52 Chapter 1. Groups I Proof. Define a function f : H × K HK by f : (h, k) hk. Clearly, f is a surjection. It suffices to show, for every x HK, that |f −1(x)| = |H K|, where f −1(x) = {(h, k) H × K : hk = x} [because H × K is the disjoint union x∈HK f −1(x)]. We claim that if x = hk, then f −1(x) = {(hd, d−1k) : d H K}. Each (hd, d−1k) f −1(x), for f(hd, d−1k) = hdd−1k = hk = x. For the reverse inclusion, let (h , k ) f −1(x), so that h k = hk. Then h−1h = kk −1 H K call this element d. Then h = hd and k = d−1k, and so (h , k ) lies in the right side. Therefore, |f −1(x)| = |{(hd, d−1k) : d H∩K}| = |H∩K|, because d (hd, d−1k) is a bijection. The next two results are consequences of the First Isomorphism Theorem. Theorem 1.80 (Second Isomorphism Theorem). If H and K are subgroups of a group G with H G, then HK is a subgroup, H K K, and K/(H K) = HK/H. Proof. Since H G, Proposition 1.72 shows that HK is a subgroup. Normality of H in HK follows from a more general fact: if H S G and H is normal in G, then H is normal in S (if ghg−1 H for every g G, then, in particular, ghg−1 H for every g S). We now show that every coset xH HK/H has the form kH for some k K. Since x HK, we have x = hk, where h H and k K hence, xH = hkH. But hk = k(k−1hk) = kh for some h H, so that hkH = kh H = kH. It follows that the function f : K HK/H, given by f : k kH, is surjective. Moreover, f is a homomorphism, for it is the restriction of the natural map π : G G/H. Since ker π = H, it follows that ker f = H K, and so H K is a normal subgroup of K. The First Isomorphism Theorem now gives K/(H K) = HK/H. The Second Isomorphism Theorem gives the product formula in the special case when one of the subgroups is normal: if K/(H ∩K) = HK/H, then |K/(H K)| = |HK/H|, and so |HK||H K| = |H||K|. Theorem 1.81 (Third Isomorphism Theorem). If H and K are normal sub- groups of a group G with K H, then H/K G/K and (G/K)/(H/K) = G/H. Proof. Define f : G/K G/H by f : aK aH. Note that f is a (well-defined) function (called enlargement of coset), for if a G and a K = aK, then a−1a K H, and so aH = a H. It is easy to see that f is a surjective homomorphism. Now ker f = H/K, for aH = H if and only if a H, and so H/K is a normal subgroup of G/K. Since f is surjective, the First Isomorphism Theorem gives (G/K)/(H/K) = G/H.
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