On Diameters of Cayley Graphs over Matrix Groups

Eitan Porat

Abstract.

We establish that for the matrix groups $G=\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)$ or $G=\mathrm{SL}_{n}\left(\mathbb{F}_{p}\right)$ , there exist absolute constants $c\in\left(0,1\right)$ and $C>0$ such that any symmetric generating set $A$ with $\left|A\right|\geq\left|G\right|^{1-c}$ has a covering number $\leq Cn^{2}$ . This result is sharp up to the value of the constant $C>0$ .

1. Introduction

For a finite group $G$ and a symmetric generating set $A\subseteq G$ (in the sense that $A^{-1}=A$ ), the Cayley graph $\mathrm{Cay}\left(G,A\right)$ is defined as a connected graph with vertex set $G$ and edge set

\left\{\left(g,ag\right)\mid g\in G,a\in A\right\}.

The diameter of this graph is the smallest integer $k$ such that every element of $G$ can be expressed as a product of at most $k$ elements from $A$ . The diameter of the group is denoted by $\mathrm{diam}\left(G\right)$ , and it represents the maximum among all Cayley graph diameters $\mathrm{Cay}\left(G,A\right)$ where $A$ runs through all generating sets of $G$ .

This becomes an intrinsic property of $G$ , rather than being tied to the Cayley graph associated with $G$ or any specific generating set.

The diameter of $\mathrm{Cay}\left(G,A\right)$ is referred to as the covering number of $A$ , denoted by $\mathrm{cn}\left(A\right)=\min\left\{k:A^{k}=G\right\}$ . Babai [1] proposed the following conjecture:

Conjecture 1.

If $G$ is a non-Abelian finite simple group, then $\mathrm{diam}\left(G\right)=\left(\log\left|G\right|\right)^{O\left(1\right)}$ .

This conjecture remains one of the major unresolved problems in combinatorics. Liebeck and Shalev [7] confirmed the conjecture when the generating set $A$ is a normal set. The conjecture was proven by Helfgott [5] for the case $G=\mathrm{SL}_{2}\left(\mathbb{F}_{p}\right)$ , and subsequently, it was verified for finite simple groups of Lie type and bounded rank independently by Pyber and Szabo [8] and Breuillard, Green, and Tao [2]. It remains to prove Babai’s conjecture for alternating groups and for classical groups of unbounded rank. A special case of this conjecture was proven by Halasi [4] for the case where $G=\mathrm{SL}_{n}\left(\mathbb{F}_{p}\right)$ and $A$ is a generating set that includes a transvection, defined as a matrix of the form $I+x$ , where $x$ is a rank 1 matrix.

In this paper, we prove a special case of this conjecture for the scenario where $G$ is the general linear group $\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)$ and $A$ is a sufficiently large generating set.

Theorem 2.

If $\left|A\right|\geq\left|\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)\right|^{1-c}$ for some $0<c<1$ , then for some absolute constant $C>0$ , $\mathrm{cn}(A)\leq Cn^{2}$ .

Remark 3.

The proof can be readily adapted to work with $G=\mathrm{SL}_{n}\left(\mathbb{F}_{p}\right)$ .

Remark 4.

Although $\mathrm{SL}_{n}\left(\mathbb{F}_{p}\right)$ is not a simple group, the theorem also holds for $\mathrm{PSL}_{n}\left(\mathbb{F}_{p}\right)$ .

This result is sharp up to the absolute constant $C$ . Our proof relies on the concept of a groumvirate, which is a conjugate of the natural embedding of $\mathrm{\mathrm{G}L}_{n-t}\left(\mathbb{F}_{p}\right)$ in $\mathrm{\mathrm{G}L}_{n}\left(\mathbb{F}_{p}\right)$ , i.e., the subgroup

\left\{\begin{pmatrix}I_{t\times t}&0\\ 0&X\end{pmatrix}\mid X\in\mathrm{\mathrm{G}L}_{n-t}\left(\mathbb{F}_{p}\right)\right\}

where $0\leq t\leq n$ . We define $\mu\left(A\right):=\frac{\left|A\right|}{\left|\mathrm{\mathrm{G}L}_{n}\left(\mathbb{F}_{p}\right)\right|}$ as the uniform measure on $\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)$ .

Theorem 5 (Evra, Kindler, and Lifshitz [3]).

There exists a $d>0$ such that for every subset $A$ of $\mathrm{SL}_{n}\left(\mathbb{F}_{p}\right)$ , the set $AA^{-1}AA^{-1}$ contains a groumvirate with density at least $\mu\left(A\right)^{d}$ .

For a given $t$ ,

\mu\left(\left\{\begin{pmatrix}I_{t\times t}&0\\ 0&X\end{pmatrix}\mid X\in\mathrm{GL}_{n-t}\left(\mathbb{F}_{p}\right)\right\}\right)\leq\frac{q^{\left(n-t\right)^{2}}}{\left|\mathrm{\mathrm{G}L}_{n}\left(\mathbb{F}_{p}\right)\right|}=\frac{q^{\left(n-t\right)^{2}}}{q^{n^{2}}\left(q-1\right)^{n}}\leq q^{\left(n-t\right)^{2}-n^{2}}

Thus, to contain a groumvirate on $n-t$ variables, the density of $A$ must exceed,

\mu\left(A\right)\geq q^{\frac{\left(n-t\right)^{2}-n^{2}}{d}}

That is

\left|A\right|\geq q^{\frac{\left(n-t\right)^{2}-n^{2}}{d}}q^{n^{2}}=q^{\left(1-\frac{1}{d}\right)n^{2}+\frac{1}{d}\left(n-t\right)^{2}}

For $t=\varepsilon n$ ,

\left|A\right|\geq q^{\left(1-\left(\frac{1-\left(1-\varepsilon\right)^{2}}{d}\right)\right)n^{2}}=q^{\left(1-c_{\varepsilon}\right)n^{2}}=\Omega\left(\left|\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)\right|^{1-c_{\varepsilon}}\right)

where $c_{\varepsilon}=\frac{1-\left(1-\varepsilon\right)^{2}}{d}$ . The bound is non-trivial if $\varepsilon>0$ . For the remainder of the paper, we take $\varepsilon=\frac{1}{4}$ and $\left|A\right|\geq\left|\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)\right|^{1-c_{\varepsilon}}$ , and we will often omit explicitly stating this assumption.

