Section 1.6. Quotient Groups 51 It is clear that im ϕ ⊆ im f. For the reverse inclusion, note that if y ∈ im f, then y = f(a) for some a ∈ G, and so y = f(a) = ϕ(aK). Thus, ϕ is surjective. Finally, we show that ϕ is injective. If ϕ(aK) = ϕ(bK), then f(a) = f(b). Hence, 1 = f(b)−1f(a) = f(b−1a), so that b−1a ∈ ker f = K. Therefore, aK = bK, by Lemma 1.46(i), and so ϕ is injective. We have proved that ϕ: G/K → im f is an isomorphism. • Note that iϕπ = f, where π : G → G/K is the natural map and i: im f → H is the inclusion, so that f can be reconstructed from G and K = ker f. Given any homomorphism f : G → H, we should immediately ask for its ker- nel and image the First Isomorphism Theorem will then provide an isomorphism G/ ker f ∼ = imf. Since there is no significant difference between isomorphic groups, the First Isomorphism Theorem also says that there is no significant difference between quotient groups and homomorphic images. Example 1.77. Let us revisit Example 1.59, which showed that any two cyclic groups of order m are isomorphic. If G = a is a cyclic group of order m, define a function f : Z → G by f(n) = an for all n ∈ Z. Now f is easily seen to be a homomorphism it is surjective (because a is a generator of G), while ker f = {n ∈ Z : an = 1} = m , by Proposition 1.26. The First Isomorphism Theorem gives an isomorphism Z/ m ∼ = G. We have shown that every cyclic group of order m is isomorphic to Z/ m , and hence that any two cyclic groups of order m are isomorphic to each other. Of course, Example 1.74 shows that Z/ m = Im, so that every finite cyclic group of order m is isomorphic to Im. The reader should have no diﬃculty proving that any two infinite cyclic groups are isomorphic to Z. Example 1.78. What is the quotient group R/Z? Take the real line and identify integer points, which amounts to taking the unit interval [0, 1] and identifying its endpoints, yielding the circle. Define f : R → S1, where S1 is the circle group, by f : x → e2πix. Now f is a homomorphism that is, f(x + y) = f(x)f(y). The map f is surjective, and ker f consists of all x ∈ R for which e2πix = cos 2πx + i sin 2πx = 1 that is, cos 2πx = 1 and sin 2πx = 0. But cos 2πx = 1 forces x to be an integer since 1 ∈ ker f, we have ker f = Z. The First Isomorphism Theorem now gives R/Z ∼ = S1. Here is a counting result. Proposition 1.79 (Product Formula). If H and K are subgroups of a finite group G, then |HK||H ∩ K| = |H||K|. Remark. The subset HK = {hk : h ∈ H and k ∈ K} need not be a subgroup of G, but see Proposition 1.72 and Exercise 1.80 on page 59.

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