Does there exist a subgroup of $GL(n,\mathbb F_2)$ that requires $n$ generators, when $n>2?$ In the answer to this previous question, we showed that, for prime powers $q>2,$ $GL(n,q)=GL(n,\mathbb F_q)$ has a subgroup $H$ that requires $n$ elements to generate it.
But over $\mathbb F_2$ there isn’t always such a subgroup. $n=3$ is known to have no subgroup requiring more than $2$ generators.
For each $n,$ the subgroup $U$ of upper-triangular matrices requires $n-1$ generators, at least.
That’s because there is an onto homomorphism: $U\mapsto \mathbb F_2^{n-1}$ with $(a_{ij})\mapsto (a_{k(k+1)})_{k=1}^{n-1}.$ Any generators for $U$ must, in the image, generate $\mathbb F_2^{n-1},$ and this requires $n-1$ generators.
But I believe the matrices with one off-diagonal, $A_k, k=1,\dots,n-1$ with:$$(A_k)_{ij}=\begin{cases}1&i=j\\1&i=k,j=k+1\\0&\text{otherwise}\end{cases}$$
generate $U.$
I found a paper behind a paywall whose title implies that the whole group $GL(n,2)$ is generated by $2$ elements for any $n.$ If true, that means we’d want proper subgroups.
Other than subgroups mapping onto abelian groups, I don’t have a lot of tools for solving this problem. I know all finite simple groups require only $2$ generators (or $1$ if abelian.) (Apparently, this is a deep result, currently requiring the classification of finite simple groups.)

Aside: When $q=p^k$ the upper triangular matrices with $1$ along the diagonal in $GL(n,q)$ require at least $k(n-1)$ generators, which can be significantly more than $n$ when $k$ is big. For example, $q=8,n=3$ these upper-triangular matrices requires at least $6$ generators.
 A: Let $m,k$ be positive integers and $n=m+k$. Let $\mathbb{F}$ be a field. Let $V$ be an $n$-dimensional vector space over $\mathbb{F}$, and let $W$ be an $m$-dimensional subspace. Extend a basis $\{e_1, \ldots, e_m \}$ of $W$ to a basis $\{e_1,\ldots, e_n\}$ of $V$. Let $G$ be the group of invertible linear transformations of $V$, $G=\operatorname{GL}_n(\mathbb{F})$ in terms of that basis. Let $P$ be the (parabolic) subgroup of $g$ consisting of those linear transformations that take $W$ into itself. In terms of matrices, these are the block matrices $\begin{bmatrix} A & B \\ 0 & C \end{bmatrix}$ acting on column vectors where $A \in \operatorname{GL}_m(\mathbb{F})$ acts on $W$, $B \in \mathbb{F}^{m \times k}$ maps vectors from $V/W$ into $W$, and $C \in \operatorname{GL}_k(\mathbb{F})$ acts on $V/W$. We take the (unipotent radical) subgroup $U$ consisting of those matrices that act as the identity on both $W$ and $V/W$. In terms of matrices, we take $A=I_m$ and $C=I_k$. So
$$U = \left\{ \begin{bmatrix} I & B \\ 0 & I \end{bmatrix} : B \in \mathbb{F}^{m\times k}\right\}$$
Consider two elements in this group,
$$g_i = \begin{bmatrix} I & B_i \\ 0 & I \end{bmatrix}\text{ for } i = 1,2$$
and their product:
$$g_1 g_2 = \begin{bmatrix} I & B_1 + B_2 \\ 0 & I \end{bmatrix}$$
which is clearly commutative. Hence $U$ is isomorphic to the additive group of $\mathbb{F}^{m \times k}$. If $\mathbb{F}$ has $p^f$ elements for $p$ prime, then this is an elementary abelian group of order $p^{fmk}$, which requires at least $fmk$ generators in any generating set (and has a generating set of size $fmk$).
In the specific case of $p^f=2^1$, and $n \geq 4$, we take $m = \lfloor n/2 \rfloor$ and $k=\lceil n/2 \rceil$ to get that $U$ is a subgroup requiring more than $(n/2)^2 = n \cdot (n/4)$ generators. Since $n \geq 4$, this is at least $n$ generators.
For $n=2$ (and $m=k=1$), this subgroup is cyclic of order 2. In fact, the only subgroup of $G=\operatorname{GL}(2,\mathbb{F}_2)\cong S_3$ that is not cyclic is $G$ itself. For $n=3$ (and $m=1, k=2$), this subgroup is elementary abelian of order $2^2$, so only requires two generators. In fact, $G=\operatorname{GL}(3,\mathbb{F}_2)$ has an exceptional isomorphism with $\operatorname{PSL}(2,7)$, and every subgroup of $\operatorname{PSL}(2,q)$ can be generated by two elements since its subgroups are cyclic, dihedral, other PSL, or subgroups isomorphic to $A_4$, $S_4$, or $A_5$.
This subgroup is well-known (thanks to @j.p. for reminding me I used to know this), and was mentioned by Schur when classifying the maximal commutative subalgebra of a full matrix algebra.