In this paper, we assume that the groumvirate is expressed in the standard basis, i.e., $\left\{\begin{pmatrix}I_{t\times t}&0\\ 0&X\end{pmatrix}\mid X\in\mathrm{GL}_{n-t}\left(\mathbb{F}_{p}\right)\right\}=I_{t}\otimes\mathrm{GL}_{n-t}\left(\mathbb{F}_{p}\right)$ for some $t$ . Since $A$ is symmetric, $AA^{-1}AA^{-1}=A^{4}$ . We may assume without loss of generality that $I\in A$ (this does not alter the asymptotics since $I\in AA^{-1}$ ).

Corollary 6.

Suppose $\left|A\right|\geq\left|\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)\right|^{1-c_{\varepsilon}}$ , $A^{4}\supseteq\left\{\begin{pmatrix}I_{\epsilon n}&0\\ 0&X\end{pmatrix}\mid X\in\mathrm{\mathrm{GL}}_{(1-\varepsilon)n}\left(\mathbb{F}_{p}\right)\right\}$ , up to a change of basis.

This groumvirate acts as $\mathrm{GL}_{n-t}\left(\mathbb{F}_{p}\right)$ on a subspace of $\mathbb{F}_{p}^{n}$ . Our goal is to upgrade it to an action on the whole space.

We define the standard basis to be $e_{1}=\begin{pmatrix}1\\ 0\\ \vdots\\ 0\end{pmatrix},e_{2}=\begin{pmatrix}0\\ 1\\ \vdots\\ 0\end{pmatrix},\dots,e_{n}=\begin{pmatrix}0\\ 0\\ \vdots\\ 1\end{pmatrix}$ . The span of a set of vectors $v_{1},\dots,v_{k}$ is denoted by $\left\langle v_{1},\dots,v_{k}\right\rangle$ . For a set $S\subseteq[n]$ , denote $e_{S}=\{e_{i}:i\in S\}$ and $e_{\bar{S}}=\{e_{i}:i\notin S\}$ . We define the projection of a vector $v$ onto $\left\langle e_{1},\dots,e_{t}\right\rangle$ by $\pi_{1}\left(v\right)$ and its projection onto $\left\langle e_{t+1},\dots,e_{n}\right\rangle$ by $\pi_{2}\left(v\right)$ . Additionally, we can modify the $\pi_{2}$ component of vectors using $M\in A^{4}$ .

Lemma 7.

Let $\{v_{1},\dots,v_{k}\}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ and $\{v^{\prime}_{1},\dots,v^{\prime}_{k}\}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ , for $k\leq n-t$ . Assume that $\{v_{1},\dots,v_{k}\}$ is a linearly independent set and $\{v_{1}^{\prime},\dots,v_{k}^{\prime}\}$ is also linearly independent; then there exists $M\in A^{4}$ such that $v_{i}=v^{\prime}_{i}$ , which fixes $\{e_{1},\dots,e_{t}\}$ .

Proof.

Since $\left\{v_{1},\dots,v_{k}\right\}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ and $\{v_{1}^{\prime},\dots,v_{k}^{\prime}\}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ are linearly independent sets, there exists an invertible linear transformation $T:\left\langle e_{t+1},\dots,e_{n}\right\rangle\to\left\langle e_{t+1},\dots,e_{n}\right\rangle$ such that $T\left(v_{i}\right)=v_{i}^{\prime}$ for every $1\leq i\leq k$ . We can extend $T$ to a transformation $T^{\prime}:\mathbb{F}_{p}^{n}\to\mathbb{F}_{p}^{n}$ such that $T\left(e_{i}\right)=e_{i}$ for every $1\leq i\leq t$ . Define $M$ to be a matrix representation of $T^{\prime}$ in the standard basis, that is $M=\left[T^{\prime}\right]_{\left\langle e_{1},\dots,e_{n}\right\rangle}^{\left\langle e_{1},\dots,e_{n}\right\rangle}$ . By construction, $M\in I_{t}\otimes\mathrm{GL}_{n-t}(\mathbb{F_{p}})\subseteq A^{4}$ . ∎

Our main contribution is the following Theorem:

Theorem 8.

Let $\mathcal{V}=\left\{(U,W)\mid U\oplus W=\mathbb{F}_{p}^{n}\right\}$ , and define an action $\cdot:A\times\mathcal{V}\to\mathcal{V}$ by $g\cdot(U,W)=(gU,gW)$ . Consider the Schreier graph corresponding to this action on the vertex set $\mathcal{V}$ . Then, any two vertices in $\mathcal{V}^{\prime}=\left\{(\langle e_{S}\rangle,\langle e_{\bar{S}}\rangle)\mid S\subseteq[n],|S|=t\right\}$ are at most $O(t^{2})$ steps apart in this graph.

Intuitively, this allows us to upgrade the action of the groumvirate to an arbitrary set of size $n-t$ of columns or rows of our choosing.

Corollary 9.

Any transformation $X\in\mathrm{GL}_{n}(\mathbb{F}_{p})$ which fixes $e_{S}$ and is invariant on $\langle e_{\bar{S}}\rangle$ for $|S|=t$ , can be constructed in $O(t^{2})$ steps.

2. Upgrading The Groumvirate’s Action

Lemma 10.

Let $V\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ be a vector space of dimension $\dim\left(V\right)=k$ . Let $\left\{u_{i}\right\}_{i=1}^{m}\cup\left\{w_{i}\right\}_{i=1}^{\ell}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ be a linearly independent set, where $m+\ell\leq k$ . There exists $T\in A^{4}$ such that $Tu_{i}=u_{i}$ for every $1\leq i\leq m$ , $Tw_{i}\in V$ for every $1\leq i\leq\ell$ , and $Te_{i}=e_{i}$ for every $1\leq i\leq t$ .

Proof.

