# Elementary Abelian $p$-Subgroups of $GL_n(\mathbb{F}_p)$

Let $p$ be a prime number. If $G$ is a finite group, an elementary abelian $p$-subgroup of $G$ of rank $r$ is any group $E\subset G$ such that $E\cong (\mathbb{Z}/p)^{\oplus r}.$

Let $\mathcal{E}_r$ be the collection of all elementary abelian $p$-subgroups of rank $r$ of the finite group $G=GL_n(\mathbb{F}_p).$ Is anything known about $\mathcal{E}_r$? I would be happy with a reference, or an answer to any of the following related questions:

• What is $|\mathcal{E}_r|$?
• How is $\mathcal{E}_r$ partitioned into orbits under the conjugation action of $G$?
• How many elements of $\mathcal{E}_r$ lie in a fixed Sylow $p$-subgroup of $G$?
• @Jared, let me know if any part of my answer interests you. It is already long, but I did not finish the literature search or the case-by-case analysis. I figured this was enough for you to tell if those sorts of answers are useful to you. – Jack Schmidt Jun 3 '13 at 16:38
• @JackSchmidt: Thanks so much for your answer. I will look at it more closely and get back to you. You are right to think it is best (for now) to only consider the case $p\ge n$. What I'd really like to know the answer to is my second question, or at least how many orbits exist. – Jared Jun 3 '13 at 16:50
• I'll look into the number of orbits for large $r$ specifically. It might take a day or two for me (I left Gorenstein-Lyons at the office). – Jack Schmidt Jun 3 '13 at 17:10
• BTW The only thing I found in G-L was Malcev's straightening trick. I still haven't written it up or figured out exactly how great it is. – Jack Schmidt Jul 12 '13 at 17:23

This is another non-answer just to record some basics and try to figure out what an answer for general $n$ could look like. I didn't bother including the corrections for p=2 since they are numerous and special. At the end I mention a way in which these sets are currently studied (the Quillen complex) which I think uses hard mathematics to answer easier questions than the ones you asked.

## n=2

$$\left|\mathcal{E}_r\right| = \begin{cases} 1 & r = 0 \\ (p+1) & r = 1 \\ 0 & r \geq 2 \end{cases}, \qquad \left|\mathcal{E}_r \cap P\right| = \begin{cases} 1 & r \\ 1 & r = 1 \\ 0 & r \geq 2 \end{cases}$$

If $n=2$, then the Sylow $p$-subgroups are order $p$ and intersect trivially. Sylow's theorem guarantees single orbits for ranks 0 and 1, and no groups of rank 2 or higher.

## n=3

$$(p \neq 2) \quad \left|\mathcal{E}_r\right| = \begin{cases} 1 & r = 0 \\ \left[(p^2+p+1)(p+1)\right] + \left[(p^3-1)(p+1)p\right] & r = 1 \\ \left[(p^2+p+1)\right] + \left[(p^2+p+1)\right] + \left[ (p^3-1)(p+1) \right] & r = 2 \\ 0 & r \geq 3 \end{cases}$$ $$(p \neq 2) \quad \left|\mathcal{E}_r \cap P\right| = \begin{cases} 1 & r = 0 \\ [1] + (p+1)[p] & r = 1 \\ [1] + [1] + (p-1)[1] & r = 2 \\ 0 & r \geq 3 \end{cases}\qquad\qquad\qquad\qquad\qquad\qquad\quad\!$$

If $n=3$, then a Sylow $p$-subgroup has order $p^3$ but is not Abelian, so the largest rank of an elementary abelian subgroup is 2. The subgroups of rank 1 divide into two classes, the ones contained in the center of a Sylow $p$-subgroup, and those that are not. The subgroups of rank 2 divide into three conjugacy classes, two of them conjugate under $\operatorname{Aut}(GL(3,p))$. The more “root” subgroups that are normal in a Borel subgroup are the two Aut-conjugate classes, and a more mixed subgroup forms the other class (in Aut(Sylow) all of these are conjugate, but in the Borel subgroup they are distinguished). When looking at the subgroups of a particular Sylow it is unclear to me where conjugacy should be considered (in P, in B, in G, in Aut(G)). The numbers in brackets are P-classes, and a coefficient on a bracketed expression describes how many P-classes fuse under GL.

