# 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 $n$ 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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# commutator grouop of sn proof

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