By Olver P.J.

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That was a touch informal, so we supply a formal definition to be on the safe side. 34 Suppose that G is a group and that X ~ G. Let Sx = {H I H ~ G,X ~ H}. Observe that Sx i= 0 since G E Sx. 8) (X) = H. 12 asserts that the intersection of subgroups is a subgroup. This group (X) is therefore the unique smallest subgroup of G which contains X. Note that for H ~ G we have H ~ G if and only if H = (H). A word on X is any finite sequence w(X) of elements of X and their inverses. Suppose that Xl, X2, ...

11 Suppose that H :S G and N ~ G. It follows that H N :S G. 11 to show that HN :S G. Now IHIN E HN so HN t 0. Next suppose that h 1,h2 E Hand nl,n2 E N, so h1nl,h2n2 are generic elements of H N. Now h1nl(h2n2)-1 = h1nln2'lh2'1 = hlh2'lh2(nln2'1)h2'1. Now hlh2'l E Hand h 2(nln2'1)h2'1 is a conjugate of an element of the normal subgroup N, and so itself is an element of N. Thus h1nl (h2n2)-1 E H N so HN:S G. 3 Factor Groups Suppose that N ~ G. The set of left cosets G / N and the set of right cosets N\G are the same because gN = Ng for each 9 E G.

11 to show that HN :S G. Now IHIN E HN so HN t 0. Next suppose that h 1,h2 E Hand nl,n2 E N, so h1nl,h2n2 are generic elements of H N. Now h1nl(h2n2)-1 = h1nln2'lh2'1 = hlh2'lh2(nln2'1)h2'1. Now hlh2'l E Hand h 2(nln2'1)h2'1 is a conjugate of an element of the normal subgroup N, and so itself is an element of N. Thus h1nl (h2n2)-1 E H N so HN:S G. 3 Factor Groups Suppose that N ~ G. The set of left cosets G / N and the set of right cosets N\G are the same because gN = Ng for each 9 E G. In this circumstance, it is usual to stick to the notation G / N.