Let $\left\{v_{1},\dots,v_{k}\right\}$ be a basis of $V$ . Since $\dim\left(\left\langle u_{1},\dots,u_{m}\right\rangle\right)=m\leq k$ , there exists a vector $v_{1}\in V$ such that $\dim\left\langle v,u_{1},\dots\right\rangle=m+1$ . We inductively add vectors $v_{1},\dots,v_{k-m}$ until we obtain a linearly independent set $\left\{u_{1},\dots,u_{m},v_{1},\dots,v_{k-m}\right\}$ . Applying Lemma 7, there exists a transformation such that $Tu_{i}=u_{i}$ for every $1\leq i\leq m$ , $Tw_{i}=v_{i}$ for every $1\leq i\leq\ell$ , and $Te_{i}=e_{i}$ for every $1\leq i\leq t$ . ∎

Lemma 11.

Let $V$ be a vector space and $a\in\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)$ . Consider the subspace $W\subseteq V$ of $v\in V$ such that $av\in V$ , then $\dim\left(W\right)\geq 2\dim\left(V\right)-n$ .

Proof.

By Grassmann’s Lemma,

	$\displaystyle\dim\left(W\right)$	$\displaystyle=\dim\left(V\cap a^{-1}V\right)=\dim\left(V\right)+\dim\left(a^{-1}V\right)-\dim\left(V+a^{-1}V\right)$
		$\displaystyle=2\dim\left(V\right)-\dim\left(V+a^{-1}V\right)\geq 2\dim\left(V\right)-n.$

∎

Lemma 12.

There exists $a\in A^{O(t^{2})}$ such that $ae_{i}=e_{t+i}$ for $1\leq i\leq t$ .

Proof.

(1)

We use Lemma 13 to construct vectors $v_{1},\dots,v_{t}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ and $a_{i}\in A^{i}$ and $b_{t}\in A^{O(t^{2})}$ such that $\left\{\pi_{1}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}$ is a basis for $\left\langle e_{1},\dots,e_{t}\right\rangle$ , $\left\{\pi_{2}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}$ is a linearly independent set, and $b_{t}a_{i}v_{i}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ for every $1\leq i\leq t$ .
(2)

We use Lemma 14 to construct a transformation $c_{t}\in A^{O(t^{2})}$ such that $\left\{\pi_{2}\left(c_{t}e_{i}\right)\right\}_{i=1}^{t}$ is a linearly independent set. There exist $\left\{z_{1},\dots z_{t}\right\}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ such that $\left\{\pi_{2}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}\cup\left\{z_{1},\dots z_{t}\right\}$ is a basis of $\left\langle e_{t+1},\dots,e_{n}\right\rangle$ . By Lemma 7, there exists $T_{1}\in A^{4}$ such that $T\pi_{2}\left(c_{t}e_{i}\right)=z_{i}$ for $1\leq i\leq t$ . Hence, we may generally assume that $\left\{\pi_{2}\left(c_{t}e_{i}\right)\right\}_{i=1}^{t}\cup\left\{\pi_{2}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}$ is a linearly independent set.
(3)

We express $c_{t}e_{i}$ as $c_{t}e_{i}=\sum_{j=1}^{t}\alpha_{ij}a_{j}v_{j}+w_{i}$ with $\alpha_{ij}\in\mathbb{F}_{p}$ and $w_{i}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ (since $\{\pi_{1}(a_{i}v_{i})\}_{i=1}^{t}$ is a basis of $\left\langle e_{1},\dots,e_{t}\right\rangle$ ). By construction $\left\{\pi_{2}\left(c_{t}e_{i}\right)\right\}_{i=1}^{t}\cup\left\{\pi_{2}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}$ is linearly independent. Since elementary operations do not change the linear independence of a set, $\left\{\pi_{2}\left(w_{i}\right)\right\}_{i=1}^{t}\cup\left\{\pi_{2}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}$ is also linearly independent. By Lemmas 10 and 11, there exists a transformation $T_{2}\in A^{4}$ such that $T_{2}w_{i}=w_{i}^{\prime}\in b_{t}^{-1}\left\langle e_{t+1},\dots,e_{n}\right\rangle\cap\left\langle e_{t+1},\dots,e_{n}\right\rangle$ , $T_{2}\pi_{2}\left(a_{i}v_{i}\right)=\pi_{2}\left(a_{i}v_{i}\right)$ , and $T_{2}e_{i}=e_{i}$ for $1\leq i\leq t$ .
(4)

Since $b_{t}a_{j}v_{j}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ and $b_{t}w_{i}^{\prime}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ ,

$b_{t}\left(\sum_{j=1}^{t}\alpha_{ij}a_{j}v_{j}+w_{i}^{\prime}\right)\in\left\langle e_{t+1},\dots,e_{n}\right\rangle.$
(5)

We apply lemma 7 to construct a transformation $T_{3}\in I_{t}\otimes\mathrm{GL}_{n}(\mathbb{F}_{p})$ such that $T_{3}b_{t}T_{2}T_{1}c_{t}e_{i}=e_{t+i}$ for every $1\leq i\leq t$ .

∎

Lemma 13.

There exist $\{v_{i}\}_{i=1}^{t}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ , $\{a_{i}\}_{i=1}^{t}\subseteq A^{t}$ and $b\in A^{O(t^{2})}$ such that the following hold:

(1)

$\left\{\pi_{1}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}$ is a basis for $\left\langle e_{1},\dots,e_{t}\right\rangle$ .
(2)

$\left\{\pi_{2}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}$ is a linearly independent set.
(3)

$\{ba_{i}v_{i}\}_{i=1}^{t}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ .

Proof.

Part 1 is proven by induction on $t$ . Since $A$ is a generating set, $\left\langle e_{t+1},\dots,e_{n}\right\rangle$ is not invariant under $A$ , thus there exists $v_{1}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ and $a_{1}\in A$ such that $\pi_{1}\left(a_{1}v_{1}\right)\neq 0$ . For the induction step, since $\left\langle a_{1}v_{1},\dots,a_{i}v_{i},e_{t+1},\dots,e_{n}\right\rangle$ is not invariant under $A$ , there exists $v\in\left\langle a_{1}v_{1},\dots,a_{i}v_{i},e_{t+1},\dots,e_{n}\right\rangle$ and $a_{i+1}\in A$ such that $a_{i+1}v\notin\left\langle a_{1}v_{1},\dots,a_{i}v_{i},e_{t+1},\dots,e_{n}\right\rangle$ . Therefore, there exists $v_{i+1}=e_{i}$ for some $t+1\leq i\leq n$ or $v_{i+1}=a_{j}v_{j}\in A^{j}\left\langle e_{t+1},\dots,e_{n}\right\rangle$ such that $a_{i+1}v_{i+1}\notin\left\langle a_{1}v_{1},\dots,a_{i}v_{i},e_{t+1},\dots,e_{n}\right\rangle$ , therefore $a_{i+1}v_{i+1}\in A^{i+1}\left\langle e_{t+1},\dots,e_{n}\right\rangle$ . Hence, $\left\{\pi_{1}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}$ is a linearly independent set.

