp2

r2 × · · · × Cptrt ,

where the pi are (not necessarily distinct) primes; the product is unique upto reordering of the factors.

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

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