Let $G$ be a finite non-abelian simple group. Question:

Does there always exist a maximal subgroup $M$ of $G$ such that $M$ has a non-trivial normal elementary abelian $2$-subgroup?

This seems to be the case for alternating groups (take $M$ a stabilizer of a $4$-set) and sporadic simple groups (look at tables in ATLAS). For groups of Lie type in characteristic $2$ it seems to me that one could choose $M$ to be a maximal parabolic subgroup.

Also if the answer to the question is no, what about the weaker question of $M$ containing a non-trivial normal abelian subgroup? And if the answer to either question is yes, is there a proof without classification?

  • 2
    $\begingroup$ I think you have answered the second question yourself (allowing use of the classification): it only remains to check simple groups of Lie type in odd characteristic p, and then any maximal subgroup containing a Borel subgroup ( Sylow p- normalizer) works $\endgroup$ Jan 26, 2018 at 14:55
  • $\begingroup$ Group of Lie type in char 2 include such groups as Suzuki groups for which the $SL_n$ intuition is possibly misleading. Is the notion of "maximal parabolic subgroup" well-defined and it is certain that they have a nontrivial abelian normal subgroup of 2-power order? $\endgroup$
    – YCor
    Jan 26, 2018 at 20:49
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    $\begingroup$ As long as a simple group has a characteristic p BN-pair, ( and the Suzuki groups are examples where p=2 and the BN-pair has rank 1), I think things are OK. All overgroups of the Borel B are parabolic subgroups. There must be some maximal subgroup containing B-possibly B itself-and this must be a parabolic ( necessarily a maximal parabolic). Any such parabolic P has a non-trivial normal p-subgroup U, and then Z(U) is a non-trivial Abelian normal p-subgroup of P. $\endgroup$ Jan 27, 2018 at 17:38
  • $\begingroup$ @GeoffRobinson: Thank you for the comments. Perhaps one could make a trivial modification and assume that $G$ has even order, even though the question might remain very difficult. For "classification-free", I would say relying on Feit-Thompson is still classification-free, as long as there is no case-by-case checking of all finite simple groups. $\endgroup$
    – spin
    Jan 29, 2018 at 10:41
  • $\begingroup$ Yes, using F-T is indeed not using the whole of CFSG, but I was using that example to illustrate the level of difficulty. In any case, I had deleted the comment. $\endgroup$ Jan 29, 2018 at 12:55

1 Answer 1


Unfortunately I made a mistake in my original answer. The answer to the question is no, but the only finite simple groups that do not have a maximal subgroup with nontrivial normal 2-subgroup are the groups $L_p(3)$ with $p$ a prime and $p \ge 5$.

Let's go through the finite nonabelian simple groups.

As you say, for $G=A_n$ we can take $M$ to be the stabilizer of a set of size $4$.

You have checked the sporadic groups in the ATLAS.

If $G$ is of Lie type in characteristic $2$, then we can take $M$ to be a maximal parabolic subgroup.

So we can assume from now on that $G$ is of Lie type in odd characteristic.

The geometric-type maximal subgroups of the classical groups are listed in the book by Kleidman and Liebeck. Some of these are not maximal in small dimensions. The complete classification in dimensions up to $12$ was completed in a book by myself, Bray and Roney-Dougal. So we have enough information to check the result in groups of odd characteristic, and in fact imprimitive groups can be used for $M$ in most cases.

In some of the lists of $M$ below, there are a few small $n$ and $q$ for which $M$ is not maximal, but in each of these cases you can check that there is some other maximal subgroup that works - more details on request.

For $G=L_2(q)$ we have a dihedral subgroup of order $(q-1)$ or $(q+1)$.

Form now on, the subgroup $M$ is described as a subgroup of the relevant quasisimple matrix group, so we have to take it modulo scalars to get the corresponding maximal of the simple group.

In general, for $G=L_n(q)$ with $n \ge 3$ and $q \ge 5$, we can take $M$ to be the imprimitive group with structure $(q-1)^{n-1}:S_n$.

For $L_n(3)$, the subgroup with this structure is not maximal (it preserves an orthogonal form). If $n=rs$ is composite with $r \ge 2$ and $s>2$, then there is an imprimitive subgroup ${\rm GL}_r(3) \wr S_s$ of ${\rm GL}_n(3)$, and its intersection with ${\rm SL}_n(3)$ projects onto a maximal subgroup of $L_n(3)$ with a nontrivial normal $2$-subgroup.

Also $L_3(3)$ has $S_4$ as maximal subgroup and $L_4(3)$ has $S_4 \times S_4$, but for primes $p \ge 5$, $L_p(3)$ has no maximal with the required property. This can be seen from the complete list of its maximal subgroups, and I checked it directly by computer for $p=5,7,11$.

$G=U_n(q)$, we can take $M$ to be the imprimitive subgroupn with structure $(q+1)^{n-1}:S_n$, and this really is maximal for all $q \ge 3$.

$G=S_{2n}(q)$, take $M={\rm Sp}_2(q)^n:S_n$.

The orthogonal groups all have reducible maximal subgroups that are stabilizers of non-degenerate subspaces of dimension $2$ with nontrivial normal $2$-subgroups.

I know very little about the maximal subgroups of the exceptional groups of Lie type, but I looked at Table 5.2 of

M.W. Liebeck, J. Saxl and G.M. Seitz, “Subgroups of maximal rank in finite exceptional groups of Lie type”, Proc. London Math. Soc. 65 (1992), 297-325.

In all cases there appears to be a maximal subgroup that looks like an imprimitive maximal of a classical groups, and has normal subgroups with structure $(q-1)^n$ or $(q+1)^n$, where $n$ is the Lie rank of the group.

  • $\begingroup$ Very nice answer. This is indeed a surprising outcome, which I for one initially would not have expected, and which makes the question a really good one. I suppose in another sense it indicates that there are maximal subgroups of some simple groups which don't reflect very much about the group itself. $\endgroup$ Jan 30, 2018 at 11:11
  • $\begingroup$ @GeoffRobinson unfortunately I failed to check the small exceptional cases correctly, and I found a mistake. But I guess the corrected answer is also interesting. $\endgroup$
    – Derek Holt
    Jan 30, 2018 at 12:55
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    $\begingroup$ Still a nice answer and a good question. $\endgroup$ Jan 30, 2018 at 17:45
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    $\begingroup$ What originally brought me to wonder about this question was Iwasawa's simplicity criterion: if $G$ is a finite group such that $G = [G,G]$ and $S$ is a primitive $G$-set such that $\operatorname{Stab}_G(x)$ has a normal solvable subgroup whose $G$-conjugates generate $G$, then $G$ is simple. So one could ask if there is a converse for this. Then we are basically asking for $G$ simple whether there exists a maximal subgroup $M$ of $G$ with a nontrivial normal solvable (equivalently nontrivial normal abelian) subgroup. This weaker statement holds, as noted in the comments by Geoff Robinson. $\endgroup$
    – spin
    Jan 31, 2018 at 19:53
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    $\begingroup$ Well ${\rm GL}(r,3)$ has an element of order $2$ in its centre, so the wreath product with $S_s$ contains an elementary abelian $2^s$ as normal subgroup. On passing to ${\rm PSL}(rs,3)$ we lose at most two of the factors from the $2^s$, so if $s \ge 3$ then there we have a nontrivial normal $2$-subgroup. $\endgroup$
    – Derek Holt
    Feb 1, 2018 at 11:48

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