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