Section 1.6. Quotient Groups 53 The Third Isomorphism Theorem is easy to remember: the Ks can be canceled in the fraction (G/K)/(H/K). We can better appreciate the First Isomorphism Theorem after having proved the third one. The quotient group (G/K)/(H/K) consists of cosets (of H/K) whose representatives are themselves cosets (of K). A direct proof of the Third Isomorphism Theorem could be nasty. The next result, which can be regarded as a fourth isomorphism theorem, de- scribes the subgroups of a quotient group G/K. It says that every subgroup of G/K is of the form S/K for a unique subgroup S ⊆ G containing K. Proposition 1.82 (Correspondence Theorem). Let G be a group, let K G, and let π : G → G/K be the natural map. Then S → π(S) = S/K is a bijection between Sub(G K), the family of all those subgroups S of G that contain K, and Sub(G/K), the family of all the subgroups of G/K. Moreover, T ⊆ S ⊆ G if and only if T/K ⊆ S/K, in which case [S : T ] = [S/K : T/K], and T S if and only if T/K S/K, in which case S/T ∼ = (S/K)/(T/K). Remark. The following diagram is a way to remember this theorem: G S G/K T S/K K T/K {1} Proof. Define Φ: Sub(G K) → Sub(G/K) by Φ: S → S/K (it is routine to check that if S is a subgroup of G containing K, then S/K is a subgroup of G/K). To see that Φ is injective, we begin by showing that if K ⊆ S ⊆ G, then π−1π(S) = S. As always, S ⊆ π−1π(S). For the reverse inclusion, let a ∈ π−1π(S), so that π(a) = π(s) for some s ∈ S. It follows that as−1 ∈ ker π = K, so that a = sk for some k ∈ K. But K ⊆ S, and so a = sk ∈ S. Assume now that π(S) = π(S ), where S and S are subgroups of G containing K. Then π−1π(S) = π−1π(S ), and so S = S as we have just proved in the preceding paragraph hence, Φ is injective. To see that Φ is surjective, let U be a subgroup of G/K. Now π−1(U) is a subgroup of G containing K = π−1({1}), and π(π−1(U)) = U. Now T ⊆ S ⊆ G implies T/K = π(T ) ⊆ π(S) = S/K. Conversely, assume that T/K ⊆ S/K. If t ∈ T , then tK ∈ T/K ⊆ S/K and so tK = sK for some s ∈ S. Hence, t = sk for some k ∈ K ⊆ S, and so t ∈ S. Let us denote S/K by S∗. To prove that [S : T ] = [S∗ : T ∗], it suﬃces to show that there is a bijection from the family of all cosets of the form sT , where s ∈ S, and the family of all cosets of the form s∗T ∗, where s∗ ∈ S∗, and the reader may

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