Adv Alegbra HW Solutions 78

Adv Alegbra HW Solutions 78 - G are 1 and G it follows that...

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78 of disjoint transpositions, and the analysis above shows that H = A 4 or H = S 4 . 2.130 Let { i 1 ,..., i r }⊆{ 1 , 2 ,... n } , and let F ={ σ A n : σ f xes all i with i ±= i 1 ,..., i r } . Prove that F = A r . Solution. Absent. 2.131 Prove that A 5 is a group of order 60 having no subgroup of order 30. Solution. If H A 5 has order 30, then it has index 2. By Proposi- tion 2.97(ii), H is a normal subgroup of A 5 , and this contradicts the fact that A 5 is a simple group. 2.132 Let X ={ 1 , 2 , 3 ,... } be the set of all positive integers, and let S X be the symmetric group on X . (i) Prove that F ={ σ S X : σ moves only f nitely many n X } is a subgroup of S X . Solution. Absent. (ii) De f ne A to be the subgroup of F generated by the 3-cycles. Prove that A is an in f nite simple group. Solution. Absent. 2.133 (i) Prove that if a simple group G has a subgroup of index n , then G is isomorphic to a subgroup of S n . Solution. If [ G : H ]= n , let X denote the family of all the cosets of H in G . The representation of G on the cosets of H gives a homomorphism ρ : G S X = S n with ker ρ H . Since ker ρ is a normal subgroup, and since the only normal subgroups of
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Unformatted text preview: G are { 1 } and G , it follows that ker ρ = { 1 } and ρ 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 represen-tation of G on the cosets of H gives a homomorphism ρ : G → S m with ker ρ ≤ H . If G is simple, we must have ker ρ = { 1 } , so that G is iso-morphic to a subgroup of S m . By Lagrange ’ s theorem, | H | | | S m | ; that is,...
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