Is there a way to prove other than using a counterexample that $\sum_4/V_4$ is not abelian?

Question

Suppose $\sum_4/V_4$ abelian, then if $a,b\in \sum_4$, then $aV_4bV_4=bV_4aV_4$, so $abV_4=baV_4$, which is only possible if there exists $ a^{-1}b^{-1}ab\in V_4$. If $a=(1,2),b=(1,2,3,4)$ then $a^{-1}b^{-1}ab=(1,3,2)$ but $(1,3,2)\not \in V_4$ so $\sum_4/V_4$ is not an abelian group.

Is there a way to show it differently?

*$\sum_4$ is the symmetric group.

*$V_4$ is Klein four-group.

*$\sum_4/V_4$ is the quotient group.

Answer 1

$V=\{e,(12)(34), (13)(24), (14)(23)\}$ is a normal subgroup of $S_4$.

$S_3=\{e,(123),(132), (12),(13), (23)\}$ is a subgroup of order $6$ such that $V\cap S_3=\{e\}$.

Hence $V S_3$ is a subgroup of $S_4$ whose order is $24$: that is $S_4=VS_3$.

Now use the $HK/K\simeq H/H\cap K$ theorem to see that $S_4/V\simeq S_3/\{e\}\simeq S_3$.

Answer 2

If $S_4/V_4$ would be abelian, the commutator subgroup $[S_4,S_4]=A_4 \subseteq V_4$, which is nonsense, since $|A_4|=12 \nmid |V_4|=4$.