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Unformatted text preview: G are { 1 } and G , it follows that ker = { 1 } and is an injection. (ii) Prove that an in f nite simple group has no subgroups of f nite index n > 1. Solution. If G is an in f nite simple group, and if G has a subgroup H of f nite index n > 1, then (i) shows that the in f nite group G is isomorphic to a subgroup of the f nite group S n , a contradiction. 2.134 Let G be a group with  G  = mp , where p is a prime and 1 < m < p . Prove that G is not simple. Solution. By Cauchy s theorem, G has an element of order p , and the (cyclic) subgroup H it generates has order p and index m . The representation of G on the cosets of H gives a homomorphism : G S m with ker H . If G is simple, we must have ker = { 1 } , so that G is isomorphic to a subgroup of S m . By Lagrange s theorem,  H    S m  ; that is,...
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 Fall '11
 KeithCornell

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