# $\operatorname{GL}(n, \mathbb k)$ 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.

• Hmm, I think $(12),(23),\dots,((n-1)n)$ are enough to generate $\mathfrak S_n.$ You don't need $(1n).$ $$(23)(12)(23)=(13),\\\vdots \\(k(k+1))(1k)(k(k+1))=(1(k+1))\\\vdots\\((n-1)n)(1(n-1))((n-1)n)=(1n)$$ Jul 7 at 18:41
• 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 $\alpha^k E_\sigma.$ Jul 7 at 18:46
• @ThomasAndrews, that is a good observation. Evidently, I hadn't noticed that. Jul 7 at 18:50
• This claim is not true when $k$ is the field of order $2$, but it is true for all other fields. Jul 7 at 22:22
• @DerekHolt Specifically, the only counterexample is $\mathbb k\cong\mathbb F_2$ and $n=3.$ Other $n$ work with $\mathbb F_2.$ Jul 12 at 16:55

$$\mathfrak S_n$$ 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 $$2.$$

• Thanks, missed that. Fixed the partial answer. @EricWofsey Jul 7 at 19:26
• The question's claim is false for GL(3,2) Jul 7 at 19:32
• @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.) Jul 7 at 19:46
• 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). Jul 7 at 20:05
• Yes, all finite simple groups are 2 generated. mathoverflow.net/questions/59213/… Jul 7 at 21:33

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.