Proof. Let $A$ is a ring which is elementarily equivalent to $\mathbb{Z}$ (i.e. satisfies the same first-order sentences as $\mathbb{Z}$ ) and let $I$ be an ideal of $A$ . Proposition 2.3 and Theorem 2.4 imply that

(1) $$\begin{eqnarray}[\operatorname{SL}_{n}(A;I^{2}):\operatorname{SL}_{n}(A;I^{2})\cap \langle U_{n}(A;I)L_{n}(A;I)\rangle ]\leqslant 2\cdot 8!.\end{eqnarray}$$

Assume the corollary is false. Then, for every $k\in \mathbb{N}$ , there are $q_{k}\in \mathbb{N}^{+}$ and matrices $g_{k,1},\ldots ,g_{k,2\cdot 8!+1}\in \operatorname{SL}_{n}(\mathbb{Z};q_{k}^{2})$ such that $(g_{k,i})^{-1}g_{k,j}\notin (U_{n}(\mathbb{Z};q_{k})L_{n}(\mathbb{Z};q_{k}))^{k}$ if $i\neq j$ .

Choose a non-principal ultrafilter ${\mathcal{U}}$ on $\mathbb{N}$ , and let $A$ be the ultrapower of $\mathbb{Z}$ over ${\mathcal{U}}$ . Then $A$ is elementarily equivalent to $\mathbb{Z}$ and $\operatorname{SL}_{n}(A)$ is isomorphic to the ultrapower of $\operatorname{SL}_{n}(\mathbb{Z})$ over ${\mathcal{U}}$ . Let $I$ be the ideal of $A$ represented by $\prod _{k}q_{k}\mathbb{Z}$ , and for every $1\leqslant i\leqslant k$ , let $g_{i}\in \operatorname{SL}_{n}(A;I^{2})$ be the element represented by $(g_{k,i})_{k}$ . Then $g_{1},\ldots ,g_{2\cdot 8!+1}$ belong to different cosets of $\langle U_{n}(A;I)L_{n}(A;I)\rangle$ , contradicting (1).◻

Proof. Denote $Y=X^{2}$ . Then $1\in Y$ and there are $d$ translates of $Y$ that cover $G$ . Since $1\in Y$ , $Y^{k}\subseteq Y^{k+1}$ for every $k$ . It is enough to show that $Y^{k}=Y^{k+1}$ for some $k\leqslant 2d+1$ . Suppose that $G=\bigcup _{i=1}^{d}g_{i}Y$ for some $g_{1},\ldots ,g_{d}\in G$ . We can assume that $g_{1}=1$ . For every $k$ , if $Y^{k}\neq Y^{k+1}$ , choose $h\in Y^{k+1}\smallsetminus Y^{k}$ . By assumption, there is $i$ such that $h\in g_{i}Y$ . Then $g_{i}\in Y^{k+2}$ but $g_{i}\notin Y^{k}$ . By induction we see that if $Y^{2k-1}\neq Y^{2k}$ for some $1\leqslant k$ , then $Y^{2k+1}$ contains at least $k$ distinct $g_{i}$ . This implies that $Y^{2d+1}=Y^{2d+2}$ .◻

Proof. Since $1\in K\subseteq X$ , the sets $(X^{n}K\cap H)\subseteq H$ are non-decreasing. Hence, there is $n\leqslant 4[H:K]-3$ such that

$$\begin{eqnarray}X^{n}K\cap H=X^{n+1}K\cap H=X^{n+2}K\cap H=X^{n+3}K\cap H=X^{n+4}K\cap H.\end{eqnarray}$$

Since $HX=G$ , we have $X^{n+3}\subseteq (X^{n+4}\cap H)X$ . Thus,

$$\begin{eqnarray}X^{n+3}\subseteq (X^{n+4}\cap H)X\subseteq (X^{n+4}K\cap H)X\subseteq (X^{n}K\cap H)X\subseteq X^{n+2},\end{eqnarray}$$

so $X^{n+2}$ is a group.◻

Proof. For any $k$ , the set $(L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q))^{k+1}$ contains the symmetric subset $(L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q))^{k}\cup (U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q))^{k}$ . Corollary 2.6 and Lemma 2.7 imply that there is a constant $D^{\prime }$ such that $(L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q))^{D^{\prime }}$ contains a subgroup $S(I)$ of $\operatorname{SL}_{n}(\mathbb{Z};q^{2})$ of index at most $2\cdot 8!$ .

Note that $\operatorname{SL}_{n}(\mathbb{Z},q)/\text{SL}_{n}(\mathbb{Z},q^{2})$ is abelian so $\operatorname{SL}_{n}(\mathbb{Z},q^{2})L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)$ is a subgroup of $\operatorname{SL}_{n}(\mathbb{Z})$ . The desired result follows by applying Lemma 2.8 to $K=S(I)$ , $H=\operatorname{SL}_{n}(\mathbb{Z};q^{2})$ , $G=\operatorname{SL}_{n}(\mathbb{Z},q^{2})L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)$ , and $X=(L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q))^{D^{\prime }}\cup (U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z}:q))^{D^{\prime }}\subseteq (L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q))^{D^{\prime }+1}$ .◻

Proof. We follow the proof of [Reference TitsTit76, Proposition 2.3].

Let $1\leqslant i\neq j\leqslant n+1$ , $1\leqslant r\neq s\leqslant n+1$ , and $a,b\in A$ . Recall the following relations:

(2) $$\begin{eqnarray}\left\{\begin{array}{@{}ll@{}}e_{r,s}(b)e_{i,j}(a)e_{r,s}(b)^{-1}=e_{i,j}(a)e_{i,s}(-ab)\quad & \text{if }j=r\text{ and }i\neq s,\\ e_{r,s}(b)e_{i,j}(a)e_{r,s}(b)^{-1}=e_{i,j}(a)e_{r,j}(ab)\quad & \text{if }j\neq r\text{ and }i=s,\\ e_{r,s}(b)e_{i,j}(a)e_{r,s}(b)^{-1}=e_{i,j}(a)\quad & \text{if }j\neq r\text{ and }i\neq s.\end{array}\right.\end{eqnarray}$$

For every $1\leqslant i\neq j\leqslant n+1$ , denote $F_{i,j}(I^{2}):=\langle e_{i,j}(a),e_{j,i}(a)\mid a\in I^{2}\rangle$ . Let $F_{i,j}^{\lhd }(I^{2})$ be the minimal normal subgroup of $F_{i,j}:=F_{i,j}(A)$ which contains $F_{i,j}(I^{2})$ . Define $F^{\lhd }(I^{2}):=\langle F_{i,j}^{\lhd }(I^{2})\mid 1\leqslant i\neq j\leqslant n+1\rangle$ . Equation (2) implies that for every $1\leqslant i\neq j\leqslant n+1$ and every $a\in A$ , $e_{i,j}(a)F^{\lhd }(I^{2})e_{i,j}(a)^{-1}=F^{\lhd }(I^{2})$ . Thus $F^{\lhd }(I^{2})$ is a normal subgroup of $E(n+1,A)$ containing all $e_{i,j}(a)$ , $a\in I^{2}$ , so it must be equal to $E^{\lhd }(n+1,A,I^{2})$ . Thus, in order to finish the proof it is enough to show that for every $1\leqslant i<j\leqslant n+1$ , $F_{i,j}^{\lhd }(I^{2})\subseteq K(I)$ .

