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Math Help - Commutator subgroup of Sn

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    Commutator subgroup of Sn

    If Sn is the symmetric group what is the form of the commutator subgroup?
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    Quote Originally Posted by morganfor View Post
    If Sn is the symmetric group what is the form of the commutator subgroup?
    It depends what n is. If n\geq 5 then (S_n)' = A_n, to see a proof go here. If n=4 then (S_4)' must be a normal subgroup. This means (S_4) ' = S_4, A_4,V where V=\{(1),(12)(34),(13)(24),(14)(23)\}. It remains for you to eliminate which of these are not possible. We begin by noting that (S_4)'\not = S_4 because S_4/A_4 is abelian which would imply that S_4\subseteq A_4 a contradiction. Now for S_4/V to be abelian it would imply it is a cyclic group of order six. But it is impossible for \sigma V\in S_4/V to have order 6 since \sigma is a cycle. Thus, S_4/V\simeq S_3 is not abelian which implies A_4 must be the commutator subgroup. Try doing it for (S_3)'.
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