Problem 20.30 in the Kourovka Notebook asks whether the maximum size of a conjugacy class of a perfect and centreless finite group $G$ is bounded below by $|G|^{\frac{1}{2}}$. Clearly, there cannot be a lower bound better than $|G|^{\frac{2}{3}}$ (this can be seen when looking at the groups ${\rm PSL}(2,q)$ for $q \rightarrow \infty$). Computational investigations — I checked the perfect groups of order $\leq 1342740$ — suggest that that bound does indeed hold. However — since the question in the Kourovka Notebook only asks for the bound $|G|^{\frac{1}{2}}$ — can the better bound be refuted?
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asked Mar 14, 2022 at 11:57
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$\begingroup$ Did you try other simple groups like $A_n$? $A_n$ does not work. But there are only a few infinite series of finite simple groups, and their conjugacy classes are known. If that will not work, try wreath products. $\endgroup$– markvsCommented Mar 14, 2022 at 14:30
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$\begingroup$ @markvs Simple groups are unlikely to be counterexamples. -- For ${\rm PSL}(2,q)$, the asymptotics is just the given bound, and for other simple groups, even the bound $|G|^{\frac{3}{4}}$ seems to hold. $\endgroup$– Stefan Kohl ♦Commented Mar 14, 2022 at 15:21
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$\begingroup$ Then try wreath products. $\endgroup$– markvsCommented Mar 14, 2022 at 17:14
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$\begingroup$ Looks like the wreath products do not work: they are too large. I do not have other suggestions except to ask the authors of the problem (Vaughan-Lee, the other author has died). $\endgroup$– markvsCommented Mar 15, 2022 at 4:54
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$\begingroup$ See Neumann, Peter M.; Vaughan-Lee, M. R. An essay on BFC groups. Proc. London Math. Soc. (3) 35 (1977), no. 2, 213–237. $\endgroup$– markvsCommented Mar 15, 2022 at 10:03
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It doesn't really answer the question, but Bob Guralnick and I proved in our 2006 paper on commuting probability that if $G$ is a finite group with $F(G) = 1$, then the number of conjugacy classes $k(G)$ is at most $|G|^{\frac{1}{2}}$, so that $G$ has a conjugacy class of size at least $|G|^{\frac{1}{2}}$ (our proof requires CFSG). The proof shows that we can't do much better in general than $k(G) \leq |G|^{0.41}$ when $F(G) = 1$, but it might be possible to do better for perfect groups. Later edit: well,direct products of copies of $A_{5}$ show that even for perfect centerless groups, we can't do much better than $k(G) \leq |G|^{0.39}$, but in those examples, it is still the case that the largest conjugacy class still has size greater than $|G|^{\frac{2}{3}}.$ But this does suggest that improving the $\frac{1}{2}$ bound to $\frac{2}{3}$ might be delicate.
answered Mar 16, 2022 at 21:20