generators of $SL(2,\mathbf{Q}_p)$

Question

I want to ask how to prove the normal subgroup generated by matrices $(1,;0,1)$ and $(1,0;,1)$ for $$ small is still the whole of $SL(2,\mathbf{Q}_p)$? This is mentioned in the answer of this question.

Thanks!

Answer 1

Let $K$ be a field of characteristic $\neq 2$ and $e$ a nonzero element of $K$. Let $e_{12}(e)$ be the matrix $\begin{pmatrix}1 & e \\ 0 & 1\end{pmatrix}$. Let $S^n$ be the set of products of $\le n$ $\mathrm{SL}_2(K)$-conjugates of $e_{12}(e)^{\pm 1}=e_{12}(\pm e)$. Note that $S^n$ is closed under inversion.

Conjugating with $\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$, we see that $e_{21}(-e)=\begin{pmatrix}1 & e \\ 0 & 1\end{pmatrix}$ also belongs to $S^1$.

Conjugating with $\begin{pmatrix}s & 0 \\ 0 & s^{-1}\end{pmatrix}$, we see that $e_{12}(es^2)=\begin{pmatrix}1 & s^2e \\ 0 & 1\end{pmatrix}$ and $e_{21}(-es^2)=\begin{pmatrix}1 & 0 \\ -s^2e & 1\end{pmatrix}$ belong to $S^1$.

Hence $e_{12}(e(s^2-t^2))=e_{12}(es^2)e_{12}(et^2)^{-1}$ is in $S^2$ for all $s,t$. Since every element can be written as $s^2-t^2$, we deduce that $e_{12}(t)\in S^2$ for all $t$, and similarly $e_{21}(t)$ does. These are called elementary matrices.

Next, every matrix of determinant 1 is a product of 4 elementary matrices. Indeed, every nondiagonal matrix of determinant 1 is product of 3 elementary matrices:

Writing $e_{12}(u)e_{21}(v)e_{12}(w)=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ for given $(a,b,c,d)$ with $ad-bc=1$, yields $(a,b,c,d)=(1+uv,u+w+uvw,v,1+uw)$, which, provided $c\neq 0$, solves as $(u,v,w)=(c^{-1}(a-1),c,c^{-1}(d-1))$. If $c=0$, non-diagonal implies $b\neq 0$ and then it can be written as $e_{21}(u)e_{12}(v)e_{21}(w)$. (It also shows that a non-identity diagonal matrix cannot be product of 3 elementary matrices.)

Hence every non-diagonal matrix of determinant 1 belongs to $S^6$. For a diagonal matrix of determinant 1, first multiplying by $e_{12}(e)$, we see that it belongs to $S^7$. Hence, $S^7=\mathrm{SL}_2(K)$.