Let $E^{+}(n,A;I)$ and $E^{-}(n,A;I)$ be the images of $E(n,A,I)$ in $\operatorname{SL}_{n+1}(A)$ under the embeddings $M\mapsto (\!\begin{smallmatrix}M & 0\\ 0 & 1\end{smallmatrix}\!)$ and $M\mapsto (\!\begin{smallmatrix}1 & 0\\ 0 & M\end{smallmatrix}\!)$ . By applying Proposition 2.3 with respect to $E^{+}(n,A;I)$ and $E^{-}(n,A;I)$ , we see that $K(I)$ contains $F_{i,j}^{\lhd }(I^{2})$ for every $1\leqslant i<j\leqslant n+1$ such that $(i,j)\neq (1,n+1)$ .

Equation (2) implies that $K(I)$ in normalized by $e_{1,n+1}(a)$ and $e_{n+1,1}(a)$ for every $a\in R$ . For every $a,b\in I$ , $e_{1,n+1}(ab)=[e_{1,2}(a),e_{2,n+1}(-b)]\in K(I)$ and $e_{n+1,1}(ab)=[e_{n+1,2}(a),e_{2,1}(-b)]\in K(I)$ . Thus, $F_{1,n+1}^{\lhd }(I^{2})\leqslant K(I)$ .◻

Proof. The proof follows [Reference Dennis and VasersteinDV88, proof of Lemma 7] and is by induction on $m$ . The base case $m=n$ is clear. It remains to show that if the claim is true for some $m\geqslant 3$ then it is also true for $m+1$ .

Let $T:=(U_{m+1}(\mathbb{Z};q)L_{m+1}(\mathbb{Z};q))^{D}$ and $H=\{g\in \operatorname{SL}_{m+1}(\mathbb{Z})\mid gT=T\}$ . Since $H$ is a group, it is enough to prove that $H$ contains both $U_{m+1}(\mathbb{Z};q)$ (which is clear) and $L_{m+1}(\mathbb{Z};q)$ .

We embed $L_{m}(\mathbb{Z};q)$ and $U_{m}(\mathbb{Z};q)$ in $\operatorname{SL}_{m+1}(\mathbb{Z};q)$ by the embedding $M\mapsto (\!\begin{smallmatrix}M & 0\\ 0 & 1\end{smallmatrix}\!)$ . We denote the abelian group $\langle e_{i,m+1}(a)\mid 1\leqslant i\leqslant m,~a\in q\mathbb{Z}\rangle$ by $C_{m+1}(\mathbb{Z};q)$ and the abelian group $\langle e_{m+1,i}(q)\mid 1\leqslant i\leqslant m,~a\in q\mathbb{Z}\rangle$ by $R_{m+1}(\mathbb{Z};q)$ . We have that $U_{m+1}(\mathbb{Z};q)=U_{m}(\mathbb{Z};q)\ltimes C_{m}(\mathbb{Z};q)$ , that $L_{m+1}(\mathbb{Z};q)=L_{m}(\mathbb{Z};q)\ltimes R_{m}(\mathbb{Z};q)$ , and that $U_{m}(\mathbb{Z};q)$ and $L_{m}(\mathbb{Z};q)$ each normalize both $C_{m}(\mathbb{Z};q)$ and $R_{m}(\mathbb{Z};q)$ . The induction hypothesis implies that

$$\begin{eqnarray}\displaystyle & & \displaystyle L_{m}(\mathbb{Z}:q)(U_{m+1}(\mathbb{Z}:q)L_{m+1}(\mathbb{Z};q))^{D}\nonumber\\ \displaystyle & & \displaystyle \quad =L_{m}(\mathbb{Z}:q)(U_{m}(\mathbb{Z}:q)C_{m}(\mathbb{Z}:q)L_{m}(\mathbb{Z}:q)R_{m}(\mathbb{Z}:q))^{D}\nonumber\\ \displaystyle & & \displaystyle \quad =L_{m}(\mathbb{Z}:q)(U_{m}(\mathbb{Z}:q)L_{m}(\mathbb{Z}:q))^{D}\cdot (C_{m}(\mathbb{Z}:q)R_{m}(\mathbb{Z}:q))^{D}\nonumber\\ \displaystyle & & \displaystyle \quad =(U_{m}(\mathbb{Z}:q)L_{m}(\mathbb{Z}:q))^{D}\cdot (C_{m}(\mathbb{Z}:q)R_{m}(\mathbb{Z}:q))^{D}\nonumber\\ \displaystyle & & \displaystyle \quad =(U_{m}(\mathbb{Z}:q)C_{m}(\mathbb{Z}:q)L_{m}(\mathbb{Z}:q)R_{m}(\mathbb{Z}:q))^{D}\nonumber\\ \displaystyle & & \displaystyle \quad =(U_{m+1}(\mathbb{Z}:q)L_{m+1}(\mathbb{Z}:q))^{D}.\nonumber\end{eqnarray}$$

Hence, $L_{m}(\mathbb{Z}:q)\subseteq H$ , that is, for every $1\leqslant i<j\leqslant m$ and $a\in I$ , we have $e_{j,i}(a)\in H$ . Arguing similarly using the embedding $M\mapsto (\!\begin{smallmatrix}1 & 0\\ 0 & M\end{smallmatrix}\!)$ , we get that $e_{j,i}(a)\in H$ , for every $2\leqslant i<j\leqslant m+1$ and $a\in I$ . It remains to show that for every $a\in I$ , $e_{m+1,1}(a)\in H$ .

The main theorem of [Reference MennickeMen65] says that $E^{\lhd }(n,\mathbb{Z},k)=\operatorname{SL}_{n}(\mathbb{Z},k)$ for every $k\in \mathbb{N}^{+}$ . Thus, Lemma 2.10 implies that $\operatorname{SL}_{m+1}(\mathbb{Z};q^{2})=E^{\lhd }(m+1,\mathbb{Z};q^{2})\subseteq H$ . Since $\operatorname{SL}_{m+1}(\mathbb{Z};q)/\text{SL}_{m+1}(\mathbb{Z};q^{2})$ is abelian, $e_{m+1,1}(a)U_{m+1}(\mathbb{Z}:q)e_{m+1,1}(a)^{-1}\subseteq \operatorname{SL}_{m+1}(\mathbb{Z};q^{2})\cdot U_{m+1}(\mathbb{Z}:q)$ , for every $a\in I$ . It follows that, for every $a\in I$ ,

$$\begin{eqnarray}\displaystyle & & \displaystyle e_{m+1,1}(a)(U_{m+1}(\mathbb{Z}:q)L_{m+1}(\mathbb{Z}:q))^{D}\nonumber\\ \displaystyle & & \displaystyle \quad =e_{m+1,1}(a)U_{m+1}(\mathbb{Z}:q)e_{m+1,1}(a)^{-1}e_{m+1,1}(a)L_{m+1}(\mathbb{Z}:q)(U_{m+1}(\mathbb{Z}:q)L_{m+1}(\mathbb{Z}:q))^{D-1}\nonumber\\ \displaystyle & & \displaystyle \quad \subseteq \operatorname{SL}_{m+1}(\mathbb{Z};q^{2})\cdot U_{m+1}(\mathbb{Z}:q)L_{m+1}(\mathbb{Z}:q)(U_{m+1}(\mathbb{Z}:q)L_{m+1}(\mathbb{Z}:q))^{D-1}\nonumber\\ \displaystyle & & \displaystyle \quad =(U_{m+1}(\mathbb{Z}:q)L_{m+1}(\mathbb{Z}:q))^{D}.\nonumber\end{eqnarray}$$

In particular, for every $a\in I$ , $e_{m+1,1}(a)\in H$ . Hence, $L_{m+1}(\mathbb{Z}:q)\subseteq H$ as desired.◻

Proof of Theorem 1.7.

