Is there any closed formula for the function $m(n)$, that is defined as the maximal $m$, such that there is $GL(n, 2)$ has a subgroup isomorphic to $\mathbb{Z}_2^m$?
The only things I know currently, is that $m(1) = 0$ (as $GL(1, 2)$ is trivial) and $m(2) = 1$ (as $GL(2, 2)$ is isomorphic to $S_3$). With $GL(3, 2)$ the things become very complicated (as it is a simple group of order 168), so $m(3)$ or any other $m(n)$ with $3 \leq n$ is unknown to me currently.
Any help will be appreciated.
Claim: $m(n)\geq\lfloor n^2/4\rfloor$.
Proof: For all $1\leq a\leq n-1$, there exists a subgroup of $GL(n,2)$ isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{a\times(n-a)}$, given by
$$ \left\{\left(\begin{matrix}I_{a\times a} & A \\ 0 & I_{(n-a)\times(n-a)}\end{matrix}\right)\middle|A\in(\mathbb{Z}/2\mathbb{Z})^{a\times(n-a)}\right\}. $$
When $a=\lfloor n/2\rfloor$, we have $a(n-a)=\lfloor n^2/4\rfloor$.
I believe $\lfloor n^2/4\rfloor$ is the best possible. This might be proved by looking at the subgroup of all upper-triangular matrices in $GL(n,2)$, which is one of the $2$-Sylow subgroups of $GL(n,2)$.