$\operatorname{GL}(n, \mathbb k)$ Contains a Finite Subgroup Minimally Generated by $n$ Elements Let $n$ be a positive integer. Let $\mathbb k$ be a field. Consider the general linear group $\operatorname{GL}(n, \mathbb k)$ that consists of invertible $n \times n$ matrices with entries in the field $\mathbb k$ under the operation of matrix multiplication.

Claim. There exists a finite subgroup $H$ of $\operatorname{GL}(n, \mathbb k)$ that is minimally generated by $n$ elements (i.e., there exist elements $h_1, \dots, h_n$ of $H$ such that $H = \langle h_1, \dots, h_n \rangle$ and no $n − 1$ elements of $\operatorname{GL}(n, \mathbb k)$ generate $H$).

Like several commenters have noted, the claim does not hold when $\mathbb k = \mathbb Z / 2 \mathbb Z$ and $n = 3.$
I had initially thought of the following. Let $I$ denote the $n \times n$ identity matrix. Given a permutation $\sigma$ of the symmetric group $\mathfrak S_n$ on $n$ letters, define the permutation matrix $E_\sigma$ that is obtained by permuting the rows of $I$ according to $\sigma,$ e.g., if $n = 3$ and $\sigma = (1, 2, 3)$ is the three-cycle, then $$E_\sigma = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}.$$ Considering that permuting the rows of a matrix only affects the sign of its determinant, it follows that $E_\sigma$ is invertible and so belongs to $\operatorname{GL}(n, \mathbb k).$ Further, it can be shown that $E_\sigma E_\tau = E_{\sigma \tau}.$ Consequently, there is an injective group homomorphism $i : \mathfrak S_n \to \operatorname{GL}(n, \mathbb k)$ defined by $i(\sigma) = E_\sigma.$
I have looked for $n$ carefully chosen permutations, but in all cases, the subgroup of $\operatorname{GL}(n, \mathbb k)$ has been isomorphic to $\mathfrak S_n$ and so minimally generated by two elements -- namely the permutation matrices $E_\sigma$ and $E_\tau$ corresponding to the two cycle $\sigma = (1, 2)$ and the $n$-cycle $\tau = (1, 2, \dots, n).$ I would appreciate any comments or observations. Thank you for your time and consideration.
 A: $\mathfrak S_n$ has a generating set of size $2,$ $\sigma=(12)$ and $\rho=(123\dots n).$ Then $\rho((k-1)k)\rho^{-1}=(k(k+1)).$ (Depending on the order that you compose permutations.) So $H\cong \mathfrak S_n$ won't do.
In a field not of characteristic $2,$  consider $H$ all the matrices of the form:
$$\operatorname{diag}\left(\pm 1,\dots,\pm 1\right)$$
Then $H$ is isomorphic to the additive group $\left(\mathbb Z/2\mathbb Z\right)^n.$ But this is a vector space over $\mathbb F_2$ and any set of group generators smaller than $n$ shows there is a basis for the $n$-dimension vector space with fewer than $n$ elements.

So you need to only deal with the case when $\mathbb k$ has characteristic $2.$
If we solve it for $\mathbb k=\mathbb F_2,$ you are done, since that will work for any $\mathbb k$ containing $\mathbb F_2.$ This thus reduces to the question for $GL(n,\mathbb F_2),$ which is a finite group.
From the comments, it is not true for $GL(3,\mathbb F_2).$
It is true for $n=2$ since the group itself is non-commutative and of order $6,$ and thus requires $2$ generators.
From a question I asked, for $n\geq 4,$ and $1\leq m\leq n$ the subgroup consisting of matrices of the form:
$$\begin{pmatrix}I_m&A\\0&I_{n-m}\end{pmatrix}$$ is isomorphic to the additive group of $m\times(n-m)$ matrices, which is a vector space over $\mathbb F_2$ of dimension $m\times(n-m).$
When $n\geq 4,$ we can take $2\leq m\leq n-2,$ then $m(n-m)\geq n,$ and take an $n$-dimensional subspace of this space.
This also  works if $n=3$ and $\mathbb k$ is of characteristic $2$ with more than $2$ elements. This is because give two distinct non-zero  $a,b\in \mathbb k$ we get an additive subgroup $V=\{0,a,b,a+b\}$ hitch is a vector space of dimension $2$ over $\mathbb F_2,$ so we can take the subgroup of matrices of the form:
$$\begin{pmatrix}1&0&v\\0&1&c\\0&0&1\end{pmatrix}$$ with $v_1\in V,c\in \mathbb F_2.$

So the only counterexample is $k\cong \mathbb F_2, n=3.$

In all the cases, we have found abelian subgroups isomorphic to $(\mathbb F_2^n,+),$ even when $\mathbb k$ is not of characteristic $2.$
A: Using the following lemma, one of my students suggested a very clever proof to me.

Lemma. Every finite subgroup of $\operatorname{GL}(1, \mathbb k)$ is cyclic.
Proof. Observe that $\operatorname{GL}(1, \mathbb k) \cong \mathbb k^\times$ by the map that sends $[a] \mapsto a.$ Every finite subgroup of the multiplicative group of units of a field is cyclic (e.g., see this proof). QED.

Credit is due to Wayne Voon Van Ng Kwing King for the following.
Proof. Recall that the direct sum of two matrices $A$ and $B$ is given by $$A \oplus B = \begin{pmatrix} A & O_A \\ O_B & B \end{pmatrix},$$ where $O_A$ is the zero matrix with the same number of rows as $A$ and the same number of columns as $B$ (and $O_B$ is defined analogously). Consequently, it follows that $\oplus_{i = 1}^n \operatorname{GL}(1, \mathbb k)$ consists of invertible $n \times n$ diagonal matrices (i.e., diagonal matrices with strictly nonzero diagonal entries). By the lemma, for each copy of $\operatorname{GL}(1, \mathbb k),$ there exists a finite subgroup $H_i$ that is generated by some invertible matrix $[a_i].$ Given that $\mathbb k$ is not the field with two elements, the unit $a_i$ can be chosen so that it is not equal to $1.$ For each integer $1 \leq i \leq n,$ define an $n \times n$ diagonal matrix $A_i$ whose $i$th diagonal entry is $a_i$ and whose other diagonal entries are $1.$ Consider the subgroup $H = \langle A_1, A_2, \dots, A_n \rangle.$ By definition, the matrices $A_i$ all satisfy $A_i^{n_i} = I$ for some positive integer $n_i.$ Consequently, we find that $H$ is a finite subgroup of $\operatorname{GL}(n, \mathbb k).$ Further, there is no way to write $A_i$ as a product of powers of the matrices of $\{A_1, \dots, A_n\} \setminus \{A_i\}$ because the $i$th diagonal entry of such a product is $1$ and $a_i$ can be chosen so that it is not equal to $1.$ Ultimately, we have found a finite subgroup of $\operatorname{GL}(n, \mathbb k)$ that is minimally generated by $n$ elements. QED.
