Let $S_3$ be the symmetric group on 3 elements, i.e., the group of bijections from $\{1,2,3\}$ to itself. Let $\sigma \in S_3$ such that $\sigma(1) = 2, \sigma(2) = 3, \sigma(3) = 1$.
I have already shown the following points:
- $|S_3| = 6$
- $\langle \sigma \rangle $ is normal in $S_3$
From shown results in the lecture I know that
$S_3/\langle \sigma \rangle $ is abelian $\implies [S_3,S_3] \subseteq \langle \sigma \rangle$. Further I know that $S_3$ is not abelian and $[S_3, S_3]$ is normal in $S_3$.
How can I conclude that $[S_3, S_3] = \langle \sigma \rangle$?
My thought:
$${\rm card}(S_3) = {\rm card}([S_3,S_3]) \cdot {\rm card}(S_3/[S_3,S_3])$$
or more precisely
$$6 ={\rm card}([S_3,S_3]) \cdot{\rm card}(S_3/[S_3,S_3])$$
but since $[S_3,S_3] \subseteq \langle \sigma \rangle$ we have that ${\rm card}([S_3,S_3]) \leq{\rm card}(\langle \sigma \rangle)$ and since $[S_3,S_3]$ is not trivial its cardinality can not be $1$. So it is either $2$ or $3$.
How can I proceed from here on?