Part 2 is proven by induction. By Lemma 13,

\dim\left(a_{i}^{-1}\left\langle e_{t+1},\dots,e_{n}\right\rangle\cap\left\langle e_{t+1},\dots,e_{n}\right\rangle\right)\geq t

for every $1\leq i\leq t$ .

Thus, there exists $u_{1}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ such that $\pi_{2}(a_{1}u_{1})\neq 0$ , set $v_{1}^{\prime}=v_{1}+u_{1}$ . Assume we have constructed $v_{1}^{\prime},\dots,v_{i}^{\prime}$ such that $\{\pi_{2}(a_{j}v_{j}^{\prime})\}_{j=1}^{i}$ is a linearly independent set. For the induction step, if $\pi_{2}(a_{i+1}v_{i+1})$ is linearly independent of $\{\pi_{2}(a_{j}v_{j}^{\prime})\}_{j\leq i}$ , set $v_{i+1}^{\prime}=v_{i+1}$ . Otherwise, set $v_{i+1}^{\prime}=v_{i+1}+u_{i+1}$ where $u_{i+1}\in a_{i+1}^{-1}\left\langle e_{t+1},\dots,e_{n}\right\rangle\cap\left\langle e_{t+1},\dots,e_{n}\right\rangle$ and $\pi_{2}(a_{i+1}u_{i+1})$ is linearly independent of $\{\pi_{2}(a_{j}v_{j}^{\prime})\}_{j\leq i}$ . We replace $\{v_{1},\dots,v_{t}\}$ with $\{v_{1}^{\prime},\dots,v_{t}^{\prime}\}$ and proceed to the last part.

For Part 3, the construction of $b=b_{t}$ is done inductively, where the induction is on the number of vectors $\{a_{1}v_{1},\dots,a_{i}v_{i}\}$ such that $\{b_{i}a_{1}v_{1},\dots,b_{i}a_{i}v_{i}\}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ . Initially, we are given a set $\left\{a_{1}v_{1},\dots,a_{t}v_{t}\right\}$ where $\{a_{1}v_{1},\dots,a_{t}v_{t}\}\cap\left\langle e_{t+1},\dots,e_{n}\right\rangle=\emptyset$ and $\left\{\pi_{2}\left(a_{i}v_{i}\right)\right\}_{i=1}^{t}$ is linearly independent. For the base case, we use Lemma 7 to add $w_{i}\in a_{1}\left\langle e_{t+1},\dots,e_{n}\right\rangle\cap\left\langle e_{t+1},\dots,e_{n}\right\rangle$ to $a_{i}v_{i}$ for every $1\leq i\leq t$ , where $\left\{w_{i}\right\}_{i=1}^{t}$ are chosen such that $\left\{\pi_{2}\left(a_{1}^{-1}\left(a_{i}v_{i}+w_{i}\right)\right)\right\}_{i=1}^{t}$ is a linearly independent set. After adding $\left\{w_{i}\right\}_{i=1}^{t}$ , we apply $a_{1}^{-1}$ to all the vectors. We are left with the set

\left\{v_{1}+a_{1}^{-1}w_{1},a_{1}^{-1}a_{2}v_{2}+a_{1}^{-1}w_{2},\dots,a_{1}^{-1}a_{t}v_{t}+a_{1}^{-1}w_{t}\right\}

We define $d_{1}=v_{1}+a_{1}^{-1}w_{1}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ . Note that $\left\{\pi_{2}\left(a_{1}^{-1}\left(a_{i}v_{i}+w_{i}\right)\right)\right\}_{i=1}^{t}$ is a linearly independent set.

Now assume we are given

\left\{d_{1},\dots,d_{i},a_{i}^{-1}a_{i+1}v_{i+1}+w_{i+1},\dots,a_{i}^{-1}a_{t}v_{t}+w_{t}\right\}

where $d_{1},\dots,d_{i},w_{i+1},\dots,w_{t}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ and

\left\{\pi_{2}\left(d_{1}\right),\dots,\pi_{2}\left(d_{i}\right)\right\}\cup\left\{\pi_{2}\left(a_{i}^{-1}a_{i+1}v_{i+1}+w_{i+1}\right),\dots,\pi_{2}\left(a_{i}^{-1}a_{t}v_{t}+w_{t}\right)\right\}

is a linearly independent set. By Lemma 10 and 11, there exists a transformation $M\in I_{t}\otimes\mathrm{GL}_{n-t}\left(\mathbb{F}_{p}\right)$ such that $Md_{j}\in a_{i}^{-1}a_{i+1}\left\langle e_{t+1},\dots,e_{n}\right\rangle\cap\left\langle e_{t+1},\dots,e_{n}\right\rangle$ for every $1\leq i\leq j$ such that $M\pi_{2}\left(a_{i}^{-1}a_{i+1}v_{j}+w_{i+1}\right)=\pi_{2}\left(a_{i}^{-1}a_{i+1}v_{j}+w_{i+1}\right)$ . This is where the $t\leq\frac{n}{4}$ assumption comes into play.

