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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$\begingroup$ I'm sure you mean a positive power. $\endgroup$– Gerry MyersonCommented Feb 9, 2011 at 4:56
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$\begingroup$ Yep! that's what I meant. $\endgroup$– SreshnaCommented Feb 9, 2011 at 5:37
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1$\begingroup$ Also: mathematik.uni-muenchen.de/~yarotsky/jarnik/… (problem 3) and mathematik.uni-muenchen.de/~yarotsky/jarnik/… . $\endgroup$– darij grinbergCommented Feb 9, 2011 at 13:24
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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 $SL(n, \mathbb{Z})$.
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1$\begingroup$ I think you can safely erase the 'should'. $\endgroup$ Commented Feb 9, 2011 at 5:20