# Abelian subgroups of $GL_n(\mathbb{F}_p)$

Let $p$ be a prime number, and let $k=\mathbb{F}_p$ be the field of $p$ elements. Let $G=GL_n(k)$. We know that

$$|G|=\prod_{i=0}^{n-1}(p^n-p^i)=p^{\binom{n}{2}}\prod_{i=0}^{n-1}(p^{n-i}-1)$$

so that the Sylow $p$-subgroups of $G$ have order $p^{\binom{n}{2}}$. One such subgroup is $U$, the upper-triangular unipotent subgroup consisting of all upper-triangular matrices with $1$'s on the diagonal. Let $A_{ij}=I_n+E_{ij}$ for $j>i$, where $E_{ij}$ is the matrix with a $1$ in the $ij$th entry and $0$'s elsewhere. If we find $i_1,\ldots,i_r,j_1,\ldots,j_r$ such that the $A_{i_kj_k}$ pairwise commute, then we have:

$$(\mathbb{Z}/p\mathbb{Z})^{\oplus r}\cong \langle A_{i_1j_1},\ldots,A_{i_rj_r}\rangle\subset U$$

Two questions:

• Is every copy of $(\mathbb{Z}/p\mathbb{Z})^{\oplus r}$ inside of $U$ conjugate under $G$ to a subgroup of the form $\langle A_{i_1j_1},\ldots,A_{i_rj_r}\rangle$?
• Do there exist distinct, conjugate subgroups of the form $\langle A_{i_1j_1},\ldots,A_{i_rj_r}\rangle$ and $\langle A_{i_1'j_1'},\ldots,A_{i_r'j_r'}\rangle$?

The largest value of $r$ for which such subgroups exist is $r=\lfloor n^2/4\rfloor$. In this case, I believe the answers to these questions are yes and no, respectively. I'd like to know if this also holds for smaller values of $r$.

• let me know if you write up your notes. I'd be interested in reading them. These are pretty important ideas in a pretty important group. – Jack Schmidt Jul 12 '13 at 17:13

For $r=1$: No and yes as long as $n \geq 4$.

For $r$ maximal and $n$ odd, the second question has the answer “yes” as well.

Typically, I think the answer to the first question is very definitely “no”. Every subgroup has a “straightening” of the same order and roughly the same embedding that is of roughly the form you describe (they are called algebra subgroups, and may not be exactly products of root subgroups; see page 109 of the 3rd book of the GLS revised CFSG; it is called Malcev's striaghtening). However, for $r$ maximal it might be “yes” anyways.

## $r=1$

For the first question: In $G=\operatorname{GL}(4,p)$, one has the following elementary abelian subgroup that is not conjugate to a root subgroup: $\langle A_{1,3} \cdot A_{1,4} \cdot A_{2,4}\rangle$. Every generator of every cyclic root subgroup has the property that $g-g^0$ has rank 1 which is invariant under GL conjugation, but this subgroup does not have that property.

For the second question: In $G=\operatorname{GL}(3,p)$, one has the following cyclic root subgroups are conjugate: $\langle A_{1,2} \rangle \cong_G \langle A_{2,3} \rangle$.

## $r$ maximal

For $r=(n-1)(n+1)/4 = \lfloor n^2/4 \rfloor$, that is, for $n$ odd and $r$ maximal, the second question's answer is no: there are two elementary abelian root subgroups of this maximal rank that are conjugate under the inverse-transpose map, but not conjugate in GL. Each are formed of block matrices: $$P_{i,j} = \left\{ \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} : A \in M_{i \times j}(\mathbb{F}_p) \right\}$$ The subgroups are $P_{(n-1)/2,(n+1)/2}$ and $P_{(n+1)/2,(n-1)/2}$.

• Hi Jack, thanks again. In another question of mine that you answered, you used the term 'root subgroups'. Can you clarify what those are? Thanks. – Jared Jun 28 '13 at 23:03
• Oh sorry, they are the subgroups generated by a $A_{i,j}$. – Jack Schmidt Jun 28 '13 at 23:05
• If $i<j$ then they are called positive root subgroups, if $j=i+1$ then they are called simple root subgroups. If you work over fields other than $\mathbb{Z}/p\mathbb{Z}$, then the root subgroups also include all the matrices with the same zero-pattern (so $I_n +\alpha E_{i,j}$ for any $\alpha$ in the field). Many of the same ideas that work for GL work for other groups of lie type, though the explicit formula for root subgroups will change. – Jack Schmidt Jun 28 '13 at 23:08
• An “algebra subgroup” is a subgroup of the form $\left\{ I_n + A : A \in \mathcal{A} \right\}$ where $\mathcal{A}$ is a set of matrices closed under addition, multiplication, and scalar multiplication. I think they are about as nice as products of root subgroups. You can use linear algebra to understand them, so that is good. – Jack Schmidt Jun 28 '13 at 23:11