Prove that $S_4 \cong V_4 \rtimes_\phi S_3$ for any isomorphism $\phi: S_3 \to \text{Aut}(V_4)$

Question

Note that $\text{Aut}(V_4) \cong S_3$. I know how to prove that $S_4$ isomorphic to some semidirect product of $V_4$ and $S_3$. I know if it works for an isomphorism it works for any isomorphism. However, I'm having trouble seeing that the $\phi$ must be an isomorphism of $\phi: S_3 \to \text{Aut}(V_4)$ (i.e. the kernel is trivial). Is there a better way to check this without doing each case (kernel cannot be 3-cycles, or all of $S_3$)?

Answer 1

To show that $S_4 \cong V_4 \rtimes S_3$, first note that $V_4$ is isomorphic to the double transpositions in $S_4$, and this $V_4$ is normal in $S_4$. Consider an isomorphic copy of $S_3$ in $S_4$ in the usual way. Note that their intersections are trivial. Denote the two subgroups as $H$ and $K$, then $HK$ is a subgroup of $S_4$ of size $\frac{|H||K|}{|H \cap K|} = 4 \cdot 6 = |S_4|$, so $HK$ is equal to $S_4$, meaning $S_4$ is a semidirect product of $V_4$ and $S_3$.

To show that $S_4 \cong V_4 \rtimes_\phi S_3$ for some isomorphism of $S_3 \to \text{Aut}(V_4)$ (rather than, more generally, some homomorphism), note that the kernel of $\phi$ must be a normal subgroup of $S_3$. We make use of Jyrki's comment. The only nontrivial normal subgroups of $S_3$ are $C_3$ and all of $S_3$, which includes $C_3$. Note that $V_4$ is abelian. If $C_3$ is in the kernel of $\phi$, then

$$\{ (h, k) \in V_4 \rtimes_\phi S_3 | h \in V_4, k \in C_3 \}$$

is an abelian subgroup of order 12. However, $V_4$ has no abelian subgroup of order 12. Therefore the kernel of $\phi$ must be trivial, i.e. it is an isomorphism of $S_3 \to \text{Aut}(V_4)$.

To show that $S_4$ is a semidirect product of $V_4$ and $S_3$ for any isomorphism of $S_3 \to \text{Aut}(V_4)$, see the answer to this question.