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.

  • $\begingroup$ Do you mean $< \epsilon$ rather than $ < G$? $\endgroup$ Aug 19, 2019 at 17:57
  • $\begingroup$ 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." $\endgroup$
    – anon
    Aug 19, 2019 at 19:09
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    $\begingroup$ 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 $\omega^{\omega}$. $\endgroup$
    – anon
    Aug 19, 2019 at 19:18
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    $\begingroup$ 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. $\endgroup$ Aug 19, 2019 at 22:25
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    $\begingroup$ Its' true for finite abelian groups. This was proved by Gary Sherman. $\endgroup$
    – James
    Oct 8, 2019 at 5:47


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