Is there a permutation group satisfying the following property?

Question

Let $n=3^m$ for some positive integer $m$. Let $G\leq S_n$ be a transitive permutation group on $n$ letters. Denote the largest normal subgroup of $G$ with odd order by $O_{2'}(G)$. My question is the following:

Does there exist $G$ such that $G/O_{2'}(G)\cong A_4$ or $S_4$ for suitable $m$?

Answer 1

Let $\Omega=\{1,2,3,4,5,6,7,8,9\}$ and let $G\leq\mathrm{Sym}(\Omega)$ be the group generated by the following permutations:

• $(1, 2, 9)$
• $(4, 5)(7, 8)$
• $(1, 4, 7)(2, 5, 8)(3, 6, 9)$
• $(3, 6)(4, 7)(5, 8)$

This is an example with $n=9$, with $G/O_{2'}\cong S_4$. If you want $G/O_{2'}\cong A_4$, remove the last generator.

To get larger examples, we can simply use direct products: let $H$ be a group of odd order acting transitively on $\Omega'$, with $|\Omega'|=3^{m-2}$ (for example, an elementary abelian $3$-group acting regularly). We can view $G\times H$ as a permutation group on $\Omega\times\Omega'$ with coordinate wise action, and it has the required properties.

Answer 2

This answers the first, unedited, version of the question. I leave it because the question may be edited again. If you take $G=S_4$ then 1)$G/O_{2'}(G)=S_4$ and 2) $G$ is inside $S_9$ and any $S_{3^m}$ for $m>1$.