50 Chapter 1. Groups I There is another way to regard quotient groups. After all, we saw, in the proof of Lemma 1.46, that the relation on G, defined by a b if b−1a K, is an equivalence relation whose equivalence classes are the cosets of K. Thus, we can view the elements of G/K as equivalence classes, with the multiplication aKbK = abK being independent of the choices of representative. We remind the reader of Lemma 1.46(i): two cosets aK and bK of a subgroup K are equal if and only if b−1a K. In particular, when b = 1, then aK = K if and only if a K. We can now prove the converse of Proposition 1.62(ii). Corollary 1.75. Every normal subgroup K G is the kernel of some homomor- phism. Proof. Define the natural map π : G G/K by π(a) = aK. With this notation, the formula aKbK = abK can be rewritten as π(a)π(b) = π(ab) thus, π is a (surjective) homomorphism. Since K is the identity element in G/K, ker π = {a G : π(a) = K} = {a G : aK = K} = K, by Lemma 1.46(i). The next theorem shows that every homomorphism gives rise to an isomorphism and that quotient groups are merely constructions of homomorphic images. Noether emphasized the fundamental importance of this fact, and this theorem is often named after her. Theorem 1.76 (First Isomorphism Theorem). If f : G H is a homomor- phism, then ker f G and G/ ker f = im f. In more detail, if ker f = K, then ϕ: G/K im f H, given by ϕ: aK f(a), is an isomorphism. Remark. The following diagram describes the proof of the First Isomorphism The- orem, where π : G G/K is the natural map a aK and i: im f H is the inclusion. G f π H G/K ϕ im f i Proof. We have already seen, in Proposition 1.62(ii), that K = ker f is a normal subgroup of G. Now ϕ is a well-defined function: if aK = bK, then a = bk for some k K, and so f(a) = f(bk) = f(b)f(k) = f(b), because f(k) = 1. Let us now see that ϕ is a homomorphism. Since f is a homomorphism and ϕ(aK) = f(a), ϕ(aKbK) = ϕ(abK) = f(ab) = f(a)f(b) = ϕ(aK)ϕ(bK).
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