With some abuse of notation, we assume $d_{1},\dots,d_{i}$ are already in $a_{i}^{-1}a_{i+1}\left\langle e_{t+1},\dots,e_{n}\right\rangle\cap\left\langle e_{t+1},\dots,e_{n}\right\rangle$ . Now we add $\left\{w_{j}^{\prime}\right\}_{j=1}^{t}\in a_{i}^{-1}a_{i+1}\left\langle e_{t+1},\dots,e_{n}\right\rangle\cap\left\langle e_{t+1},\dots,e_{n}\right\rangle$ to all the vectors such that

\displaystyle\left\{\pi_{2}\left(a_{i+1}^{-1}a_{i}\left(d_{j}+w_{j}^{\prime}\right)\right)\right\}_{j=1}^{i}\cup

\displaystyle\left\{\pi_{2}\left(a_{i+1}^{-1}a_{i}\left(a_{i}^{-1}a_{j}v_{j}+w_{j}+w_{j}^{\prime}\right)\right)\right\}_{j=i+1}^{t}

is a linearly independent set. We set $w_{j}^{\prime\prime}\leftarrow a_{i+1}^{-1}a_{i}\left(w_{j}+w_{j}^{\prime}\right)$ , and $d_{j}\leftarrow d_{j}+w_{j}^{\prime\prime}$ . Thus $d_{j}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle$ and

a_{i+1}^{-1}a_{i}\left(a_{i}^{-1}a_{i+1}v_{i+1}+w_{j}+w_{j}^{\prime}\right)=v_{i+1}+w_{j}^{\prime\prime}\in\left\langle e_{t+1},\dots,e_{n}\right\rangle

and for $j\leq i+1\leq n$ ,

a_{i+1}^{-1}a_{i}\left(a_{i}^{-1}a_{j}v_{j}+w_{j}+w_{j}^{\prime}\right)=a_{i+1}^{-1}a_{j}v_{j}+w_{j}^{\prime\prime}.

We set $d_{i+1}=v_{i+1}+w_{j}^{\prime\prime}$ , and proceed inductively. As each step takes $O\left(t\right)$ operations in $A$ and there are $t$ steps, the number of operations required to construct $b$ is $O\left(t^{2}\right)$ . ∎

Lemma 14.

There exists $c_{t}\in A^{O\left(t^{2}\right)}$ such that $\left\{\pi_{2}\left(c_{t}e_{i}\right)\right\}_{i=1}^{t}$ is a linearly independent set.

Proof.

We prove this lemma by induction on $i$ , where $i$ is the maximum number such that $\left\{\pi_{2}\left(c_{i}e_{j}\right)\right\}_{j=1}^{i}$ is linearly independent. For the base case $i=1$ , since $A$ is a generating set, $\dim\left(A^{t}\left\langle e_{1}\right\rangle\right)>t$ , there exists $c_{1}\in A^{t}$ such that $\pi_{2}\left(c_{1}e_{1}\right)\neq 0$ .

For the induction step, assume we have constructed $c_{i}$ such that $\left\{\pi_{2}\left(c_{i}e_{j}\right)\right\}_{j=1}^{i}$ is linearly independent. If $\pi_{2}\left(c_{i}e_{i+1}\right)$ is linearly independent of $\left\{\pi_{2}\left(c_{i}e_{j}\right)\right\}_{j=1}^{i}$ , then we are done. Otherwise, $\pi_{2}\left(c_{i}e_{i+1}\right)=\sum_{j=1}^{i}\alpha_{j}\pi_{2}\left(c_{i}e_{j}\right)$ for some $\left\{\alpha_{j}\right\}_{j=1}^{i}\in\mathbb{F}_{p}^{n}$ . As $c_{i}e_{i+1}$ is linearly independent of $\left\{c_{i}e_{j}\right\}_{j=1}^{i}$ , $\pi_{1}\left(c_{i}e_{i+1}\right)\neq\pi_{1}\left(c_{i}\sum_{j=1}^{i}\alpha_{j}e_{j}\right)$ .

Let $z=\pi_{1}\left(c_{i}e_{i+1}\right)-\pi_{1}\left(c_{i}\sum_{j=1}^{i}\alpha_{j}e_{j}\right)$ . Since $\dim\left(A^{t}\left\langle z\right\rangle\right)\geq t$ , there exists $\tilde{c}_{i+1}\in A^{t}$ such that $\pi_{2}\left(\tilde{c}_{i+1}z\right)\neq 0$ . Consider the vector space $V\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ spanned by the set $\left\{\pi_{2}\left(\tilde{c}_{i+1}e_{1}\right),\dots,\pi_{2}\left(\tilde{c}_{i+1}e_{t}\right)\right\}$ which has a basis $\left\{v_{1},\dots,v_{\ell}\right\}$ .

Let $W=\left\langle e_{t+1},\dots,e_{n}\right\rangle\cap\tilde{c}_{i+1}^{-1}\left\langle e_{t+1},\dots,e_{n}\right\rangle$ , by Lemma 11, $\dim\left(W\right)\geq 2n-t\geq 2t$ . So there exist $\left\{w_{1},\dots,w_{i}\right\}\subseteq W$ such that $\left\{v_{1},\dots,v_{\ell},w_{1},\dots,w_{i}\right\}\subseteq\left\langle e_{t+1},\dots,e_{n}\right\rangle$ is a linearly independent set.

Let $M\in I_{t}\otimes\mathrm{GL}_{n-t}\left(\mathbb{F}_{p}\right)$ be a linear transformation such that $M\pi_{2}\left(c_{i}e_{j}\right)=\tilde{c}_{i+1}^{-1}w_{j}$ for every $1\leq j\leq i$ . For $1\leq j\leq i$ ,

	$\displaystyle\pi_{2}\left(\tilde{c}_{i+1}Mc_{i}e_{j}\right)$	$\displaystyle=\pi_{2}\left(\tilde{c}_{i+1}\left(\pi_{1}\left(Mc_{i}e_{j}\right)+\pi_{2}\left(Mc_{i}e_{j}\right)\right)\right)$
		$\displaystyle=\pi_{2}\tilde{c}_{i+1}\pi_{1}Mc_{i}e_{j}+\pi_{2}\tilde{c}_{i+1}\pi_{2}Mc_{i}e_{j}$

Since $M\in I_{t}\otimes\mathrm{GL}_{n-t}\left(\mathbb{F}_{p}\right)$ ,

\pi_{2}\left(\tilde{c}_{i+1}Mc_{i}e_{j}\right)=\pi_{2}\tilde{c}_{i+1}\pi_{1}c_{i}e_{j}+\pi_{2}\tilde{c}_{i+1}\pi_{2}Mc_{i}e_{j}.

