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Prove that any element of $SL(2,\mathbb{Z})$ can be represented by a finite product of matrices of the following form. $$\begin{pmatrix}1-ab & a^2\\ -b^2 & 1+ab\end{pmatrix}.$$ We are given that $SL(2,\mathbb{Z})$ is generated by $\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$ and $ \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$. When $a=1$, $b=0$ we get the first generator, not sure how to find the second.

Its probably very easy but I am having trouble.

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@scott: Are you allowed inverses of those matrices as well? –  Jim Apr 24 '13 at 23:50
    
I don't believe that I am. –  scott Apr 24 '13 at 23:56
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@Sanchez I have $S^2T^{-1}=$$$\begin{pmatrix} -2 & 1\\ -1 & 0\end{pmatrix}$$ Im not sure how that helps. –  scott Apr 25 '13 at 0:05
    
@scott, sorry, I meant to say $T^{-1}S^{-2}$. This ideally is of the form you mentioned, which allows you to get $T^{-1}$. If you are not allowed to use inverse though, I'm not sure. –  user27126 Apr 25 '13 at 0:13
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The cube of $T^{-1}$ is $T$. –  Jim Apr 25 '13 at 0:17

1 Answer 1

So $a=1$, $b=0$ gives $X := \left(\begin{array}{cc}1&1\\0&1\end{array}\right)$, and $a=b=1$ gives $Y := \left(\begin{array}{cc}0&1\\-1&2\end{array}\right)$.

Now $YX^2 = \left(\begin{array}{cc}0&1\\-1&0\end{array}\right)$, which is the inverse of your second generator of ${\rm SL}(2,{\mathbb Z})$, but that has finite order, so that's OK.

Now, to solve the problem, we still need to get $X^{-1}$ as a product of matrices of the required form. But $X^{-1} = (YX^2)^{-1}YX = (YX^2)^3YX$.

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