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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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I'm sure you mean a positive power. –  Gerry Myerson Feb 9 '11 at 4:56
    
Yep! that's what I meant. –  Sreshna Feb 9 '11 at 5:37
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up vote 4 down vote accepted

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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I think you can safely erase the 'should'. –  Torsten Ekedahl Feb 9 '11 at 5:20
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