Since $M\pi_{2}c_{i}e_{j}=\tilde{c}_{i+1}^{-1}w_{j}$ ,

	$\displaystyle\pi_{2}\left(\tilde{c}_{i+1}Mc_{i}e_{j}\right)$	$\displaystyle=\pi_{2}\tilde{c}_{i+1}\pi_{1}c_{i}e_{j}+\pi_{2}w_{j}$
		$\displaystyle=\pi_{2}\tilde{c}_{i+1}\pi_{1}c_{i}e_{j}+w_{j}\in V\oplus W$

On the other hand,

\pi_{2}\left(\tilde{c}_{i+1}Mc_{i}e_{i+1}\right)=\pi_{2}\tilde{c}_{i+1}\pi_{1}c_{i}e_{i+1}+\sum_{j=1}^{i}\alpha_{j}w_{j}\in V\oplus W

Since $w_{j}$ are linearly independent and $V\cap W=\left\{0\right\}$ then $\left\{\pi_{2}\left(\tilde{c}_{i+1}Mc_{i}e_{j}\right)\right\}_{j=1}^{i}$ is a linearly independent set.

Suppose towards contradiction that $\pi_{2}\left(\tilde{c}_{i+1}Mc_{i}e_{i+1}\right)$ is linearly dependent on $\left\{\pi_{2}\left(\tilde{c}_{i+1}Mc_{i}e_{j}\right)\right\}_{j=1}^{i}$ , then there exist $\left\{\beta_{j}\right\}_{j=1}^{i}\subseteq\mathbb{F}_{p}$ such that

\pi_{2}\left(\tilde{c}_{i+1}Mc_{i}e_{i+1}\right)=\sum_{j=1}^{i}\beta_{j}\pi_{2}\left(\tilde{c}_{i+1}Mc_{i}e_{j}\right)

In particular, since $V\oplus W$ is a direct sum of vector spaces

\sum_{j=1}^{i}\beta_{j}w_{j}=\sum_{j=1}^{i}\alpha_{j}w_{j}

thus $\beta_{j}=w_{j}$ . On the other hand,

\pi_{2}\tilde{c}_{i+1}\pi_{1}c_{i}e_{i+1}=\sum_{j=1}^{i}\beta_{j}\pi_{2}\tilde{c}_{i+1}\pi_{1}c_{i}e_{j}

thus

	$\displaystyle 0$	$\displaystyle=\pi_{2}\left(\tilde{c}_{i+1}\pi_{1}c_{i}e_{i+1}-\sum_{j=1}^{i}\beta_{j}\tilde{c}_{i+1}\pi_{1}c_{i}e_{j}\right)$
		$\displaystyle=\pi_{2}\left(\tilde{c}_{i+1}\pi_{1}c_{i}e_{i+1}-\sum_{j=1}^{i}\alpha_{j}\tilde{c}_{i+1}\pi_{1}c_{i}e_{j}\right)$
		$\displaystyle=\pi_{2}\left(\tilde{c}_{i+1}z\right)\neq 0$

So for $c_{i+1}=\tilde{c}_{i+1}Mc_{i}$ , the set $\left\{\pi_{2}\left(c_{i+1}e_{j}\right)\right\}_{j=1}^{i+1}$ is linearly independent. As every iteration takes $O(t)$ steps and there are $t$ such iterations, it takes $O(t^{2})$ steps to construct $c_{t}$ . ∎

Our goal is to modify $a$ so that in addition to $ae_{i}=e_{t+i}$ for $1\leq i\leq t$ , we will also have $ae_{t+i}=e_{i}$ . To achieve this, we first prove the following lemma,

Lemma 15.

Let $A\in\mathrm{Mat}_{t\times t}\left(\mathbb{F}_{p}\right)$ and $B\in\mathrm{Mat}_{t\times n-2t}\left(\mathbb{F}_{p}\right)$ , where $\begin{pmatrix}A&B\end{pmatrix}$ has full column rank; then there exists $J\in\mathrm{Mat}_{n-t\times t}\left(\mathbb{F}_{p}\right)$ such that $A+BJ$ has full column rank.

Proof.

We prove by induction on $t-\dim\left(\mathrm{col}\left(A\right)\right)$ . If $\dim\left(\mathrm{col}\left(A\right)\right)=t$ , then we are done. Otherwise, there exists $A_{i}$ such that $A_{i}\in\mathrm{Span}\left\{A_{j}\right\}_{j\neq i}$ ; without loss of generality, assume $i=1$ . Since $\dim\left(\mathrm{col}\left(A\right)\right)<t$ , there exists $B_{k}$ for $1\leq k\leq n-2t$ such that $\dim\left(\mathrm{col}\left(A\right)+\left\langle B_{k}\right\rangle\right)=\dim\left(\mathrm{col}\left(A\right)\right)+1$ . Thus for $J=E_{k1}$ ,

\dim\mathrm{col}\begin{pmatrix}A_{1}+B_{k}&A_{2}&\dots&A_{t}\end{pmatrix}=\dim\mathrm{col}\left(A+BJ\right)=\dim\left(\mathrm{col}\left(A\right)\right)+1.

∎

Lemma 16 (Swapping Lemma).

A_{\mathrm{swap}}=\begin{pmatrix}0&I_{t\times t}&0\\ I_{t\times t}&0&0\\ 0&0&I_{n-2t\times n-2t}\end{pmatrix}\in A^{O\left(t^{2}\right)}

Proof.

In Lemma 12, we constructed $a\in A^{O(t^{2})}$ which such that $ae_{i}=e_{t+i}$ for every $1\leq i\leq t$ . Thus the matrix $a$ is of the form

a=\begin{pmatrix}0_{t\times t}&A_{t\times t}&B_{t\times n-2t}\\ I_{t\times t}&C_{t\times t}&D_{t\times n-2t}\\ 0_{n-2t\times t}&E_{n-2t\times t}&F_{n-2t\times n-2t}\end{pmatrix}

where $\begin{pmatrix}A&B\end{pmatrix}$ has full column rank and $\begin{pmatrix}E&F\end{pmatrix}$ also has full column rank. Since $\begin{pmatrix}A&B\end{pmatrix}$ has full column rank there exists $X_{1}\in\mathrm{Mat}_{n-t\times t}\left(\mathbb{F}_{p}\right)$ such that $A+BX_{1}$ has rank $t$ . Applying $\begin{pmatrix}I_{t}&0&0\\ 0&I_{t}&0\\ 0&X_{1}&I_{n-2t}\end{pmatrix}$ from the right