Proposition 2.9 implies that there is a constant $C=D(3)$ such that $(U_{3}(\mathbb{Z};q)L_{3}(\mathbb{Z};q))^{C}=\langle U_{3}(\mathbb{Z};q)L_{3}(\mathbb{Z};q)\rangle$ . Lemma 2.11 implies that $(U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q))^{C}=\langle U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q)\rangle$ for every $n\geqslant 3$ . Proposition 2.3 implies that $(U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q))^{C}$ contains the congruence subgroup $\operatorname{SL}_{n}(\mathbb{Z};q^{2})$ and this subgroup has a finite index in $\operatorname{SL}(n,\mathbb{Z})$ . ◻

Proof of Theorem 1.4 without an explicit bound.

Identify $\operatorname{SL}_{2}(\mathbb{Z})$ with its image in $\operatorname{SL}_{3}(\mathbb{Z})$ under the embedding $M\mapsto (\!\begin{smallmatrix}M & 0\\ 0 & 1\end{smallmatrix}\!)$ . Since $\operatorname{SL}_{2}(\mathbb{Z})$ contains a non-abelian free group there exists $\pm I_{2}\neq g\in w(\operatorname{SL}_{2}(\mathbb{Z}))$ . There exists $h\in \langle e_{1,3}(1),e_{2,3}(1)\rangle$ such that $[g,h]=g^{-1}h^{-1}gh$ is a non-trivial element and this element is conjugate to $e_{1,3}(q)$ for some positive $q\in \mathbb{N}$ . For the chosen $g$ and $h$ , we have $[g,h]^{n}=[g,h^{n}]\in w(\operatorname{SL}_{3}(\mathbb{Z}))^{2}$ . Since $w(\operatorname{SL}_{3}(\mathbb{Z}))^{2}$ is a normal subset, $\langle e_{1,3}(q)\rangle \subseteq w(\operatorname{SL}_{3}(\mathbb{Z}))^{2}$ . We will show that the statement of Theorem 1.4 holds with respect to $d=q^{2}$ .

We claim that for any integers $a_{1},\ldots ,a_{n-1}$ , there is $g\in w(\operatorname{SL}_{n}(\mathbb{Z}))^{8}\cap U_{n}(\mathbb{Z};q)$ such that for every $i$ , $g_{i,i+1}=qa_{i}$ . Using two different embeddings of the group $\operatorname{SL}_{3}(\mathbb{Z})\times \cdots \times \operatorname{SL}_{3}(\mathbb{Z})$ ( $\lfloor n/3\rfloor$ times) into $\operatorname{SL}_{n}(\mathbb{Z})$ as block-diagonal matrices, we get that there is a matrix $g^{1}\in w(\operatorname{SL}_{n}(\mathbb{Z}))^{4}\cap U_{n}(\mathbb{Z};q)$ such that $g_{i,i+1}^{1}=qa_{i}$ if $i\equiv 1\text{ (mod 3)}$ and $g_{i,i+1}^{1}=0$ otherwise. Using one embedding of the group $\operatorname{SL}_{3}(\mathbb{Z})\times \cdots \times \operatorname{SL}_{3}(\mathbb{Z})$ ( $\lfloor n/3\rfloor$ times) into $\operatorname{SL}_{n}(\mathbb{Z})$ as block-diagonal matrices, we get that there is a matrix $g^{2}\in w(\operatorname{SL}_{n}(\mathbb{Z}))^{2}\cap U_{n}(\mathbb{Z};q)$ such that $g_{i,i+1}^{2}=qa_{i}$ if $i\equiv 2\text{ (mod 3)}$ and $g_{i,i+1}^{2}=0$ otherwise. Similarly, there is $g^{3}\in w(\operatorname{SL}_{n}(\mathbb{Z}))^{2}\cap U_{n}(\mathbb{Z};q)$ such that $g_{i,i+1}^{3}=qa_{i}$ if $i\equiv 3\text{ (mod 3)}$ and $g_{i,i+1}^{3}=0$ otherwise. The matrix $g=g^{0}g^{1}g^{2}\in w(\operatorname{SL}_{n}(\mathbb{Z}))^{8}\cap U_{n}(\mathbb{Z};q)$ satisfies $g_{i,i+1}=qa_{i}$ . The proof of the claim in now complete.

It follows from Lemma 3.1 that $w(\operatorname{SL}_{n}(\mathbb{Z}))^{8}$ contains all elements $g\in U_{n}(\mathbb{Z};q)$ such that $g_{i,i+1}=q$ for every $i$ .

Next, we claim that $U_{n}(\mathbb{Z};q)\subseteq w(\operatorname{SL}_{n}(\mathbb{Z}))^{16}$ . Indeed, let $h\in U_{n}(\mathbb{Z};q)$ . There is an element $f\in w(\operatorname{SL}_{n}(\mathbb{Z}))^{8}\cap U_{n}(\mathbb{Z};q)$ such that for every $i$ , $f_{i,i+1}=q-h_{i,i+1}$ . Then $hf\in U_{n}(\mathbb{Z};q)$ and, for every $i$ , $(fh)_{i,i+1}=q$ , so $fh\in w(\operatorname{SL}_{n}(\mathbb{Z}))^{8}$ . Since $w(\operatorname{SL}_{n}(\mathbb{Z}))^{8}$ is symmetric, it follows that $h\in w(\operatorname{SL}_{n}(\mathbb{Z}))^{16}$ . Similarly, $L_{n}(\mathbb{Z};q)\subseteq w(\operatorname{SL}_{n}(\mathbb{Z}))^{16}$ .

