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Let $\begin{bmatrix} a & b \\ c & d\end{bmatrix} \in \mathrm{SL}_2(\mathbb{Z})$. Let $S$ be the set $(\mathbb{Z} \times \mathbb{Z}) \setminus \{(0,0)\}$. Show that the map $\varphi: S \rightarrow S$ given by $\varphi((m,n)) = (am + cn, bm + dn)$ is a bijection. Hint: try to construct the inverse map.

I need to show that $\varphi$ is injective and surjective.

Injectivity: I tried to show this part by showing that if $(m_1,n_1) \neq (m_2, n_2)$, then $\varphi(m_1,n_1) \neq \varphi(m_2, n_2)$. I think this is one way to start the proof.

Surjectivity: I am not sure where to start, but I believe the hint works for this problem.

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Why not use the hint? – vadim123 May 3 '14 at 1:51

2 Answers 2

Since $\begin{bmatrix} a & b \\ c & d\end{bmatrix} \in \mathrm{SL}_2(\mathbb{Z})$ , also $\begin{bmatrix} a & c \\ b & d\end{bmatrix} \in \mathrm{SL}_2(\mathbb{Z})$. Now the inverse of this matrix is given by $\begin{bmatrix} d & -c \\ -b & a\end{bmatrix} \in \mathrm{SL}_2(\mathbb{Z})$. Note in all this, I used the fact that we have $ad-bc=1$. I think you can work your way now.

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$$\phi(m,n)=\begin{bmatrix}m & n\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

$$\phi^{-1}(m,n)=\begin{bmatrix}m & n\end{bmatrix}\begin{bmatrix}d&-c\\-b&a\end{bmatrix}$$

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I made that mistake too, your $ \phi (m,n) $ is wrong. – The very fluffy Panda May 3 '14 at 12:05
Already figured out. – NasuSama May 3 '14 at 14:46
@PandaBear Thank you for pointing out this. – gaoxinge May 4 '14 at 2:22

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