Suppose that $$1\to A\xrightarrow{f} B\xrightarrow{g}C\to 1$$ is an exact sequence of groups, where $A$ has order $85$ and $C$ has order $9$. Show that $B$ is abelian.
I have proved earlier that for a Sylow $3$-subgroup $S$ of $B$, we have that $g$ maps $S$ isomorphically to $C$. Further, I know that $|\text{Aut}(A)|=64$. I am interested in a solution that actually uses one or both of these facts in what I presume is the spirit of the problem.
I think I am able to sketch a solution in my head involving tracing out cases for the nature of the maps $f$ and $g$ and then showing that $C$ must be in the center of $B$ (which would allow us to conclude instantly), but I can't imagine that this is the morally correct perspective. As such, I ask that any solution try to use the aforesaid facts, if possible.
Any and all help would be appreciated.
You have already said that you can pretend that $C$ lies in $B$, so that $B$ has a normal subgroup $A$, a subgroup $C$, and $A\cap C=1$ with $B=AC$. Notice that as $A$ is normal, $C$ acts on $A$ by conjugation.
As $A$ has automorphism group a $2$-group, and $C$ is a $3$-group, the only map from $C$ to $\mathrm{Aut}(A)$ is the trivial map. Thus the conjugation of $C$ on $A$ is trivial, i.e., $C$ centralizes $A$. Thus $A\leq C_B(A)$ and $C\leq C_B(A)$, and thus $C$ is central in $B$. Thus $CA=C\times A$, and $B$ is abelian.