Is $GL_n(\mathbb{F}_2)$ generated by a Jordan cell of size $n$ and its transpose?

Question

Let $n\geq 3$. Consider the group $GL_n(\mathbb{F}_2)$, and let $J_n=\pmatrix{1 & 1 & & \cr & 1 & 1 & & \cr & & 1 & 1 \cr & & & \ddots&\ddots}$ the Jordan cell of size $n$ (wrt to the eigenvalue $1$)

Since $GL_n(\mathbb{F}_2)$ is a simple group, it is known that it can be generated by two elements. If I understood correctly, if you pick two matrices "randomly", the two matrices will be generators, but I was curious to see if I could find two "not-too-complicated" generators.

Few tests with a CAS suggest that $GL_n(\mathbb{F}_2)$ is generated by $J_n$ and $J_n^t$ for all $n\geq 3, n\neq 4$.

I have no idea if this is true in general and, if so, how to prove it, but if true, it should certainly be well-known. I would appreciate if anyone could point me out a reference, or could provide me with a proof.

As a side question, if this is true, it would be nice to have a conceptual reason for the fact that it does not for $n=4$.

Finally, if there is a nice family of two generators for which it is easy/easier to prove it is indeed generating, I would happy to see it.

Ideally, the proofs should not use the fact that $GL_n(\mathbb{F}_2)$ is simple, if possible.

Answer 1

The answer is yes, and follows from the main result of the following paper.

R. Gow and M. C. Tamburini, Generation of $\operatorname{SL}(n,\textbf{Z})$ by a Jordan unipotent matrix and its transpose. Linear Algebra Appl. 181 (1993), 63–71.

More generally for $p$ prime and $n \geq 2$, the group $\operatorname{SL}(n,p)$ is generated by a unipotent Jordan block and its transpose, except when $(n,p) = (4,2)$.