Question

9) A group G is called solvable if there is a sequence of subgroups such that each quotient Gi/Gi-1 is abelian. Here Gi-1 Gi means Gi-1 is a normal subgroup of Gi. For example, any abelian group is solvable: If G s abelian, take Go f1), Gi- G. Then G1/Go G is abelian and hence G is solvable (a) Show that S3 is solvable Suggestion: Let Go- [l),Gı-(123)), and G2 -G. Here (123)) is the subgroup generated by the 3-cycle (123). (b) Let n 2 5. Let H be a subgroup of the symmetric group Sn that contains all the 3-cycles. Suppose that K is a normal subgroup of H such that the quotient H/K is abelian. Show that K also contains all the 3-cycles. Suggestion: Consider the quotient homomorphism φ : H HJK and use question (8) of HW1. Also, it may be useful to show that an arbitrary 3-cycle (ijk) E Sn can be expressed as στο"11" i for some 3-cycles σ, τ ε Sn. For example (412) can be expressed as (123) (145) (123)-(145)-1. (c) Prove that for n 5, Sn is not solvable.2 Suggestion: Assume by contradiction that Sn is solvable and use part (b) repeatedly to end up with a contradiction

Answer 1

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