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. •

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2010 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.