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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.

Derek Holt
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Thomas Andrews
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  • Never mind, the idea doesn’t work for $G\times H.$ $GL(2,2)$ has generators $a,b$ with $a^2=b^3=1.$ Then $(a,b)^3=(a,1)$ and $(a,b)^4=(1,b).$ So $(a,b)$ and $(b,a)$ generate all of $GL(2,2)^2.$ Deleted that section. – Thomas Andrews Jul 09 '21 at 23:36
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    I calculated that one of the maximal subgroups of a Sylow 2-subgroup of GL(n,2) requires n generators for n in {4,5,6,7,8,9}. I didn't have a chance to see if it was easy to describe the subgroup or prove there was a nice family. – Jack Schmidt Jul 10 '21 at 03:36
  • I feel like it should be well-known what the maximal size of an elementary abelian 2-group of GL(n,2) is, but I don't personally know (and it's bed time, so I haven't looked it up). But this should sort of be fundamental to defining the rank of the group (which should be n as long as n isn't too small). Thompson's subgroup, or an overview of the classification of finite simple groups might mention the result. Since we are in the defining characteristic Sylow, we can freely switch between GL, SL, and PSL as needed. – Jack Schmidt Jul 10 '21 at 03:51
  • The upper triangular matrices are a $2$-Sylow subgroup. When $n=4,$ there are $64$ of them. When $n=5$ there are $2^{10}$ of them. @JackSchmidt – Thomas Andrews Jul 10 '21 at 03:53
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    @JackSchmidt: A good candidate for your maximal elementary abelian $2$-subgroup might the subgroup acting trivial on both $W$ and $V/W$ for an about $\frac{n}{2}$-dimensional subspace $W$ of $V=\mathbb{F}_2^n$. For $n=2m$ you get an elementary abelian $2$-group of rank $m^2$, and for $n=2m+1$ of rank $m(m+1)$ (which happens to be $\ge n$ unless $n<4$). – j.p. Jul 10 '21 at 07:47
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    @j.p. thanks! that will answer this question. The group of matrices of the form $\begin{bmatrix} I & A \ 0 & I \end{bmatrix}$ form an abelian group isomorphic to additive group of $\mathbb{F}^{m(n-m)}$, which needs $km(n-m)$ generators if $\mathbb{F}$ is finite of order $p^k$ for $p$ prime. – Jack Schmidt Jul 12 '21 at 14:31
  • If one of you could write up the answer, that would be great. @JackSchmidt – Thomas Andrews Jul 12 '21 at 16:05
  • Just an FYI that the title can be easily misconstrued. – Cameron Williams Jul 14 '21 at 19:02
  • @CameronWilliams Yes, it's that ridiculous word "any" again. I wish people would realize that it can mean either "some" or "all" and so it is completely unsuitable for use in formal mathematics. I will edit the title and try and capture the intended meaning! – Derek Holt Jul 14 '21 at 19:05
  • shakes fist at the word "any" – Cameron Williams Jul 14 '21 at 19:15

1 Answers1

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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.

Jack Schmidt
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