In my book they give the following proof for $S_4 \cong V_4 \rtimes S_3$ :
Let $j: S_3 \rightarrow S_4: p \mapsto \left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ p(1) & p(2) & p(3) & 4 \end{array} \right)$
Clearly, $j(S_3)$ is a subgroup $S_4$ isomorphic with $S_3$, hence $j $ is injective. We identify $S_3 $ with $ j(S_3$). Also $V_4 \triangleleft S_4$ and clearly $V_4 \cap S_3 = \{I\}$. We now only have to show that $S_4 = V_4S_3$. Hence $V_4\cap S_3 = \{I\}$, we know that $\#(V_4S_3) = \#V_4 \#S_3 = 4 \cdot 6 = 24 = \#S_4$ thus $S_4 = V_4S_3$, which implies that $S_4 \cong V_4 \rtimes S_3$.
However, I am wondering what the function $j$ is actually used for in the proof? (I do not see the connection.)