## Remarks

The collection of the $\mathcal{E}_r$ form a graded poset called the Quillen complex which should contain enough information to reconstruct all of GL once $n$ is large enough ($n=4$ or so). The geometry and cohomology of such posets is of current interest to finite group theorists and combinatorialists. I suspect a fair amount is known in the case of general linear groups, but I worry much of it cannot answer simple questions like “how many things are we talking about?”

I haven't looked much at GL(4,p), but I believe this is approximately where things start to work “correctly,” as in, $\mathcal{E}_n$ is non-empty, and $\mathcal{E}_n \cap P$ has a single element on which GL acts as GL, or something similar.

Quillen's 1978 paper has inspired quite a lot of work (see the articles linked from its math review) and the paper itself proves a number of nice results: in general the poset of elementary abelian groups of rank at least 2 is homotopy equivalent (respecting conjugation) to the poset of all non-trivial p-groups. In the case of the general linear group, it is homotopy equivalent to the standard Grassmanian (or flag) complex. You can see how simple questions about “how many are there?” are replaced by more complicated ideas like “if we wanted to count things, taking into account inclusion and exclusion, what is the remainder of that (weighted) count mod p?” (this is Brown's work alluded to in Quillen's paper).

If you are interested in this sort of thing, chapter 4 of these online notes from Casolo (2000) are pretty reasonable. The book Subgroup Complexes, Smith, 2011 is quite good.

• Quillen, Daniel. “Homotopy properties of the poset of nontrivial p-subgroups of a group.” Adv. in Math. 28 (1978), no. 2, 101–128. MR493916 DOI:10.1016/0001-8708(78)90058-0
• From GLS no 3 page 109ish: The largest $r$ is the floor of $(\tfrac n2)^2$. $\mathcal{E}_{\left\lfloor (n/2)^2 \right\rfloor}$ contains a subgroup generated by root subgroups. I think the number of orbits of such subgroups is likely 1 or 2 based on the parity of n, but I am not yet sure what other subgroups of the same rank there will be. – Jack Schmidt Jun 3 '13 at 18:52
• Yes, this result is correct, and I believe that for even $n$, there is one orbit, and for odd $n$, there are $2$ orbits. For even $n$, an example of such a subgroup is the upper right $n/2\times n/2$ block. – Jared Jun 3 '13 at 21:18

Every element of $GL_n(F)$ of order a power of $p=\operatorname{char} F$ is unipotent, meaning its characterisitc polynomial is $(X-1)^n$. It is therefore conjugate in $GL_n(F)$ to an upper triangular matrix with entries $1$ on the main diagonal, and conjugacy classes of such elements are characterised by their common Jordan type, a partition of $n$. The actual order of an element is the least power of $p$ that is no less than the size of the largest Jordan block, so to have order $p$ all Jordan blocks should be of size at most$~p$. (In particular if $n\leq p$, then the set of all unitpotent upper triangular matrices, which is always a Sylow $p$ subgroup, contains non-identity elements of order $p$ only.)

This was for individual elements, but for two $p$-elements to commute, they must be simultaneously trigonalisable; therefore every elementary $p$-subgroup is conjugate to an elementary $p$-subgroup of the upper triangular matrices (alternatively use that every $p$-subgroup is contained in a Sylow $p$-subgroup). This still answers none of your questions, but should help getting under way.

• I think this is important to emphasize how aspects of the question only occur in the Sylow (though conjugacy, overlap, and double counting are tricky, I believe). Also when $p \geq n$ it points out that rank 1 subgroups are very plentiful, and I hope it is clear, behave fairly differently than their rank 2 colleagues. I also suggest focussing on $p \geq n$ (or at least $p \geq n-1$) to avoid some bizarre special cases. (+1) – Jack Schmidt Jun 3 '13 at 16:43
• I am aware of this result that every $E$ conjugates to a subgroup of the upper triangular matrices. Given some arbitrary $E$ of rank $r$, how many subgroups of upper triangular groups exist to which $E$ can be conjugated? Perhaps we can answer my question by finding the result for upper triangular subgroups, and relating this to what I want to know. – Jared Jun 3 '13 at 16:46