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.)
Best Answer
In order for a group $G$ to be isomorphic to a semidirect product $G\cong H\rtimes K$, you must have:
In order to show that $S_4$ is isomorphic to $V_4\rtimes S_3$, you need, inter alia, to produce a subgroup of $S_4$ that is isomorphic to $S_3$; formally, $S_3$ is not a subgroup of $S_4$, because the elements of $S_4$ are bijections from $\{1,2,3,4\}$ to itself, while the elements of $S_3$ are bijections of $\{1,2,3\}$ to itself. The map $j$ is used to find a subgroup $\mathcal{K}$ that is isomorphic to $S_3$, but is inside $S_4$.