Let $G$ be a finite group, with $|G|=n$.
$\mbox{¿}$Are there explicit expressions, in terms of $n$, for the two indexes:
- $k:=[\operatorname{Sym}(G):\operatorname{Aut}(G)]=k(n)$
- $l:=[\operatorname{Aut}(G):\operatorname{Inn}(G)]$ $(=|\operatorname{Out}(G)|)=l(n)$
Why am I asking this: $\operatorname{Inn}(G)\cong G/Z(G)$, so that $n!=|\operatorname{Sym}(G)|=k|\operatorname{Aut}(G)|=kl|\operatorname{Inn}(G)|=kln/|Z(G)|$, i.e. $(n-1)!=kl/|Z(G)|$. In particular, if $G$ is centerless we get:
$$(n-1)!=kl \tag 1$$
We know that if $n=p^N$, $p$ prime, then $G$ can't be centerless (application of the Class Equation). So, I expect that if $n=p^N$, $p$ prime, $(1)$ can't hold. Unless I'm mistaken in the logic, I was hoping to show it explicitely once I had $k(n)$ and $l(n)$ in hands.