Is $PSL(2,13)$ a chief factor of the automorphism group of a $\{2,3\}$-group? Does there exists a group $H$ of order $2^7\cdot 3^4$, such that $\mathrm{PSL}(2, 13)$ is a chief factor of $\mathrm{Aut}(H)$?
 A: Here is a sketch of a proof that the answer is no.
Lemma If $N$ is an elementary abelian normal subgroup of a group $G$, then the subgroup of ${\rm Aut}(G)$ consiting of automorphisms that fix $N$ and induce the indentity on $N$ and $G/N$ is an elementary abelian $p$-group. (It is actually the group $Z^1(G/N,N)$ of $1$-cocycles of $G/N$ on $N$.)
Corollary If $1=N_0 < N_1 < N_2 < \cdots N_k =G$ with all $N_i \lhd G$ and each $N_{i+1}/N_i$ elementary abelian, then the subgroup of ${\rm Aut}(G)$ of automorphisms inducing the identity on all $N_{i+1}/N_i$ is solvable.
Now any group $G$ of order $2^7 \times 3^4$ is solvable and hence has a sequence of characteristic subgroups $1=N_0 < N_1 < N_2 < \cdots N_k =G$ with each $N_{i+1}/N_i$ elementary abelian.
By the corollary above, if ${\rm PSL}(2,13)$ occurs as a chief factor of ${\rm Aut}(G)$, then it must be a chief factor of a subgroup of ${\rm Aut}(N_{i+1}/N_i)$ for some $i$, and ${\rm Aut}(N_{i+1}/N_i) \cong {\rm PSL}(k,2)$ for some $k \le 7$ or ${\rm PSL}(k,3)$ for some $k \le 4$.
But, you can check easily that  ${\rm GL}(7,2)$ does not have order divisible by $13$, and  ${\rm GL}(4,3)$ does not have order divisible by $7$, whereas  $13$ and $7$ both divide $|{\rm PSL}(2,13)|$, so this cannot happen.
