Let $A\in SL(n,\mathbb{Z})$ and $B\in\mathcal{M}_n(\mathbb{Z})$ s.t. $\det(B)\ne 0$. Is it possible to find a power of $B^{1}AB$ in $SL(n,\mathbb{Z})$?
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Yes. Write $N = det(B)$. Then, the group $SL(n, \mathbb{Z} / N\mathbb{Z})$ is finiteorder, say of order $k$. If we raise $B^{1} A B$ to the power of $k$, then it should lie in $SL(n, \mathbb{Z})$. 