By Theorem 1.7, there is a constant $C$ (independent of $q$ ) such that

$$\begin{eqnarray}\langle U_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)\rangle =(U_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q))^{C}\subseteq w(\operatorname{SL}_{n}(\mathbb{Z}))^{32C}.\end{eqnarray}$$

Propositon 2.3 implies that $\operatorname{SL}_{n}(\mathbb{Z},q^{2})\leqslant \langle U_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)\rangle$ .◻

Proof. Let $\unicode[STIX]{x1D713}_{i}$ be as in the definition of smoothness. After permutation of the indices, we can assume that the $c\times c$ matrix $(\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D713}_{i}}{\unicode[STIX]{x2202}x_{j}}(a))$ is invertible over $\mathbb{Z}_{p}$ . For any $f\in \mathbb{Z}_{p}[x_{1},\ldots ,x_{n}]$ and any $a,b\in \mathbb{Z}_{p}^{n}$ with $0<d(a,b)<1$ , we have $|f(a)-f(b)-\langle \unicode[STIX]{x1D6FB}f(a),a-b\rangle |\leqslant \Vert a-b\Vert ^{2}<\Vert a-b\Vert$ . If $a,b\in X(\mathbb{Z}_{p})$ and $d(a,b)<1$ , we have $\unicode[STIX]{x1D713}_{i}(a)=\unicode[STIX]{x1D713}_{i}(b)=0$ , so $|\langle \unicode[STIX]{x1D6FB}\unicode[STIX]{x1D713}_{i}(a),a-b\rangle |<\Vert a-b\Vert$ . If, in addition, $\unicode[STIX]{x1D70B}(a)=\unicode[STIX]{x1D70B}(b)$ , write $a-b=(v,0)$ , where $v\in p\mathbb{Z}_{p}^{c}$ and then

$$\begin{eqnarray}\left\Vert \left(\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D713}_{i}}{\unicode[STIX]{x2202}x_{j}}(a)\right)v\right\Vert =\max \{|\langle \unicode[STIX]{x1D6FB}\unicode[STIX]{x1D713}_{i}(a),a-b\rangle |\}<\Vert v\Vert .\end{eqnarray}$$

Since invertible matrices do not decrease norm, this is a contradiction. This completes the proof of statement (1). Denoting $S:=\{c+1,\ldots ,n\}$ , statement (2) readily follows form the assumption that $(\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D713}_{i}}{\unicode[STIX]{x2202}x_{j}}(a))$ is invertible.◻

Proof. We first reduce the claim to the case where $X$ is an affine space. Suppose that $X\subset \mathbb{A}^{n}$ is $d$ -dimensional. By smoothness, it is given as the zero locus of $\unicode[STIX]{x1D711}_{1},\ldots ,\unicode[STIX]{x1D711}_{n-d}\in \mathbb{Z}_{p}[x_{1},\ldots ,x_{n}]$ such that the reductions modulo $p$ of $\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}_{i}(a)$ are linearly independent. Consider the map $F:\mathbb{A}^{n}\rightarrow Y\times \mathbb{A}^{n-d}$ given by $x\mapsto (f(x),p^{k}\unicode[STIX]{x1D711}_{1}(x),\ldots ,p^{k}\unicode[STIX]{x1D711}_{n-d}(x))$ . Then $F$ satisfies the conditions of the lemma. If the claim holds for $F$ , then it holds for  $f$ .

Next, we reduce the claim to the case where $X$ and $Y$ are affine spaces. Indeed, let $e$ be the dimension of $Y$ at $f(a)$ . Item (1) of Lemma 4.1 allows us to assume that the coordinate projection $\unicode[STIX]{x1D70B}:Y\rightarrow \mathbb{A}^{e}$ is one-to-one on $B(f(a),p^{-1})$ . Item (2) of Lemma 4.1 implies that the function $\unicode[STIX]{x1D70B}\circ f$ satisfies the conditions of the lemma, and the claim for $\unicode[STIX]{x1D70B}\circ f$ implies the claim for  $f$ .

Finally, we prove the claim in the case $X=\mathbb{A}^{n}$ and $Y=\mathbb{A}^{m}$ . We can assume that $a=0$ and $f(a)=0$ . Since the coefficients of $f$ are in $\mathbb{Z}_{p}$ , we have that $df(a^{\prime })(\mathbb{Z}_{p}^{n})\supseteq p^{k}\mathbb{Z}_{p}^{m}$ , for any $a^{\prime }\in p\mathbb{Z}_{p}^{n}$ . Let $b\in p^{k+1}\mathbb{Z}_{p}^{m}$ . We will construct a sequence $a_{\ell }\in p\mathbb{Z}_{p}^{n}$ such that $\Vert f(a_{\ell })-b\Vert <p^{-k-\ell }$ . Taking a limit point of the $a_{\ell }$ , we get that $b\in f(\mathbb{Z}_{p}^{n})$ .

The sequence $a_{\ell }$ is defined by recursion starting from $a_{0}=0$ . Given $a_{\ell }$ , the assumptions imply that there is $\unicode[STIX]{x1D716}\in p^{\ell +1}\mathbb{Z}_{p}^{n}$ such that $df(a_{\ell })(\unicode[STIX]{x1D716})=b-f(a_{\ell })$ . We have

$$\begin{eqnarray}\displaystyle \Vert f(a_{\ell }+\unicode[STIX]{x1D716})-b\Vert & = & \displaystyle \Vert f(a_{\ell }+\unicode[STIX]{x1D716})-f(a_{\ell })-df(a_{\ell })(\unicode[STIX]{x1D716})+df(a_{\ell })(\unicode[STIX]{x1D716})+f(a_{\ell })-b\Vert \nonumber\\ \displaystyle & = & \displaystyle \Vert f(a_{\ell }+\unicode[STIX]{x1D716})-f(a_{\ell })-df(a_{\ell })(\unicode[STIX]{x1D716})\Vert \leqslant p^{-k}\Vert \unicode[STIX]{x1D716}\Vert ^{2}<p^{-k-\ell -1},\nonumber\end{eqnarray}$$

since the function $x\mapsto f(a_{\ell }+x)-f(a_{\ell })-df(a_{\ell })(x)$ is a polynomial without constant or linear term and its coefficients are divisible by $p^{k}$ .◻

Proof. After identifying $T_{ab}\operatorname{SL}_{n}=ab+ab\mathfrak{sl}_{n}$ and $\mathfrak{sl}_{n}$ , the differential of $\unicode[STIX]{x1D6F7}_{a,b}$ is $(X,Y)\mapsto (X-X^{a})^{b}+(Y-Y^{b})$ . Let $\unicode[STIX]{x1D711}\in \operatorname{M}_{n}(\mathbb{F}_{q})^{\ast }$ and assume it vanishes on the image of $d\unicode[STIX]{x1D6F7}_{a,b}$ . Then there is $A\in \operatorname{M}_{n}(\mathbb{F}_{q})$ such that $\unicode[STIX]{x1D711}(X)=\operatorname{tr}(AX)$ . For every $Y\in \mathfrak{sl}_{n}(\mathbb{F}_{p})$ , $\unicode[STIX]{x1D711}(Y-Y^{b})=0$ , so $\operatorname{tr}(Y\cdot (A^{b^{-1}}-A))=\operatorname{tr}(A(Y-Y^{b}))=0$ . Thus, $A^{b^{-1}}-A$ is a scalar. Similarly, using the assumption that $\unicode[STIX]{x1D711}((X-X^{a})^{b})=0$ , we get that $(A^{b^{-1}})^{a^{-1}}-A^{b^{-1}}$ is a scalar. Using the fact that $A^{b^{-1}}-A$ is a scalar, we get that $A^{a^{-1}}-A$ is also a scalar. The set $X=\{g\in \operatorname{SL}_{n}(\mathbb{F}_{q})\mid A^{g}-A\text{ is a scalar}\}$ is closed under multiplication. Since $a^{-1},b^{-1}\in X$ , we get $X=\operatorname{SL}_{n}(\mathbb{F}_{q})$ . Since $\operatorname{SL}_{n}(\mathbb{F}_{p})$ is perfect and the function $g\mapsto A^{g}-A$ is a homomorphism between $\operatorname{SL}_{n}(\mathbb{F}_{q})$ and $\mathbb{F}_{q}\operatorname{Id}$ , we get that this homomorphism must be trivial. Hence, $A$ commutes with $\operatorname{SL}_{n}(\mathbb{F}_{q})$ , so it must be scalar. It follows that the restriction of $\unicode[STIX]{x1D711}$ to $\mathfrak{sl}_{n}(\mathbb{F}_{q})$ is zero.◻

