Does asymmetric fraction of finite groups tend to $0$?

Question

Let’s define asymmetric fraction of a finite group $G$ as the number $$\mathrm{af}(G) = \frac{|\{(g, a) \in G \times \mathrm{Aut}(G)\mid a(g) = g\}|}{|G|\cdot|\mathrm{Aut}(G)|}.$$ Equivalently it can be defined as $\mathrm{P}(A(X) = X)$, where $A$ and $X$ are independent uniformly distributed random elements of $\mathrm{Aut}(G)$ and $G$ respectively.

Is it true, that $$\forall \epsilon > 0,\; \exists N \in \mathbb{N},\; \forall G \big((\lvert\,G\,\rvert > n) \to (\mathrm{af}(G) < \epsilon)\big)?$$

I know, that $$\mathrm{af}(C_{p^n}) = \dfrac{p^n + \Sigma_{i = 1}^n p^ip^{n - 1 - i}(p - 1)}{p^{2n - 1}(p - 1)} = \dfrac{(np - n + 1)}{p^n(p - 1)}$$ and, that $$\mathrm{af}(G) \leq \frac{1}{2} + \dfrac{|\{g \in G\mid \forall a \in \mathrm{Aut}(G), \; a(g) = g\}|}{2|G|}.$$ However this is clearly not enough to prove the statement.

This question was already asked by me on MSE, but received no answers

Also, if we restrict the question to abelian case, then the statement will be true. It was proved by Gary Sherman in "What is the Probability an Automorphism Fixes a Group Element?"

Answer 1

This statement becomes correct when restricted to groups $G$ with trivial Fitting subgroup (ie with $F(G) = 1$), but at present I don't know how to prove it in general. The quantity $af(G)$ you consider is the number of ${\rm Aut}(G)$ orbits on $G$ divided by $|G|$.

When $Z(G) = 1$, $G \cong {\rm Inn}(G)$ embeds isomorphically into ${\rm Aut}(G)$, as a normal subgroup. In that case, your quantity $af(G)$ is certainly at most $\frac{k(G)}{|G|}$, where $k(G)$ is the number of conjugacy classes of $G$. The ratio $\frac{k(G)}{|G|}$ is the commuting probability of $G$. In the case that $F(G) = 1,$ Bob Guralnick and I proved that the commuting probability tends to zero as $|G| \to \infty$ (this does not require CFSG, though the convergence rate can be shown to be much faster using CFSG. There is also a more elementary proof of that fact due to P.M. Neumann, of which Guralnick and I were unaware at the time).

Later edit: Perhaps it would have been better to say at the beginning that it is true that ${\rm af}(G) \to 0$ as $[G:F(G)] \to \infty$, which is really what the quoted results imply.

Answer 2

Here is a possible road map to a solution. Let $G$ be a finite group such that $\def\af{\operatorname{af}}\def\Aut{\operatorname{Aut}}$$\af(G)$ is bounded away from $0$. We want to show that $|G|$ is bounded. By the theorem of Peter Neumann that Geoff Robinson mentioned, $G$ is bounded-by-abelian-by-bounded. In fact, $G$ has a bounded-index $2$-step nilpotent normal subgroup $H$ such that $|H'|$ is bounded. We can moreover assume that every $h \in H$ has a bounded-size $\Aut G$-orbit.

This almost reduces us to considering $H$ in place of $G$, except that automorphisms of $H$ need not extend to automorphisms of $G$. In any case, $G = H$ is an excellent model problem (i.e., $2$-step nilpotent with bounded commutator). These groups tend to have loads of outer automorphisms, so $H$ should have plenty of large $\Aut H$-orbits. (The archetypal example here is the extraspecial $p$-group of order $p^{2n+1}$ and exponent $p$, which has outer automorphism group containing $\operatorname{Sp}(2n, p)$.)

Finally we have an extension problem. Given that $H$ has bounded index in $G$, we need to show that at least one of the large $\Aut H$-orbits contains a large $\Aut G$-orbit. For example, what if $G$ is virtually abelian? (Split extensions such as a dihedral group are not interesting, because it's easy to extend automorphisms in that case.)