Sanov subgroup consists of matrices of the form

Question

I have tried taking some combination of the generating matrices but couldn't get the same result. How can I show such thing?

Answer 1

Let $G = {\rm SL}_n({\mathbb Z})$, $H$ as in your question, and let $K$ be the subgroup of $G$ consisting of all matrices of the specified form. Then Igor Rivin and you have already proved that $H \le K$, so it remains to prove $K \le H$.

I will assume the well-known result (which is not hard to prove) that $G$ is generated by the two matrices $$a := \left(\begin{array}{rr}0&1\\-1&0\end{array}\right),\ \ \ b := \left(\begin{array}{rr}0&1\\-1&1\end{array}\right).$$ It can be checked that $a^2=b^3$ and $a^4=1$ (in fact they are defining relations of $G$), where $a^2=-I_2$ is central in $G$. You can also check that the generators of $H$ are $\alpha = (b^{-1}a)^2$ and $\beta = (ba^{-1})^2$.

It is now straightforward to verify that $G$ is the union of the $12$ cosets $C := \{ Hb^ia^j : 0 \le i < 3,\, 0 \le j < 4 \}.$ To check that, it is enough to show that for any $Hg \in C$, $Hga$ and $Hgb$ lie in $C$. This is clear for $Hga$ and not hard for $Hgb$. For example $b^2ab = (b^{-1}a)^2(ab^{-1})^2ba=\alpha \beta ba$, so $Hb^2ab = Hba$.

So we have $|G:H| \le 12$. (We have not yet shown that the twelve cosets are distinct.) As I pointed out in my comment, $K$ is contained in the kernel of the natural projection $G \to {\rm SL}_2(2)$, which has order $6$, but $K$ is properly contained in this kernel, because it does not contain the elements in the kernel whose diagonal entries have the form $4k+3$. So $|G:K| \ge 12$ and, since we know that $H \le K$, we have $H=K$ and $|G:H|=12$.

Answer 2

The proof is by induction on the length of a word representing your matrix. Obviously, the result is true for length $1.$

Now, if you multiply $$ \left( \begin{array}{cc} 4 j+1 & 2 k \\ 2 l & 4 m+1 \\ \end{array} \right) $$ On the right by $\alpha$ you get: $$ \left( \begin{array}{cc} 4 j+1 & 2 (4 j+1)+2 k \\ 2 l & 4 l+4 m+1 \\ \end{array} \right) $$ and if you multiply it by $\beta,$ you get $$ \left( \begin{array}{cc} 4 j+4 k+1 & 2 k \\ 2 l+2 (4 m+1) & 4 m+1 \\ \end{array} \right) $$ Both of which satisfy your condition. You also need to multiply by $\alpha^{-1}$ and $\beta^{-1},$ which have $-2$ in place of $2.$ (exercise).

Answer 3

I also proved it by induction. see the below proof.