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Let $\mathbf{F}_q$ be a finite field of order $q$ ($q$, an odd prime, or a power of the same). I know that for the matrix algebra $M(n,\mathbf{F}_q)$ (or $gl(n,q)$), the maximal dimension for a commutative subalgebra is $\left[\frac{n^2}{4}\right] + 1$ (this is a result of Schur (for $\mathbb{C}$) and Jacobson for any field).

So, now I move to the matrix group $GL_n(\mathbf{F}_q)$, and ask if there is a formula for the order of an abelian subgroup of $GL_n(\mathbf{F}_q)$ of maximal order?

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    $\begingroup$ This question is related. I think the answer for $n \ge 4$ is $(q-1)q^{n^2/4}$ when $n$ is even and $(q-1)q^{(n^2-1)/4}$ when $n$ is odd. For $n \le 3$ it is $q^n-1$. $\endgroup$
    – Derek Holt
    Commented Apr 7, 2020 at 9:06
  • $\begingroup$ Can one argue that the max abelian subgroup must be a $p$-group where $q$ ia a power of $p$, so it must be contained (up to conjugation) in the group of upper triangular matrices which is a Sylow $p$-subgroup. Then use induction on the upper central ceries of that Sylow subgroup? $\endgroup$
    – user6976
    Commented Apr 7, 2020 at 14:18
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    $\begingroup$ @MarkSapir It's slightly larger than a $p$-group, because you can include the scalar matrices. But yes I believe you can argue like that. I am sure this result is in the literature somewhere. $\endgroup$
    – Derek Holt
    Commented Apr 7, 2020 at 15:25
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    $\begingroup$ If $A$ is an abelian subgroup of the upper triangular group $U$, then $|A|\le q^{n^2/4}$ or $q^{(n^2-1)/4}$ according as $n$ is even or odd, and these upper bounds are attained -- even by elementary abelian "upper right-hand block" subgroups. If you order the above-diagonal spots $(i,j)$ lexicographically by $(j-i,j)$, and replace each element of an abelian subgroup $A$ of $U$ by just its "least" nonzero entry -- zeroes elsewhere -- I believe you obtain an abelian subalgebra $B$ of $gl(n,q)$. Combinatorial arguments show that $|A|=|B|\le q^{n^2/4}$ or $q^{(n^2-1)/4}$ as desired. $\endgroup$
    – Richard Lyons
    Commented Apr 7, 2020 at 18:29
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    $\begingroup$ This "homogenization" trick was used by Malcev in determining the largest abelian subalgebras of all the semisimple complex Lie algebras. A. I. Malcev, “Commutative subalgebras of semi-simple Lie algebras,” Izv. Akad. Nauk SSSR, Ser. Mat., 9, No. 4, 291–300 (1945). $\endgroup$
    – Richard Lyons
    Commented Apr 7, 2020 at 18:46

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