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 $m$?



Is there a permutation group satisfying the following property?
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.