Proof. By [Reference Larsen and ShalevLS09], there is $n_{0}$ such that, if $n\geqslant n_{0}$ and $p$ is any prime, then $w(\operatorname{SL}_{n}(\mathbb{F}_{p}))^{2}$ contains all non-scalar matrices. In particular, $w(\operatorname{SL}_{n}(\mathbb{F}_{p}))^{3}=\operatorname{SL}_{n}(\mathbb{F}_{p})$ . Fix a prime $p$ . Choosing generators $a,b\in \operatorname{SL}_{n}(\mathbb{F}_{p})$ (which are not scalars), there are $g,h\in w(\operatorname{SL}_{n}(\mathbb{Z}_{p}))^{2}$ such that the reductions of $g,h$ modulo $p$ are $a,b$ respectively. We get that $w(\operatorname{SL}_{n}(\mathbb{Z}_{p}))^{4}\supset \unicode[STIX]{x1D6F7}_{g,h}(\operatorname{SL}_{n}(\mathbb{Z}_{p})\times \operatorname{SL}_{n}(\mathbb{Z}_{p}))$ . It is well known that $\operatorname{SL}_{n}$ and thus also $\operatorname{SL}_{n}\times \operatorname{SL}_{n}$ are smooth at every point. Lemmas 4.4 and 4.2 imply that $w(\operatorname{SL}_{n}(\mathbb{Z}_{p}))^{4}$ contains the coset $gh\operatorname{SL}_{n}(\mathbb{Z}_{p};p)$ . Hence, $w(\operatorname{SL}_{n}(\mathbb{Z}_{p}))^{7}=\operatorname{SL}_{n}(\mathbb{Z}_{p})$ .

Since $w(\operatorname{SL}_{n}(\widehat{\mathbb{Z}}))=\prod _{p}w(\operatorname{SL}_{n}(\mathbb{Z}_{p}))$ , the claim follows.◻

Proof of Theorem 1.1 (without an explicit bound).

By Theorem 1.4 and Lemma 4.5. ◻

Proof. It is well known that the only non-trivial $\operatorname{SL}_{n}(\mathbb{F}_{p})$ -invariant subspace of $\mathfrak{sl}_{n}(\mathbb{F}_{p})$ is the subset consisting of scalar matrices. Hence, for every $p$ , there is a constant $C_{p}$ such that every element of $\mathfrak{sl}_{n}(\mathbb{F}_{p})$ is equal to the sum of at most $C_{p}$ elements of $\{x^{-1}Ax\mid x\in \operatorname{SL}_{n}(\mathbb{F}_{p})\}$ . Therefore, in order to find a uniform $C$ , we can and will assume that $p$ is large. In particular, we assume that $p\neq 2$ .

For $1\leqslant i\neq j\leqslant n$ , let $E_{i,j}(a)$ be the matrix whose $(i,j)$ th entry is $a$ and with all other entries zero. Note that $E_{1,2}(a)$ is conjugate to $E_{1,2}(ab^{2})$ , for every $b\in \mathbb{F}_{p}$ . Since every element in $\mathbb{F}_{p}$ is a sum of two squares, we get that, for any $a\in \mathbb{F}_{p}^{\times }$ , any element of the form $E_{1,2}(b)$ is the sum of at most two conjugates of $E_{1,2}(a)$ . In particular, there exists a one-dimensional linear subspace of $\mathfrak{sl}_{n}(\mathbb{F}_{p})$ such that all its elements are sums of two conjugates of $E_{1,2}(a)$ . Using the fact that the only $\operatorname{SL}_{n}(\mathbb{F}_{p})$ -invariant subspace of $\mathfrak{sl}_{n}(\mathbb{F}_{p})$ is the subset consisting of scalar matrices once again, we see that if $a\neq 0$ , then every matrix in $\mathfrak{sl}_{n}(\mathbb{F}_{p})$ is the sum of at most $2(n^{2}-1)$ conjugates of $E_{1,2}(a)$ . Therefore, it is enough to prove that there is a constant $C$ such that, for some $a\in \mathbb{F}_{p}$ , the matrix $E_{1,2}(a)$ is a sum of $C$ conjugates of $A$ . We divide the proof into several steps.

Step A. Assume that $A$ is nilpotent. By using Jordan’s normal form we see that $A$ is conjugate to a block-diagonal matrix and each block-diagonal matrix has the from

(3) $$\begin{eqnarray}\left(\begin{array}{@{}ccccc@{}}0 & a & & & \\ & 0 & 1 & & \\ & & \ddots & \ddots & \\ & & & 0 & 1\\ & & & & 0\end{array}\right),\end{eqnarray}$$

where $0\neq a\in \mathbb{F}_{p}$ (we cannot assume that $a=1$ since we are conjugating with a matrix in $\operatorname{SL}_{n}(\mathbb{F}_{p})$ and not $\operatorname{GL}_{n}(\mathbb{F}_{p})$ ). A straightforward argument implies that it is enough to deal with the case where there is just one block. Clearly, we can assume that the dimension of this block is at least 3. Then there exists $\unicode[STIX]{x1D700}\in \{-1,1\}$ such that the diagonal matrix $\operatorname{diag}(\unicode[STIX]{x1D700},1,-1,\ldots ,(-1)^{n})$ belongs to $\operatorname{SL}_{n}(\mathbb{F}_{p})$ . Denote $B:=\operatorname{diag}(\unicode[STIX]{x1D700},1,-1,\ldots ,(-1)^{n})+E_{2,n}(1)$ . Then the $n$ th coordinate of the first row of $A+B^{-1}AB$ is non-zero while all the other rows equal zero. Thus, $A+B^{-1}AB$ is conjugate to $E_{1,2}(a)$ for some non-zero $a$ .

