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 $G$?

 A: 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.
A: 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
