Let $G$ be a transitive subgroup of $S_4$.
By the orbit-stabilizer theorem, $4\mid|G|$.
Hence the possible order of $G$ is $4,8,12,24$.
So it is possible to list all the possible structure of $G$ by considering all the subgroups of $S_4$ with those orders.

However in An Introduction of Theory of Groups by Rotman page $69$ exercise $3.51$, it is done by considering the index $[G:G\cap V]$ where $V$ is the Klein $4$-group.

(i) If $m=[G:G\cap V]$, then $m\mid 6$
(ii) If $m=6$, then $G\cong S_4$; if $m=3$, then $G=A_4$; if $m=1$, then $G=V$; if $m=2$, then either $G\cong \Bbb{Z}_4$ or $G\cong D_8$.

I don't have idea to solve part (i) yet.

For part (ii), if $m=6$ then $G=S_4$ or $A_4$ by order consideration. But $A_4\cap V=V$, then $m=3$; a contradiction. Hence $G=S_4$.

If $m=3$, then similarly $G=S_4$ or $A_4$. But $S_4\cap V=V$, then $m=6$; a contradiction. Hence $G=A_4$.

If $m=1$,then $G=G\cap V\leq V$, hence $|G|\mid 4$. Thus $|G|=4$ and $G=V$.

So I am stuck in proving for the case $m=2$. So far I can note that $G\cap V \lhd G$.

  • 1
    $\begingroup$ Part (i) follows from the fact that $|S_4/V|=6$ and $G/G \cap V \cong GV/V \le S_4/V$. $\endgroup$ – Derek Holt Mar 16 '17 at 9:10

Though this is a question from last year, I'm giving a complete answer since there is no answer yet. I'm just doing the same problem. :)

Since $G$ is transitive, by the orbit-stabilizer theorem we know $4$ divides $|G|$. Therefore, the only possible candidates for $|G|$ are $4,8,12,24$. It's easy to prove that $|G|=24$ iff $G=S_4$ iff $m=6$, and $|G|=12$ iff $G=A_4$ iff $m=3$. If $m=1$, then $G\leq V$, but $|G|=|V|=4$, hence $G=V$. The remaining case is $m=2$.

As we noted above, if $m=[G:G\cap V]=2$, then $|G|=4$ or $8$.

Case 1. If $|G|=8$, then $|G\cap V|=4$ and $V\leq G$. Let $g\in G\backslash V$. Since $G\cap V\lhd G$ (by the second isomorphism theorem), we have $g^2\in V$. But this can happen iff $g$ is a $4$-cycle, say $g=(1\ 2\ 3\ 4)$. Set $a=(1\ 2)(3\ 4)$. Then $g^4=a^2=1$ and $aga=g^{-1}$, so $G\cong D_8$.

Case 2. If $|G|=4$, then $|G\cap V|=2$. Say $G\cap V=\left<(1\ 2)(3\ 4)\right>$. If $g\in G\backslash V$, then by the same argument we know $g$ is a $4$-cycle and thus $G=\left<(1\ 2\ 3\ 4)\right>\cong\mathbb{Z}_4$.

  • $\begingroup$ How can we get $g^2 \in V$ from $V \triangleleft G$? $\endgroup$ – Andrews Mar 25 '20 at 3:24

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