Step B. Assume that $n=2$ . If $A$ is nilpotent, it is conjugate to $E_{1,2}(a)$ , for some $a$ , and the claim holds. In general, since $A$ is not scalar, it is conjugate to $(\!\begin{smallmatrix}0 & a\\ b & 0\end{smallmatrix}\!)$ . We claim that, if $p\geqslant 11$ , there are $x,y,z\in \mathbb{F}_{p}^{\times }$ such that $x^{2}+y^{2}+z^{2}=0$ and $x^{-2}+y^{-2}+z^{-2}\neq 0$ . If this claim holds, then

$$\begin{eqnarray}\displaystyle & & \displaystyle \left(\begin{array}{@{}cc@{}}x & \\ & x^{-1}\end{array}\right)A\left(\begin{array}{@{}cc@{}}x^{-1} & \\ & x\end{array}\right)+\left(\begin{array}{@{}cc@{}}y & \\ & y^{-1}\end{array}\right)A\left(\begin{array}{@{}cc@{}}y^{-1} & \\ & y\end{array}\right)+\left(\begin{array}{@{}cc@{}}z & \\ & z^{-1}\end{array}\right)A\left(\begin{array}{@{}cc@{}}z^{-1} & \\ & z\end{array}\right)\nonumber\\ \displaystyle & & \displaystyle \quad =\left(\begin{array}{@{}cc@{}}0 & 0\\ b(x^{-2}+y^{-2}+z^{-2}) & 0\end{array}\right),\nonumber\end{eqnarray}$$

which is nilpotent.

To prove the claim, let $X$ be the projective curve defined by $x^{2}+y^{2}+z^{2}=0$ . Then $X$ has $p+1$ points over $\mathbb{F}_{p}$ , and at most six of them have a zero in some coordinate. At most four of the points of $X$ satisfy the equation $x^{-2}+y^{-2}+z^{-2}=0$ (because these points satisfy $1=(y^{2}+z^{2})(y^{-2}+z^{-2})$ ). In particular, if $p\geqslant 11$ , the claim is true.

Step C. Assume $n>2$ and the claim is true for all numbers smaller than $n$ . We consider the following cases.

Case C1. Assume that $\det A=0$ . By conjugating $A$ we can assume that it is of the form

(4) $$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}0 & \ast \\ 0 & B\end{array}\right),\end{eqnarray}$$

where $B\in \mathfrak{sl}_{n-1}(\mathbb{F}_{p})$ . If $B=0$ then $A$ is a nilpotent matrix and we are done by step 1. Otherwise, we can assume that $p>n-1$ so $B$ is a non-scalar matrix since its trace is equal to zero. Then by the induction hypothesis the sum of a bounded number of conjugates of $A$ is a non-zero nilpotent matrix and we are done by step 1.

Case C2. Assume now that $\det A\neq 0$ . By using the rational canonical normal form we see that there exist non-zero $a,b\in \mathbb{F}_{p}$ such that $A$ is conjugate to a block-diagonal matrix and one of the blocks of $A$ (for notational ease, assume it is the first) is of the form

(5) $$\begin{eqnarray}\left(\begin{array}{@{}ccccc@{}}0 & a & & & \\ & 0 & 1 & & \\ & & \ddots & \ddots & \\ & & & 0 & 1\\ b & \ast & \cdots \, & \ast & \ast \end{array}\right).\end{eqnarray}$$

Denote $B:=\operatorname{diag}(-1,-1,1,\ldots ,1)$ . Then $A+B^{-1}AB$ is a non-zero singular matrix and we are done by case C1.◻

Proof. (1) Let $k=\min \{i\mid (\exists g\in X)g\text{ is not central modulo }p^{k+1}\}$ . Clearly, $\langle X\rangle \subseteq Z(\operatorname{SL}_{n}(\mathbb{Z}_{p}))\cdot \operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k})$ . We will show that there is $C$ such that $\operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k})\subseteq X^{C}$ , and it will follow that $X^{C+|Z(\operatorname{SL}_{n}(\mathbb{Z}_{p}))|}=\langle X\rangle$ , which proves the claim since $|Z(\operatorname{SL}_{n}(\mathbb{Z}_{p}))|\leqslant n$ .

Case 1: $k=0$ . Let $\overline{X}\subseteq \operatorname{SL}_{n}(\mathbb{F}_{p})$ be the reduction of $X$ modulo $p$ . By assumption, $\overline{X}$ is non-central, so [Reference Liebeck and ShalevLS01, Corollary 1.9] implies that there is $C_{1}$ , depending only on $n$ , such that $\overline{X}^{C_{1}}=\operatorname{SL}_{n}(\mathbb{F}_{p})$ . Let $a,b\in X^{C_{1}}$ such that their reductions modulo $p$ generate $\operatorname{SL}_{n}(\mathbb{F}_{p})$ . By Lemma 4.4, the differential of the map $\unicode[STIX]{x1D6F7}_{\overline{a},\overline{b}}$ at $(1,1)$ is onto, which implies that $d\unicode[STIX]{x1D6F7}_{a,b}(\mathfrak{sl}_{n}^{2}(\mathbb{Z}_{p}))=\mathfrak{sl}_{n}(\mathbb{Z}_{p})$ , since $d\unicode[STIX]{x1D6F7}_{a,b}$ is $\mathbb{Z}_{p}$ -linear. By Lemma 4.2, $ab\operatorname{SL}_{n}(\mathbb{Z}_{p};p)\subseteq \unicode[STIX]{x1D6F7}_{a,b}(\operatorname{SL}_{n}(\mathbb{Z}_{p})\times \operatorname{SL}_{n}(\mathbb{Z}_{p}))\subseteq X^{2C_{1}}$ . It follows that $X^{3C_{1}}=\operatorname{SL}_{n}(\mathbb{Z}_{p})$ .

Case 2: $k>0$ . Let $g\in X$ be such that $g$ is not a scalar modulo $p^{k+1}$ . Since $g$ is a scalar modulo $p^{k}$ , there exists $h\in \operatorname{SL}_{n}(\mathbb{Z}_{p})$ such that $g^{|Z(\operatorname{SL}_{n}(\mathbb{Z}_{p}))|-1}h^{-1}gh\in X^{|Z(\operatorname{SL}_{n}(\mathbb{Z}_{p}))|}$ belongs to $\operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k})$ and is not a scalar modulo $p^{k+1}$ . Since $\operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k})/\text{SL}_{n}(\mathbb{Z}_{p};p^{k+1})=\mathfrak{sl}_{n}(\mathbb{F}_{p})$ as $\operatorname{SL}_{n}(\mathbb{Z}_{p})$ -modules, Lemma 4.6 implies that there is a constant $C$ , independent of $X$ , such that $X^{C}\cdot \operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k+1})\supseteq \operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k})$ . Let $\overline{a}$ be a maximal nilpotent Jordan block and let $\overline{b}=\overline{a}^{T}$ . Note that the intersection of the centralizers of $\overline{a},\overline{b}$ in $\operatorname{M}_{n}$ is the collection of scalar matrices. Choose $a,b\in X^{C}\cap \operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k})$ whose images in $\operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k})/\text{SL}_{n}(\mathbb{Z}_{p};p^{k+1})=\mathfrak{sl}_{n}(\mathbb{F}_{p})$ are $\overline{a}$ and  $\overline{b}$ . We will show that $\unicode[STIX]{x1D6F7}_{a,b}:\operatorname{SL}_{n}\times \operatorname{SL}_{n}\rightarrow \operatorname{SL}_{n}$ satisfies the conditions of Lemma 4.2.

