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Let $G \neq 1$ be a finite perfect group which is not simple. Is it true that $G$ necessarily has a maximal subgroup whose derived subgroup has nontrivial core in $G$?

Remark 1: This holds for all such $G$ of order less than 100000.

Remark 2: In case the answer is negative, I would mainly be interested in a counterexample with nontrivial Frattini subgroup.

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  • $\begingroup$ There are not that many perfect groups of order less than $100000$. $\endgroup$
    – markvs
    Commented Apr 11, 2022 at 9:57
  • $\begingroup$ @markvs More precisely, there are 444 perfect groups of order less than 100000. Of these, 31 are simple and 1 is trivial -- so in this range of orders, there are 412 nontrivial perfect groups which are not simple. $\endgroup$
    – Stefan Kohl ♦
    Commented Apr 11, 2022 at 14:31
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    $\begingroup$ @markvs I checked that the assertion is true for all non-simple perfect groups of order less than 1000000. $\endgroup$
    – Stefan Kohl ♦
    Commented Apr 12, 2022 at 9:01
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    $\begingroup$ There are no counterexamples among quasisimple groups. If $G$ is quasisimple, let $p$ be a prime divisor of $Z(G)$, and $P$ a Sylow $p$-subgroup of $G$. Then by a transfer result, $P\cap Z(G)\le[P,P]$. Any maximal subgroup $M$ containing $P$ will then satisfy $P\cap Z(G)\le[M,M]$. $\endgroup$
    – Richard Lyons
    Commented Apr 12, 2022 at 14:47
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    $\begingroup$ Does there exist a finite simple group $S$, a field of prime order $K$ and a nontrivial irreducible $KS$-module $V$ such that for every proper subgroup $M$ of $S$ one has $V^M\neq 0$? (if so, $S\ltimes V$ is an example for the question). Even the answer for $K=\mathbf{C}$ would be of interest (and probably enough), and could be checked with a computer for groups $S$ for which maximal subgroups are known and character tables are known for $S$ and its maximal subgroups (I have only checked $|S|=60$: there's no $V$ then). $\endgroup$
    – YCor
    Commented Apr 17, 2022 at 15:16

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The answer to the title question is 'No.' Following YCor's comment, an example is furnished by $S=J_1$, the smallest Janko sporadic group, and its complex irreducible character $\chi$ of degree 76 (in Atlas notation, 76a). I have checked by hand (hopefully correctly), using the Atlas, that $(\chi\downarrow M,1_M)>0$ for all maximal subgroups $M$ of $J_1$. For any prime $p$ not dividing $|J_1|=2^3.3.5.7.11.19$, let $N_p$ be an $F_pS$-module affording the mod-$p$ reduction of $\chi$. Then, as YCor points out, the semidirect product $N_pS$ has the property that for every maximal subgroup $H\le N_pS$, $[H,H]$ does not contain $N_p$.

However, this example has trivial Frattini subgroup.

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  • $\begingroup$ If the character is not defined over $\mathbf{Q}$, the set of possible $p$ might be smaller than a subset of finite complement of the set of primes. However it should be a set of positive density in any case (by Chebotarev). $\endgroup$
    – YCor
    Commented Apr 18, 2022 at 16:04
  • $\begingroup$ @YCor: Good point, but the character is in fact rational-valued, and Schur indices are trivial over finite fields. $\endgroup$
    – Richard Lyons
    Commented Apr 18, 2022 at 16:09
  • $\begingroup$ Thanks a lot for your counterexample! Can you tell whether the answer to the question would get positive if one replaces non-simple perfect group by non-simple minimal non-solvable group? $\endgroup$
    – Leyli Jafari
    Commented Apr 20, 2022 at 9:04
  • $\begingroup$ Well, the top composition factor would be one of Thompson's $N$-groups, and their character tables are known, so if there's a similar counterexample, one could in principle find it. $\endgroup$
    – Richard Lyons
    Commented Apr 20, 2022 at 19:52

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