	$\displaystyle a$	$\displaystyle\leftarrow\begin{pmatrix}0_{t\times t}&A_{t\times t}+B_{t\times n-2t}X_{1}&B_{t\times n-2t}\\ I_{t\times t}&C_{t\times t}+D_{t\times n-2t}X_{1}&D_{t\times n-2t}\\ 0_{n-2t\times t}&E_{n-2t\times t}+F_{n-2t\times n-2t}X_{1}&F_{n-2t\times n-2t}\end{pmatrix}$
		$\displaystyle=\begin{pmatrix}0_{t\times t}&A_{t\times t}^{\prime}&B_{t\times n-2t}\\ I_{t\times t}&C_{t\times t}^{\prime}&D_{t\times n-2t}\\ 0_{n-2t\times t}&E_{n-2t\times t}^{\prime}&F_{n-2t\times n-2t}\end{pmatrix}$

Now $A_{t\times t}^{\prime}$ is full rank. We apply $\begin{pmatrix}I_{t\times t}&0&0\\ 0&\left(A_{t\times t}^{\prime}\right)^{-1}&0\\ 0&0&I_{n-2t\times n-2t}\end{pmatrix}$ from the right.

a\leftarrow\begin{pmatrix}0_{t\times t}&I_{t\times t}&B_{t\times n-2t}\\ I_{t\times t}&C_{t\times t}^{\prime}\left(A_{t\times t}^{\prime}\right)^{-1}&D_{t\times n-2t}\\ 0_{n-2t\times t}&E_{n-2t\times t}^{\prime}\left(A_{t\times t}^{\prime}\right)^{-1}&F_{n-2t\times n-2t}\end{pmatrix}

renaming

	$\displaystyle C^{\prime\prime}$	$\displaystyle=C^{\prime}A^{\prime-1}$
	$\displaystyle E^{\prime\prime}$	$\displaystyle=E^{\prime}A^{\prime-1}.$

We have

a\leftarrow\begin{pmatrix}0_{t\times t}&I_{t\times t}&B_{t\times n-2t}\\ I_{t\times t}&C_{t\times t}^{\prime\prime}&D_{t\times n-2t}\\ 0_{n-2t\times t}&E_{n-2t\times t}^{\prime\prime}&F_{n-2t\times n-2t}\end{pmatrix}

Now we apply

\begin{pmatrix}I_{t\times t}&0&0\\ 0&I_{t\times t}&-B_{t\times n-2t}^{\prime}\\ 0&0&I_{n-2t\times n-2t}\end{pmatrix}

from the right.

	$\displaystyle a$	$\displaystyle\leftarrow\begin{pmatrix}0_{t\times t}&I_{t\times t}&B_{t\times n-2t}-B_{t\times n-2t}\\ I_{t\times t}&C_{t\times t}^{\prime\prime}&D_{t\times n-2t}-C_{t\times t}^{\prime\prime}B_{t\times n-2t}\\ 0_{n-2t\times t}&E_{n-2t\times t}^{\prime\prime}&F_{n-2t\times n-2t}-E_{t\times t}^{\prime\prime}B_{t\times n-2t}\end{pmatrix}$
		$\displaystyle=\begin{pmatrix}0_{t\times t}&I_{t\times t}&0_{t\times t}\\ I_{t\times t}&C_{t\times t}^{\prime\prime}&D_{t\times n-2t}-C_{t\times t}^{\prime\prime}B_{t\times n-2t}\\ 0_{n-2t\times t}&E_{n-2t\times t}^{\prime\prime}&F_{n-2t\times n-2t}-E_{t\times t}^{\prime\prime}B_{t\times n-2t}\end{pmatrix}$

As $Y=\begin{pmatrix}I_{t\times t}&D_{t\times n-2t}-C_{t\times t}^{\prime\prime}B_{t\times n-2t}\\ 0_{n-2t\times t}&F_{n-2t\times n-2t}-E_{t\times t}^{\prime\prime}B_{t\times n-2t}\end{pmatrix}$ has full row rank we can apply $\begin{pmatrix}I&0\\ 0&Y_{n-t\times n-2t}^{-1}\end{pmatrix}$ from the left to get

a\leftarrow\begin{pmatrix}0_{t\times t}&I_{t\times t}&0_{t\times t}\\ I_{t\times t}&X&0_{t\times n-2t}\\ 0_{n-2t\times t}&Z&I_{n-2t\times n-2t}\end{pmatrix}

from some $X$ and $Z$ . Apply

\begin{pmatrix}I_{t}&0&0\\ 0&I_{t}&0\\ 0&-Z&I_{n-2t}\end{pmatrix}

from the right to obtain

a\leftarrow\begin{pmatrix}0_{t\times t}&I_{t\times t}&0_{t\times t}\\ I_{t\times t}&X&0_{t\times n-2t}\\ 0_{n-2t\times t}&0&I_{n-2t\times n-2t}\end{pmatrix}

Restricting our focus on the first $3t$ indices of the matrix

\tilde{a}=\begin{pmatrix}0&I_{t\times t}&0\\ I_{t\times t}&X&0\\ 0&0&I_{t\times t}\end{pmatrix}

as we could construct the matrix $\tilde{a}$ , we can also construct $\tilde{a}^{-1}$ by applying the inverse transformations in reverse order

\tilde{a}^{-1}=\begin{pmatrix}-X&I_{t\times t}&0\\ I_{t\times t}&0&0\\ 0&0&I_{t\times t}\end{pmatrix}.

Conjugating the constructable matrix (as it is in the groumvirate)

b=\begin{pmatrix}I_{t\times t}&0&0\\ 0&I_{t\times t}&-X\\ 0&0&I_{t\times t}\end{pmatrix}

by the matrix $\tilde{a}^{-1}$ ,

\tilde{a}^{-1}b\tilde{a}=\begin{pmatrix}I_{t\times t}&0&-X\\ 0&I_{t\times t}&0\\ 0&0&I_{t\times t}\end{pmatrix}.

