# Transitive subgroup of $S_4$

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 $$G\cap V \lhd G$$.

• Part (i) follows from the fact that $|S_4/V|=6$ and $G/G \cap V \cong GV/V \le S_4/V$. Mar 16, 2017 at 9:10

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 $G=\left<(1\ 2\ 3\ 4)\right>\cong\mathbb{Z}_4$.
• How can we get $g^2 \in V$ from $V \triangleleft G$? Mar 25, 2020 at 3:24
• @Andrews $|G/V|=2$, so $\widehat g^2=\widehat e$ in the factor group $G/V$. Apr 2 at 18:54