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

Solution 1:

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 $P_{(n+1)/2,(n-1)/2}$.