# $GL_{2}(\mathbb{Z})$

Let $M_{2}(\mathbb{Z})$ be the ring of $2\textrm{x}2$ matrices with integer entries, and define $GL_{2}(\mathbb{Z})$ to be the subring of matrices invertible   over $\mathbb{Z}$. Thus for $M\in M_{2}(\mathbb{Z})$,

 $M\in GL_{2}(\mathbb{Z})\iff\det M=\pm 1$

To see this, we demonstrate a natural correspondence between endomorphisms of $\mathbb{Z}\oplus\mathbb{Z}$ and $M_{2}(\mathbb{Z})$ and show that invertible endomorphisms correspond to invertible matrices. Let $\varphi:\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$ be any ring homomorphism  . It is clear that $\varphi$ is determined by its action on $(1,0)$ and $(0,1)$, since

 $\varphi(x,y)=\varphi(x(1,0)+y(0,1))=x\varphi(1,0)+y\varphi(0,1)$

Suppose then that $\varphi(1,0)=(a,b)$ and $\varphi(0,1)=(c,d)$. Then

 $\varphi(x,y)=(ax,bx)+(cy,dy)=(ax+cy,bx+dy)=\begin{pmatrix}a&c\\ b&d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}$

Now, $\varphi$ is surjective  if both $(1,0)$ and $(0,1)$ are in its image. But $(1,0)\in\textrm{im}\ \varphi$ if and only if there is some $(x,y)$ such that

 $\displaystyle ax+cy$ $\displaystyle=1$ $\displaystyle bx+dy$ $\displaystyle=0$

Solving this pair of equations for $y$ we see that we must have $y(bc-ad)=1$ and thus $bc-ad=\pm 1$. Similarly, $(0,1)\in\textrm{im}\ \varphi$ if and only if $y(ad-bc)=\pm 1$. Thus $\varphi$ is surjective precisely when $ad-bc=\pm 1$, i.e. precisely when the matrix representation  of $\varphi$ has determininant $\pm 1$. This then gives a map from $Aut_{\mathbb{Z}}(\mathbb{Z}\oplus\mathbb{Z})$ to $GL_{2}(\mathbb{Z})$ that is obviously a ring isomorphism. This concludes the proof.

$A=Aut_{\mathbb{Z}}(\mathbb{Z}\oplus\mathbb{Z})$ has a simple and well-known set of generators     as a group:

 $\displaystyle r$ $\displaystyle=(x,y)\mapsto(y,x)$ $\displaystyle s$ $\displaystyle=(x,y)\mapsto(x,x+y)$

Note that $s^{m}=(x,y)\mapsto(x,mx+y)$ for any integer $m$. We now prove this fact.

Define the subgroup   $A^{\prime}\subset A$ by $A^{\prime}=$, the subgroup of $A$ generated by $r$ and $s$. If $\varphi_{1},\varphi_{2}\in A$, define $\varphi_{1}\sim\varphi_{2}$ if $\varphi_{1}$ and $\varphi_{2}$ are in the same $A^{\prime}$-coset.

Our objective is to show that $A^{\prime}=A$, which we can do by showing that each $\varphi\sim e$, where $e$ is the identity transformation of $A$. This demonstration is essentially an application of the Euclidean algorithm  . For suppose

 $\varphi(x,y)=(ax+cy,bx+dy)$

Assume, by applying $r$ if necessary, that $a\leq b$, and choose $m$ such that $b=am+q,0\leq q. Then $rs^{-m}\varphi(x,y)=r(ax+cy,(b-am)x+(d-cm)y)=r(ax+cy,qx+dy)=(qx+dy,ax+cy)$, so that

 $\varphi\sim(x,y)\mapsto(qx+dy,ax+cy)$

Continuing this process, we eventually see that

 $\varphi\sim(x,y)\mapsto(cy,bx+dy)$

But $ad-bc=\pm 1$, so we have $bc=\pm 1$. Applying either $s^{d}$ or $s^{-d}$ as appropriate, we get

 $\varphi\sim(x,y)\mapsto(cy,bx)\sim(x,y)\mapsto(bx,cy)$

Thus, we are done if we show that all such forms $(bx,cy)$ with $b,c=\pm 1$ are in the same $A^{\prime}$-coset as $e$. The case where $b=c=1$ is obvious. For the other cases, note that

 $\displaystyle(x,y)\mapsto(x,-y)$ $\displaystyle=s^{-1}rsrs^{-1}r$ $\displaystyle(x,y)\mapsto(-x,y)$ $\displaystyle=rs^{-1}rsrs^{-1}r$

and $(x,y)\mapsto(-x,-y)$ is obviously the composition of these two.

This result is often phrased by saying that the matrices

 $\begin{pmatrix}1&0\\ 1&1\end{pmatrix},\quad\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$

generate $GL_{2}(\mathbb{Z})$ as a multiplicative group  .

Title $GL_{2}(\mathbb{Z})$ GL2mathbbZ 2013-03-22 16:31:38 2013-03-22 16:31:38 rm50 (10146) rm50 (10146) 7 rm50 (10146) Application msc 20G15