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 $SL(n,\mathbb{Z})$?



On some type of conjugate of elements of SL(n,Z)
Yes. Write $N = det(B)$. Then, the group $SL(n, \mathbb{Z} / N\mathbb{Z})$ is finite-order, say of order $k$. If we raise $B^{-1} A B$ to the power of $k$, then it should lie in $SL(n, \mathbb{Z})$.