Since $a-1$ is divisible by $p^{k}$ , we have $\operatorname{val}_{p}(x\mapsto x^{-1}ax-a)\geqslant k$ . It follows that the derivative of this map also has $p$ -valuation at least $k$ . Similarly, $\unicode[STIX]{x1D6F7}_{a,b}$ satisfies the first condition of Lemma 4.2.

Note that $d\unicode[STIX]{x1D6F7}_{a,b}(\mathfrak{sl}(\mathbb{Z}_{p})^{2})\subset p^{k}\mathfrak{sl}(\mathbb{Z}_{p})$ . In order to show the reverse containment, it is enough to show that the composition of $d\unicode[STIX]{x1D6F7}_{a,b}$ and the reduction map $p^{k}\mathfrak{sl}_{n}(\mathbb{Z}_{p})\rightarrow p^{k}\mathfrak{sl}_{n}(\mathbb{Z}_{p})/p^{k+1}\mathfrak{sl}_{n}(\mathbb{Z}_{p})$ is onto. This composition is the map $(X,Y)\mapsto [\overline{X},\overline{a}]+[\overline{Y},\overline{b}]$ (where $[x,y]$ is the Lie bracket), so we need to show that there is no non-zero linear functional that vanishes on all elements of the form $[\overline{X},\overline{a}]$ and $[\overline{X},\overline{b}]$ , for $\overline{X}\in \mathfrak{sl}_{n}(\mathbb{F}_{p})$ . Any such functional has the form $\operatorname{tr}(A\cdot )$ for some $A\in \mathfrak{sl}_{n}(\mathbb{F}_{p})$ . Since $\operatorname{tr}(A[B,C])=\operatorname{tr}([A,B]C)$ for every three matrices $A$ , $B$ and $C$ , the assumption that $\operatorname{tr}(A[\overline{a},\overline{X}])=0$ for all $\overline{X}\in \mathfrak{sl}_{n}(\mathbb{F}_{p})$ implies that $[A,\overline{a}]=\unicode[STIX]{x1D6FC}I$ , for some $\unicode[STIX]{x1D6FC}$ . Similarly, $[A,\overline{b}]=\unicode[STIX]{x1D6FD}I$ , for some $\unicode[STIX]{x1D6FD}$ . From $[A,\overline{a}]=\unicode[STIX]{x1D6FC}I$ we get (by induction) that $A_{i+1,i}=-i\unicode[STIX]{x1D6FC}$ , whereas from $[A,\overline{b}]=\unicode[STIX]{x1D6FD}I$ we get that $A_{i+1,i}=A_{i+2,i+1}$ . Since $n\geqslant 3$ , we get $\unicode[STIX]{x1D6FC}=0$ . Similarly, $\unicode[STIX]{x1D6FD}=0$ . Consequently, $A$ commutes with $\overline{a}$ and $\overline{b}$ , so $A=0$ , a contradiction.

Applying Lemma 4.2 to $\unicode[STIX]{x1D6F7}_{a,b}$ , we get that any element in $ab\operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k+1})$ is in $\unicode[STIX]{x1D6F7}_{a,b}(\operatorname{SL}_{n}(\mathbb{Z}_{p})^{2})$ , so, in particular, $ab\operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k+1})\subset X^{2C}$ and $\operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k+1})\subset X^{4C}$ . Since $X^{C}\operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k+1})\supseteq \operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k})$ , we get that $X^{5C}\supseteq \operatorname{SL}_{n}(\mathbb{Z}_{p};p^{k})$ , proving the claim in this case.

(2) Since $w(\operatorname{SL}_{n}(\widehat{\mathbb{Z}}))=\prod w(\operatorname{SL}_{n}(\mathbb{Z}_{p}))$ , the claim follows from the first claim.◻

Proof of Theorem 1.2.

By Theorem 1.4 and Lemma 4.8. ◻

Proof. Let $I_{3}$ be the identity matrix of $\operatorname{SL}_{3}(\mathbb{Z})$ and identify $M_{n}(\mathbb{Z})$ with $M_{m}(M_{3}(\mathbb{Z}))$ , where $M_{k}(R)$ is the ring of $k\times k$ matrices over the ring $R$ . Let $l_{1}$ be the matrix of $M_{m}(M_{3}(\mathbb{Z}))$ with $I_{3}$ on the diagonal, $g_{i}^{-1}$ on the $(i+1,i)$ th entry for every $1\leqslant i\leqslant m-1$ , and zero elsewhere. Let $l_{2}$ be the matrix of $M_{m}(M_{3}(\mathbb{Z}))$ with $I_{3}$ on the diagonal, $I_{3}$ on the $(i+1,i)$ th entry for every $1\leqslant i\leqslant m-1$ , and zero elsewhere. Let $u_{1}$ be the matrix of $M_{m}(M_{3}(\mathbb{Z}))$ with $I_{3}$ on the diagonal, $1-g_{1}\cdots g_{i}$ on the $(i,i+1)$ th entry for every $1\leqslant i\leqslant m-1$ , and zero elsewhere. Let $u_{2}$ be the matrix of $M_{m}(M_{3}(\mathbb{Z}))$ with $I_{3}$ on the diagonal, $(1-g_{1}\cdots g_{i})g_{i+1}$ on the $(i,i+1)$ th entry for every $1\leqslant i\leqslant m-1$ , and zero elsewhere. Then $g=l_{1}^{-1}u_{1}^{-1}l_{2}u_{2}=(l_{1}^{-1}l_{2})(l_{2}^{-1}u_{1}^{-1}l_{2})u_{2}\in L_{n}(\mathbb{Z};q)\tilde{U} _{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)$ .◻

Proof. Define $h:=\operatorname{diag}(g_{m},g_{m-1},\ldots ,g_{1})\in U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q)$ . Lemma 5.1 implies that $gh^{-1}\in L_{n}(\mathbb{Z};q)\tilde{U} _{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)$ . Thus,

$$\begin{eqnarray}g\in L_{n}(\mathbb{Z};q)\tilde{U} _{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)h\subseteq L_{n}(\mathbb{Z};q)\tilde{U} _{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q).\square\end{eqnarray}$$

Proof. Let $q\geqslant 1$ . The proof is by induction on $n$ . The base case $n=m$ is clear. Assume that the statement is true for some $n\geqslant m$ . We have to show that the statement is true also for $n+1$ . Let $U_{n}^{-}(\mathbb{Z};q)$ and $L_{n}^{-}(\mathbb{Z};q)$ be the images in $\operatorname{SL}_{n+1}(\mathbb{Z})$ of $U_{n}(\mathbb{Z};q)$ and $L_{n}(\mathbb{Z};q)$ under the map $M\mapsto \operatorname{diag}(1,M)$ . Denote $C_{n}^{-}(q):=\langle e_{j,1}(q)\mid 2\leqslant j\leqslant n+1\rangle$ and $R_{n}^{-}(q):=\langle e_{1,j}(q)\mid 2\leqslant j\leqslant n+1\rangle$ . Finally, recall that the main theorem of [Reference MennickeMen65] implies that for every $k\geqslant 3$ ,

$$\begin{eqnarray}E(k,\mathbb{Z};q)=\{g\in \operatorname{SL}_{k}(\mathbb{Z};q)\mid \forall 1\leqslant i\leqslant k,~g_{i,i}=1(\text{mod}~q^{2})\}.\end{eqnarray}$$

