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

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

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

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

• Do you mean $< \epsilon$ rather than $< G$? Aug 19, 2019 at 17:57
• Since there are only finitely many finite groups of a given order, the conjecture can be stated more plainly as "all but finitely many finite groups have asymmetric fraction less than any given positive real," or equivalently as "$0$ is the only limit point of the set of asymmetric fractions of finite groups."
– anon
Aug 19, 2019 at 19:09
• If we replace ${\rm Aut}(G)$ with ${\rm Inn}(G)$, this is the commuting probability, which famously must be $\le 5/8$ if it is not $1$. IIRC the set of commuting probabilities of finite groups (maybe conjecturally) has order type the reverse of the ordinal $\omega^{\omega}$.
– anon
Aug 19, 2019 at 19:18
• This isn't quite the commuting probability, even for $\operatorname{Inn}(G)$, unless the group is also centerless, since the commuting probability counts every element of the group, whereas this concept doesn't distinguish between elements that correspond to the same inner automorphism. Aug 19, 2019 at 22:25
• Its' true for finite abelian groups. This was proved by Gary Sherman. Oct 8, 2019 at 5:47