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Dear all,

I have the following question:

Let $M$ be a subgroup of $GL_n(p)$ where $p$ is a prime and let $M'$ denotes the derived subgroup of $M$. Can we conclude that $|M:M'|\leq p^n-1$?

Any counterexample or reference is very much appreciated. Thank you in advance.

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up vote 7 down vote accepted

REVISION: In fact, it is a consequence of the (now proved) so-called $k(GV)$-problem that the answer is indeed affirmative when $M$ has order prime to $p,$ and something rather stronger holds. There is a book about the $k(GV)$-problem by Peter Schmid. The $k(GV)$-problem is a special case of a problem of R. Brauer. The $k(GV)$-problem is to prove that if $G$ is a group of order prime to $p$ and $V$ is a faithful finite-dimensional ${\rm GF}(p)G$-module, then the semidirect product $GV$ has at most $|V|$ conjugacy classes. This has now been proved by a combination of authors, the final cases being handled by D.Gluck,K.Magaard, U.Riese and P.Schmid. There are examples when the bound is attained. Its proof does require the classification of finite simple groups. Note that one consequence of the $k(GV)$-problem is that under those hypotheses, $k(G) < |V|,$ where $k(G)$ denotes the number of conjugacy classes of $G.$ In particular, this yields $[G:G^{\prime}] < |V|,$ since $k(G)$ is also the number of complex irreducible characters of $G$, and $[G:G^{\prime}]$ is the number of complex linear characters of $G.$
Hence this does imply that $[M:M^{\prime}] \leq (p^{n}-1)$ when $M$ is a subgroup of order prime to $p$ of ${\rm GL}(n,p)$ (in fact, we can even conclude that $M$ has at most $p^{n}-1$ conjugacy classes).

The answer ( to this MO question) may still be affirmative if you stick to completely reducible subgroups $M$ of ${\rm GL}(n,p),$ though I am not certain of that. In this situation, these are essentially the groups with no non-identity normal $p$-subgroups- it is necessary to take a little care here, because the underlying module for $M$ may need to be replaced by the direct sum of its composition factors under the action of $M.$ However, if the subgroups $M$ with $O_{p}(M) = 1$ are understood, then that covers all necessary groups at least up to isomorphism. By looking at such subgroups $M,$ you eliminate counterexamples to the question such as those arising in Peter Mueller's answer. However, there are completely reducible subgroups of ${\rm GL}(n,p)$ which are not of order prime to $p,$ such as ${\rm GL}(n,p)$ itself. The completely reducible case reduces almost immediately to the case when $M$ is irreducible. After that, there is work to do: I don't see an immediate elementary argument, but Clifford theory begins to come into play( for example, it may be possibe to reduce to the case where the undelying module for $M^{\prime}$ is a direct sum of isomorphic irreducible modules). This may well be a difficult queston when $M^{\prime}$ is non-Abelian simple, or when $M$ itself is almost simple. The authors I would check out for relevant results here would be people like : M.Aschbacher,R.Guralnick, P.Kleidman, M.Liebeck, P.Tiep. Guralnick and Tiep and others have been trying to prove variants of the $k(GV)$-problem when $G$ acts completely reducibly on $V$. The general bounds are necessarily somewhat weaker than $|V|$. I do not know at present whether they would provide an affirmative answer to this MO question in the case that $M$ is a completely reducible subgroup of ${\rm GL}(n,p)$, or whether this special case has been considered by authors such as Guralnick and Tiep.

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Thank you very much for your very detailed explanation. Indeed, I do have the hypothesis that $M$ is an irreducible subgroup of $GL_n(p)$ but I didn't put it in my first post. – Uep Mar 17 '13 at 14:18

No, not in general. For instance $GL_4(2)$ is isomorphic to the alternating group $A_8$, and the direct product of the two Klein four groups acting regularly on $1,2,3,4$ and $5,6,7,8$, respectively, is abelian of order $16\gt 2^4-1$.

Actually, for any prime $p$, there is an abelian subgroup of order $p^4$ in $GL_4(p)$, just take the matrices of the form $\begin{pmatrix} E & A\\ 0 & E\end{pmatrix}$, where $E$ is the $2\times 2$ identity matrix, and $A$ an arbitrary $2\times 2$ matrix.

Generalizing to bigger $n$, there are more drastic counterexamples.

I would expect your result to be true if $M$ is a $p'$-group. At ant rate, Maschke + Schur easily show that $\lvert M\rvert\le p^n-1$ when $M$ is an abelian $p'$-group.

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In general ${\rm GL}_{2n}(q)$ has an abelian subgroup of order $(q-1)q^{n^2}$. – Derek Holt Mar 17 '13 at 11:42

Let $U\leq GL_n(p)$ be the subgroup of upper unitriangular matrices, and let $M\leq U$ be the subgroup with zeros everywhere except in first row and last column. Then $[M\colon M']=p^{2(n-2)}$. For $n\geq 4$, this subgroup $M$ is counter-example.

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