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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.

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    $\begingroup$ Hmm, I think $(12),(23),\dots,((n-1)n)$ are enough to generate $\mathfrak S_n.$ You don't need $(1n).$ $$(23)(12)(23)=(13),\\\vdots \\(k(k+1))(1k)(k(k+1))=(1(k+1))\\\vdots\\((n-1)n)(1(n-1))((n-1)n)=(1n)$$ $\endgroup$ Jul 7 at 18:41
  • $\begingroup$ However, it might work if you pick $\alpha\in\mathbb k\setminus\{0,1\}$ and add to the generators $\alpha I.$ Basically, then $H$ would be the set of all $\alpha^k E_\sigma.$ $\endgroup$ Jul 7 at 18:46
  • $\begingroup$ @ThomasAndrews, that is a good observation. Evidently, I hadn't noticed that. $\endgroup$ Jul 7 at 18:50
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    $\begingroup$ This claim is not true when $k$ is the field of order $2$, but it is true for all other fields. $\endgroup$
    – Derek Holt
    Jul 7 at 22:22
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    $\begingroup$ @DerekHolt Specifically, the only counterexample is $\mathbb k\cong\mathbb F_2$ and $n=3.$ Other $n$ work with $\mathbb F_2.$ $\endgroup$ Jul 12 at 16:55
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$\mathfrak S_n$ 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 $2.$

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  • $\begingroup$ Thanks, missed that. Fixed the partial answer. @EricWofsey $\endgroup$ Jul 7 at 19:26
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    $\begingroup$ The question's claim is false for GL(3,2) $\endgroup$ Jul 7 at 19:32
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    $\begingroup$ @Carlo $GL(n,q)$ is shorthand for $GL(n,\mathbb F_q)$ where $\mathbb F_q$ is the finite field of size $q.$ (There is only such a field if $q$ is a power of a prime, and it is unique up to isomorphism.) $\endgroup$ Jul 7 at 19:46
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    $\begingroup$ GL(3,2) is also PSL(2,7), and PSL(2,q) subgroups are pretty well known. For this particular case, the maximal subgroups are S4, S4, and C7 x| C3. The subgroups of S4 are A4, D8, S3, K4, C4, A3, C2, and 1 (all at most 2-generated). The subgroups of C7 x| C3 are C7, C3, 1 (all at most 2-generated). $\endgroup$ Jul 7 at 20:05
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    $\begingroup$ Yes, all finite simple groups are 2 generated. mathoverflow.net/questions/59213/… $\endgroup$ Jul 7 at 21:33
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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.

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