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If two matrices from GL(2,Z) have the same determinant, will there always be a matrix from SL(2,Z) which transforms one matrix to the other?

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    @Qiaochu: The user does not seem to have enough reputation to leave comments, though. I guess he could go around the site trying to get enough reputation so that he can finally write down his comment on this question, but still...2011-02-19

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If by "transform" you mean conjugate, no. Both $\left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right)\qquad\text{and}\qquad \left(\begin{array}{cc}1 & 0\\0 & 1 \end{array}\right)$ are in $\mathrm{GL}(2,\mathbb{Z})$, have determinant $1$, but they aren't conjugate even in $\mathrm{GL}(2,\mathbb{R})$.

But you don't mean "conjugate"; you mean, given $A$ and $B$ with same determinant and in $\mathrm{GL}(2,\mathbb{Z})$, is there a matrix $C$ from $\mathrm{SL}(2,\mathbb{Z})$ that will multiply $A$ into $B$, i.e., $CA=B$?

Well, if $CA=B$, and $C$, $A$, and $B$ are all in invertible, what is $C$? Is it in $\mathrm{SL}(2,\mathbb{Z})$?