<s>[INST] Slightly confused about the commutator subgroup Let's look at the group of all permutations of a set $\left\{ 1,2,3 \right\}$. There are exactly 6 elements in this group: $\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 2 & 3  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 3 & 2  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 1 & 3  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 3 & 1  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   3 & 1 & 2  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   3 & 2 & 1  \\
\end{matrix} \right)$. It is usually called the symmetric group $S_{3}$. Just out of curiosity I tried to find its commutator subgroup $\left[ {{S}_{3}},{{S}_{3}} \right]$. As a home assignment, I've already proven that the commutator subgroup is a normal subgroup, hence $\forall x\in {{S}_{3}}:x\left[ {{S}_{3}},{{S}_{3}} \right]=\left[ {{S}_{3}},{{S}_{3}} \right]x$. Except for the neutral element $\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 2 & 3  \\
\end{matrix} \right)$, I couldn't find a single permutation $p$ with the property: $\forall x\in {{S}_{3}}:xp=px$. For instance, $\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 3 & 2  \\
\end{matrix} \right)\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 1 & 3  \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 3 & 1  \\
\end{matrix} \right)\ne \left( \begin{matrix}
   1 & 2 & 3  \\
   3 & 1 & 2  \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 1 & 3  \\
\end{matrix} \right)\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 3 & 2  \\
\end{matrix} \right)$.
Am I missing something?
 [/INST] HINT:
$$\forall x\in {{S}_{3}}:x\left[ {{S}_{3}},{{S}_{3}} \right]=\left[ {{S}_{3}},{{S}_{3}} \right]x$$
does not mean that $xp = px$ for $p \in \left[ {{S}_{3}},{{S}_{3}} \right]$. It means, for $p \in \left[ {{S}_{3}},{{S}_{3}} \right]$, we have $xp = qx$ for some $q \in \left[ {{S}_{3}},{{S}_{3}} \right]$, not necessarily $q = p$.
</s>