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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.)

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If $\times_{\sigma}$ is meant to denote the semidirect product, you can use \rtimes to produce the semidirect product symbol $\rtimes$. – Arturo Magidin Jun 23 '11 at 19:55
Thanks for the hint! – user12205 Jun 23 '11 at 20:17
up vote 3 down vote accepted

In order for a group $G$ to be isomorphic to a semidirect product $G\cong H\rtimes K$, you must have:

  • A subgroup $\mathcal{H}$ of $G$ that is isomorphic to $H$;
  • A subgroup $\mathcal{K}$ of $G$ that is isomorphic to $K$;
  • Normality of $\mathcal{H}$, $\mathcal{H}\triangleleft G$;
  • Trivial intersection of $\mathcal{H}$ and $\mathcal{K}$; and
  • $G = \mathcal{H}\mathcal{K}$.

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 $S_4$.

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A maybe interesting addendum is that for finite groups you would only have to check two out of the following three conditions (instead of the last two bullets) (i) $\mathcal H\cap\mathcal K=1$ (ii) $\mathcal H\mathcal K=G$ (iii) $|G|=|\mathcal H|\cdot|\mathcal K|$. – Myself Jun 23 '11 at 21:18

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.

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