Let $g\in E(n+1,\mathbb{Z};q)$ . Then $\gcd (g_{1,1},g_{2,1},\ldots ,g_{n,1})=1$ and $\gcd (qg_{1,1},g_{2,1},\ldots ,g_{n,1})=q$ . Recall that $\mathbb{Z}$ satisfies the following stable range condition: if $m\geqslant 3$ and $a_{1},\ldots ,a_{m}\in \mathbb{Z}$ then there exist $t_{2},\ldots ,t_{m}\in \mathbb{Z}$ such that $\gcd (a_{1},\ldots ,a_{m})=\gcd (a_{2}-t_{2}a_{1},\ldots ,a_{n}-t_{n}a_{1})$ . Thus, there exists $h\in C_{n}^{-}(\mathbb{Z};q)g$ such that $\gcd (h_{2,1},\ldots ,h_{n,1})=q$ . Since $h\in E(n,\mathbb{Z};q)$ , we have $h_{1,1}=1$ modulo $q^{2}$ so there exists $h^{\prime }\in R_{n}^{-}(\mathbb{Z};q)h$ such that $h_{1,1}^{\prime }=1$ . Finally, there exists $h^{\prime \prime }\in C_{n}^{-}(q)h^{\prime }R_{n}^{-}(q)$ such that $h^{\prime \prime }=\operatorname{diag}(1,g^{\prime })$ for some $g^{\prime }\in \operatorname{SL}_{n}(\mathbb{Z};q)$ . Note that $g^{\prime }\in E(n,\mathbb{Z};q)$ since its diagonal entries are equal to 1 modulo $q^{2}$ . Thus, the induction hypothesis implies that $h^{\prime \prime }\in L_{n}^{-}(\mathbb{Z};q)U_{n}^{-}(\mathbb{Z};q)L_{n}^{-}(\mathbb{Z};q)E^{\ast }(m,\mathbb{Z};q)U_{n}^{-}(\mathbb{Z};q)$ . It follows that $g$ belongs to

$$\begin{eqnarray}C_{n}^{-}(\mathbb{Z};q)R_{n}^{-}(\mathbb{Z};q)C_{n}^{-}(\mathbb{Z};q)L_{n}^{-}(\mathbb{Z};q)U_{n}^{-}(\mathbb{Z};q)L_{n}^{-}(\mathbb{Z};q)E^{\ast }(m,\mathbb{Z};q)U_{n}^{-}(\mathbb{Z};q)R_{n}^{-}(\mathbb{Z};q).\end{eqnarray}$$

Since both $U_{n}^{-}(\mathbb{Z};q)$ and $L_{n}^{-}(\mathbb{Z};q)$ normalize $C_{n}^{-}(\mathbb{Z};q)$ and $R_{n}^{-}(\mathbb{Z};q)$ , we have

$$\begin{eqnarray}\displaystyle & & \displaystyle C_{n}^{-}(\mathbb{Z};q)R_{n}^{-}(\mathbb{Z};q)C_{n}^{-}(\mathbb{Z};q)L_{n}^{-}(\mathbb{Z};q)U_{n}^{-}(\mathbb{Z};q)L_{n}^{-}(\mathbb{Z};q)E^{\ast }(m,\mathbb{Z};q)U_{n}^{-}(\mathbb{Z};q)R_{n}^{-}(\mathbb{Z};q)\nonumber\\ \displaystyle & & \displaystyle \quad =C_{n}^{-}(\mathbb{Z};q)L_{n}^{-}(\mathbb{Z};q)R_{n}^{-}(\mathbb{Z};q)U_{n}^{-}(\mathbb{Z};q)C_{n}^{-}(\mathbb{Z};q)L_{n}^{-}(\mathbb{Z};q)E^{\ast }(m,\mathbb{Z};q)U_{n}^{-}(\mathbb{Z};q)R_{n}^{-}(\mathbb{Z};q)\nonumber\\ \displaystyle & & \displaystyle \quad =L_{n+1}(\mathbb{Z};q)U_{n+1}(\mathbb{Z};q)L_{n+1}(\mathbb{Z};q)E^{\ast }(m,\mathbb{Z};q)U_{n+1}(\mathbb{Z};q).\square\nonumber\end{eqnarray}$$

Proof. Proposition 2.9 implies that there exists a constant $D$ such that for every $q\in \mathbb{Z}$ , $(U_{3}(\mathbb{Z};q)L_{3}(\mathbb{Z};q))^{D}=\langle U_{3}(\mathbb{Z};q)L_{3}(\mathbb{Z};q)\rangle$ . Denote $E^{\ast }(3,\mathbb{Z};q):=\{\operatorname{diag}(1,\ldots ,1,g)\in \operatorname{SL}_{3D}(\mathbb{Z})\mid g\in E(3,\mathbb{Z};q)\}$ . Lemma 5.2 shows that, for any $q$ , $E^{\ast }(3,\mathbb{Z};q)\subseteq L_{3D}(\mathbb{Z};q)\widetilde{U}_{3D}(\mathbb{Z};q)U_{3D}(\mathbb{Z};q)L_{3D}(\mathbb{Z};q)$ . Lemma 5.3 implies that

$$\begin{eqnarray}E(n,\mathbb{Z};q)\subseteq L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q)\tilde{U} _{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q).\end{eqnarray}$$

Since $L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q)\subset L_{n}(\mathbb{Z};q)\tilde{U} _{n}(\mathbb{Z};q)$ and $U_{n}(\mathbb{Z};q)L_{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)\subset \tilde{U} _{n}(\mathbb{Z};q)U_{n}(\mathbb{Z};q)$ , we get the result.◻

Proof of Theorems 1.1 and 1.4 (with explicit bounds).

Let $n\geqslant 3$ . The proof of Theorem 1.4 shows that there is a non-zero $q\in \mathbb{Z}$ such that $U_{n}(\mathbb{Z};q)$ and $L_{n}(\mathbb{Z};q)$ are contained in $w(\operatorname{SL}_{n}(\mathbb{Z}))^{16}$ . Since $w(\operatorname{SL}_{n}(\mathbb{Z}))$ is a normal subset, the set $\tilde{U} _{n}(\mathbb{Z};q):=\{hkh^{-1}\mid k\in U_{n}(\mathbb{Z};q)\wedge h\in \operatorname{SL}_{n}(\mathbb{Z})\}$ is contained in $w(\operatorname{SL}_{n}(\mathbb{Z}))^{16}$ . Corollary 5.4 implies that if $n$ is large enough then $E(n,\mathbb{Z};q)\subseteq w(\operatorname{SL}_{n}(\mathbb{Z}))^{16\cdot 5}$ . Since $E(n,\mathbb{Z};q)$ contains a congruence subgroup, we have proved the bound in Theorem 1.4. Finally, Lemma 4.5 implies that if $n$ is large enough then $\operatorname{SL}_{n}(\mathbb{Z})\subseteq w(\operatorname{SL}_{n}(\mathbb{Z}))^{16\cdot 5+7}$ .◻