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By Joseph J. Rotman

Fourth Edition

J.J. Rotman

An creation to the idea of Groups

"Rotman has given us a truly readable and helpful textual content, and has proven us many attractive vistas alongside his selected route."—MATHEMATICAL REVIEWS

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Proof. Define a function cp: S x T --+ ST by (s, t) H st. Since cp is a surjection, it suffices to show that if x E ST, then Icp-1(x)1 = IS n We show that cp-1(X) = {(sd,d- 1t): dES n T}. It is clear that cp-1(X) contains the right side. For the reverse inclusion, let (s, t), (a, f) E cp-1(X); that is, s, a E S, t, f E T, and st = x = af. Thus, s-la = tf- 1 E S n T; let d = s-la = tf- 1 denote their common value. Then a = S(S-l a) = sd and d- 1t = felt = f, as desired. • n There is one kind of subgroup that is especially interesting because it is intimately related to homomorphisms.

A right coset St has many representatives; every element of the form st for s E S is a representative of St. The next lemma gives a criterion for Lagrange's Theorem 25 determining whether two right co sets of S are the same when a representative of each is known. 8. If S :-;::; G, then Sa ifb- 1 aES). = Sb if and only if ab- 1 E S (as = bS if and only Proof. If Sa = Sb, then a = 1a E Sa = Sb, and so there is s E S with a = sb', -1 hence, ab = s E S. Conversely, assume that ab- 1 = rr E S; hence, a = rrb.

58. Let M be a maximal subgroup of G; that is, there is no subgroup S with M < S < G. 59 (Schur). Let f: G ..... H be a homomorphism that does not send every element of G into 1. If G is simple, then f must be an injection. 2. The Isomorphism Theorems 40 Direct Products Definition. If Hand K are groups, then their direct product, denoted by H x K, is the group with elements all ordered pairs (h, k), where h e Hand k e K, and with operation (h, k)(h', k') = (hh', kk'). It is easy to check that H x K is a group: the identity is (1, 1); the inverse (h, kti is (h- I , k- I ).

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