40 Chapter 1. Groups I Proof. (i) 1 · 1 = 1 implies f(1)f(1) = f(1). Now use Exercise 1.30 on page 27. (ii) 1 = x−1x implies 1 = f(1) = f(x−1)f(x). (iii) Use induction to show that f(xn) = f(x)n for all n ≥ 0. Then observe that x−n = (x−1)n, and use part (ii). • Example 1.59. If G and H are cyclic groups of the same order m, then G and H are isomorphic. (It follows from Corollary 1.51 that any two groups of prime order p are isomorphic.) Although this is not diﬃcult, it requires some care. We have G = {1,a,a2,...,am−1} and H = {1,b,b2,...,bm−1}, and the obvious choice for an isomorphism is the bijection f : G → H given by f(ai) = bi. Checking that f is a homomorphism, that is, f(aiaj) = bibj = bi+j, involves two cases: i + j ≤ m − 1, so that aiaj = ai+j, and i + j ≥ m, so that aiaj = ai+j−m. We give a less computational proof in Example 1.77. A property of a group G that is shared by all other groups isomorphic to it is called an invariant of G. For example, the order |G| is an invariant of G, for isomorphic groups have the same order. Being abelian is an invariant. In fact, if f is an isomorphism and a and b commute, then ab = ba and f(a)f(b) = f(ab) = f(ba) = f(b)f(a) that is, f(a) and f(b) commute. The groups I6 and S3 have the same order, yet are not isomorphic (I6 is abelian and S3 is not). In general, it is a challenge to decide whether two given groups are isomorphic. See Exercise 1.49 on page 44 for more examples of invariants. Example 1.60. We present two nonisomorphic abelian groups of the same order. Let V = (1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3) be the four-group, and let μ4 = i = {1,i, −1, −i} be the multiplicative cyclic group of fourth roots of unity, where i2 = −1. If there were an isomorphism f : V → μ4, then surjectivity of f would provide some x ∈ V with i = f(x). But x2 = (1) for all x ∈ V, so that i2 = f(x)2 = f(x2) = f((1)) = 1, contradicting i2 = −1. Therefore, V and μ4 are not isomorphic. There are other ways to prove this result. For example, μ4 is cyclic and V is not μ4 has an element of order 4 and V does not μ4 has a unique element of order 2, but V has 3 elements of order 2. At this stage, you should really believe that μ4 and V are not isomorphic! Definition. If f : G → H is a homomorphism, define kernel 24f = {x ∈ G : f(x) = 1} and image f = {h ∈ H : h = f(x) for some x ∈ G}. We usually abbreviate kernel f to ker f and image f to im f. 24Kernel comes from the German word meaning “grain” or “seed” (corn comes from the same word). Its usage here indicates an important ingredient of a homomorphism.

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