# Thread: Commutator subgroup of Sn

1. ## Commutator subgroup of Sn

If Sn is the symmetric group what is the form of the commutator subgroup?

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

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# commutator grouop of sn proof

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