Question about proof for $S_4 \cong V_4 \rtimes S_3$ 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.)


It is only used to identify the subgroup S3 of S4, and is only needed as a technicality.

If you view S3 as bijections from {1,2,3} to {1,2,3} and S4 as bijections from {1,2,3,4} to {1,2,3,4}, and you view functions as having domains and ranges (not just rules), then no element of S3 is an element of S4.  The function j allows you to view elements of S3 as bijections of {1,2,3,4} that happen to leave 4 alone.  Then the elements of S3 (really j(S3)) are elements of S4, and so you can talk about it being a subgroup.
The statement of the theorem appears to mention external semi-direct products, but the proof uses internal semi-direct products. To use an internal semi-direct product, you need subgroups.