Applying a permutation matrix $P$ from both sides which swaps the $\left\{t+1,\dots,2t\right\}$ columns with the $\left\{2t+1,\dots,3t\right\}$ , we get a matrix

P\tilde{A}^{-1}B\tilde{A}P=\begin{pmatrix}I_{t\times t}&-X&0\\ 0&I_{t\times t}&0\\ 0&0&I_{t\times t}\end{pmatrix}

and

\tilde{A}P\tilde{A}^{-1}B\tilde{A}P=\begin{pmatrix}0&I_{t\times t}&0\\ I_{t\times t}&X&0\\ 0&0&I_{t\times t}\end{pmatrix}\begin{pmatrix}I_{t\times t}&-X&0\\ 0&I_{t\times t}&0\\ 0&0&I_{t\times t}\end{pmatrix}=\begin{pmatrix}0&I_{t\times t}&0\\ I_{t\times t}&0&0\\ 0&0&I_{t\times t}\end{pmatrix}

which is the desired swap matrix. ∎

Proof of Theorem 8.

By Lemma 16, the matrix $A_{\mathrm{swap}}$ can be constructed in $O\left(t^{2}\right)$ steps. Combining the swap transformation and any permutation on the last $n-t$ vectors, we obtain a transitive action on $V^{\prime}$ . The distance between any two vertices in $V^{\prime}$ is $O(t^{2})$ since each step in the proof requires at most $O(t^{2})$ multiplications. ∎

3. Constructing the Matrix Group via Its Bruhat Decomposition

Our proof relies on the following decomposition of $\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)$ known in the literature as the Bruhat decomposition (a reference for the case $\mathrm{SL}_{n}\left(\mathbb{F}_{p}\right)$ is proven in [6]).

Theorem 17.

$\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)=BWB$ where $B\subseteq\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)$ is the group of lower-triangular matrices and $W\subseteq\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)$ is the set of monomial matrices (the set of matrices in which each row and column contains exactly one non-zero element).

To construct a matrix $M\in\mathrm{GL}_{n}\left(\mathbb{F}_{p}\right)$ , we construct its Bruhat decomposition $b_{1},b_{2}\in B$ and $w\in W$ and we compose all elements to get $M=b_{1}wb_{2}$ .

Lemma 18.

The triangular matrices can be constructed in $O\left(n^{2}\right)$ steps, i.e., $B\subseteq A^{O\left(n^{2}\right)}$ .

Proof.

Let $L\in B$ . Using Theorem 8, we construct a primitive which modifies the $\left\{1,\dots,\frac{n}{3}\right\}$ columns for row indices $J\subseteq\left\{\frac{n}{3}+1,\dots,n\right\}$ where $\left|J\right|=\frac{n}{3}$ . Applying this primitive 2 times enables us to construct the $\left\{1,\dots,\frac{n}{3}\right\}$ columns of the matrix. Using Lemma 7, we can construct the rest of the columns using a matrix in $A^{4}$ . ∎

Lemma 19.

The monomial matrices can be constructed in $O\left(n^{2}\right)$ steps, i.e., $W\subseteq A^{O\left(n^{2}\right)}$ .

Proof.

Using Theorem 8, we construct a matrix which swaps the $\{1,\dots,t\}$ columns with the $\{\sigma(1),\dots,\sigma(t)\}$ columns in $A^{O(t^{2})}$ steps. Now all of the first $t$ columns are fixed in position. All that remains is to set the remaining $n-t$ columns in place. By Corollary 9, any permutation of the last $n-t$ columns can be constructed in $O(t^{2})$ steps. ∎

4. A Matching Lower Bound on the Diameter

Let $t=Cn$ for $0<C<1$ . We define the generating set

A=\left\{\begin{pmatrix}I_{t\times t}&0\\ 0&X\end{pmatrix}:X\in\mathrm{GL}_{n-t}\left(\mathbb{F}_{p}\right)\right\}\bigcup\left\{e_{i}\leftrightarrow e_{i+1}:i=1,\dots,t\right\},

where $e_{i}\leftrightarrow e_{i+1}$ is the swap between $e_{i}$ and $e_{i+1}$ which fixes the other basis vectors. This is a generating set because it generates the swap matrix (as transpositions generate the permutations). Moreover,

\left|A\right|=\Theta\left(p^{\left(n-t\right)^{2}}\right)=\Theta\left(\left(p^{n^{2}}\right)^{\left(\frac{n-t}{n}\right)^{2}}\right)\leq|G|^{1-c}

for an appropriately chosen $c$ . Let $B=\begin{pmatrix}0&I_{t\times t}&0\\ I_{t\times t}&0&0\\ 0&0&I_{n-2t\times n-2t}\end{pmatrix}$ . Assume $B=a_{k}a_{k-1}\dots a_{1}$ , where $a_{i}\in A$ . Define $A_{0}=e$ , and $A_{l}=a_{l}a_{l-1}\dots a_{1}$ for $1\leq l\leq k$ . We set $\mathcal{F}_{0}=\left\{1,\dots,t\right\}$ and $\mathcal{F}_{l}=\left\{i\in\left\{1,\dots,t\right\}:A_{j}e_{i}\in\left\{e_{1},\dots,e_{t}\right\}\forall 1\leq j\leq l\right\}$ . Let

d_{i,l}=\begin{cases}t+1-j&\text{if }A_{l}e_{i}=e_{j}\text{ for }j\in\left\{1,\dots,t+1\right\}\wedge i\in\mathcal{F}_{l}\\ 0&\text{otherwise}\end{cases}

and

d_{l}=\sum_{i=1}^{t}d_{i,l}.

We prove that $d_{l+1}\geq d_{l}-1$ for every $1\leq l\leq k-1$ . Assume $a_{l+1}$ is a swap between $e_{i}$ and $e_{i+1}$ then $d_{l}\neq d_{l+1}$ iff $a_{l+1}$ swaps $t$ and $t+1$ , thus $d_{l+1}=d_{l}-1$ . Otherwise $a_{l+1}$ acts on $\left\{e_{t+1},\dots,e_{n}\right\}$ so it cannot change $d_{i,l}$ for any $i$ . Therefore

d_{k}\geq d_{0}-k=\sum_{i=1}^{t}i-k=\begin{pmatrix}t+1\\ 2\end{pmatrix}-k.

But $d_{k}=0$ since $B$ is the matrix that swaps $e_{1},\dots,e_{t}$ with $e_{t+1},\dots,e_{2t}$ , and thus

k\geq\begin{pmatrix}t+1\\ 2\end{pmatrix}=\Omega\left(n^{2}\right).

Acknowledgments

We would like to thank Noam Lifshitz for his guidance, Elad Tzalik for helpful discussions, and Edan Orzech for reviewing the manuscript.

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