Problem 5.4.4 (Commutator subgroups of $S_4$ and $A_4$)

Question

Find the commutator subgroups of $S_4$ and $A_4$.

Richard Ganaye · Accepted Answer

\begin{proof} 
\begin{enumerate}
\item[(a)] {\bf Commutator subgroup of $S_4$.}

Let $S'_4$ be the commutator subgroup of $S_4$.

Since $\mathrm{sgn}(a^{-1}b^{-1} ab ) = 1$ for all $a,b \in S_4$, every commutator is in $A_4$, thus the subgroup generated by the commutators is contained in $A_4$:
$$ S'_4 \leq A_4.$$
(Alternatively, since $S_4/A_4 \simeq Z_2$ is abelian, then $ S'_4\leq A_4$ by Proposition 7 (4).)

Conversely, for all $a,b,c$,
$$(a\ b\ c) = (a\ c)^{-1}(b\ c)^{-1} (a\ c)(b\ c),$$
therefore every $3$-cycle is a commutator. Since $A_4$ is generated by the $3$-cycles, $A_4 \leq S'_4$, so
$$ S'_4 = A_4.$$

\item[(b)]  {\bf Commutator subgroup of $A_4$.}

Let $H$ be the subgroup of $A_4$ defined by
$$H =\langle (1\ 2)(3\ 4), (1\ 3)(2\ 4) \rangle = \{(),  (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}.$$
Then $H \unlhd A_4$, and $G/H \simeq Z_3$ is abelian, therefore, by Proposition 7 (4),
$$A'_4 \leq H.$$
\end{enumerate}
Conversely, for all $a,b,c,d$,
$$(a\ b) (c \ d) = (a\ b\ c)^{-1} (a\ b\ d)^{-1} (a\ b\ c) (a\ b\ d),$$
so every double transposition $(a\ b) (c \ d)$ is a commutator, so $H \leq A'_4$. In conclusion,
$$A'_4 = \langle (1\ 2)(3\ 4), (1\ 3)(2\ 4) \rangle.$$

\end{proof}
With Sagemath
\begin{verbatim}
sage:  G = SymmetricGroup(4)
sage:  C = G.commutator(); C
Permutation Group with generators [(2,3,4), (1,2,3)]
sage: C.is_isomorphic(AlternatingGroup(4))
True
sage: A = AlternatingGroup(4)
sage: D = A.commutator(); D
Permutation Group with generators [(1,2)(3,4), (1,4)(2,3)]
sage: D.list()
[(), (1,2)(3,4), (1,4)(2,3), (1,3)(2,4)]
\end{verbatim}

Problem 5.4.4 (Commutator subgroups of $S_4$ and $A_4$)

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Problem 5.4.4 (Commutator subgroups of $S_4$ and $A_4$)

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