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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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@bghngh121: By "transform", do you mean "conjugate"? That is, does "$M$ transforms $A$ to $B$" mean $B=M^{-1}AM$? – Arturo Magidin Feb 18 '11 at 5:45
Not conjugate- if matrices A and B from GL(2,Z) have equal determinant, will there always be a matrix C from SL(2,Z) such that A = C B ? – 24601 Feb 18 '11 at 5:48
Upon reading this question, my interpretation of "$M$ transforms $A$ to $B$" was "$MA = B$". Under this interpretation, the answer is trivially yes: take $M = $... – Pete L. Clark Feb 18 '11 at 5:50
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@p791i78: Since $A$ and $B$ are both invertible over $\mathbb{Z}$ the product $BA^{-1}$ lies in $\mathrm{GL}(2,\mathbb{Z})$; the determinant is $1$ (since $\det(A)=\det(B)$, and they are both either $1$ or $-1$, so $\det(A^{-1}) = \det(A)$), and $(BA^{-1})A = B$. So $C=BA^{-1}$ is the matrix sought, it always lies in $\mathrm{SL}(2,\mathbb{Z})$. – Arturo Magidin Feb 19 '11 at 9:08
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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... – Arturo Magidin Feb 19 '11 at 20:23
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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 $\mathrm{SL}(2,\mathbb{Z})$?

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