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Given a finite group $G$, let $p(G)$ denote the largest prime factor of the order of $G$. For the purpose of this question, we say that the group $G$ has smooth order if its order exceeds the order of the largest alternating group ${\rm A}_n$ for which $p({\rm A}_n) = p(G)$.

Question 1: Is it true that the only finite simple groups whose order is smooth in this sense are ${\rm PSU}(4,2)$, ${\rm PSU}(4,3)$, ${\rm O}(5,7)$, ${\rm O}^+(8,2)$, ${\rm McL}$, ${\rm PSU}(6,2)$, ${\rm Fi}_{22}$, ${\rm PSL}(6,3)$, ${\rm F}_4(2)$, ${\rm O}^+(8,4)$ and $^2{\rm E}_6(2)$?

Question 2: Is it true that for every finite simple group $G$ of sufficiently large order, we either have $G \cong {\rm A}_n$ for some $n$ or $p(G) > \log_2(|G|)$? — And if yes, is $|G| > |{\rm E}_7(4)| \approx 10^{80}$ sufficiently large in this sense?

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    $\begingroup$ The Hall-Janko sporadic group HJ is a fairly near miss, but does not quite make it ( as you probably checked). If it is relevant to the question at the end, I just put a paper on aRxiV which (in its most recent incarnation) prove that if a finite group $G$ has a faithful complex character of degree $n$, then $G$ has an Abelian normal subgroup $A$ such that either $[G:A] \leq 60^{n-1}$ or else $[G:A] \leq m!^{\frac{n-1}{m-2}},$ where $m > 151$ is the largest integer such that $A_{m}$ is a composition factor of $G$. It might be relevant to your question. $\endgroup$
    – Geoff Robinson
    Commented May 20, 2023 at 17:34
  • $\begingroup$ Your Question 2 does not use the concept of smooth order. Is that intentional? $\endgroup$
    – LSpice
    Commented May 20, 2023 at 22:48
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    $\begingroup$ @LSpice Yes, the term is used only in Question 1. $\endgroup$
    – Stefan Kohl ♦
    Commented May 21, 2023 at 6:39

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