$G{L}_2(\mathbb{Z})$

Let $M_2(\mathbb{Z})$ be the ring of $2\text{x}2$ matrices with integer entries, and define $G{L}_2(\mathbb{Z})$ to be the subring of matrices invertible over $\mathbb{Z}$. Thus for $M\in M_2(\mathbb{Z})$,

$$M\in G{L}_2(\mathbb{Z})\iff detM=\pm 1$$

Let $Au{t}_{\mathbb{Z}}(\mathbb{Z}\oplus \mathbb{Z})$ be the ring of automorphisms of $\mathbb{Z}\oplus \mathbb{Z}$ as a $\mathbb{Z}$-module. Then $G{L}_2(\mathbb{Z})\cong Au{t}_{\mathbb{Z}}(\mathbb{Z}\oplus \mathbb{Z})$ as rings, under the obvious operations.

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 $\phi :\mathbb{Z}\oplus \mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$ be any ring homomorphism. It is clear that $\phi$ is determined by its action on $(1,0)$ and $(0,1)$, since

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

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

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

$ax+cy$ $=1$

$bx+dy$ $=0$

Solving this pair of equations for $y$ we see that we must have $y(bc-ad)=1$ and thus $bc-ad=\pm 1$. Similarly, $(0,1)\in \text{im}\phi$ if and only if $y(ad-bc)=\pm 1$. Thus $\phi$ is surjective precisely when $ad-bc=\pm 1$, i.e. precisely when the matrix representation of $\phi$ has determininant $\pm 1$. This then gives a map from $Au{t}_{\mathbb{Z}}(\mathbb{Z}\oplus \mathbb{Z})$ to $G{L}_2(\mathbb{Z})$ that is obviously a ring isomorphism. This concludes the proof.

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

$r$ $=(x,y)\mapsto (y,x)$

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

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 , the subgroup of $A$ generated by $r$ and $s$. If ${\phi }_1,{\phi }_2\in A$, define ${\phi }_1\sim {\phi }_2$ if ${\phi }_1$ and ${\phi }_2$ are in the same $A^{\prime }$-coset.

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

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

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

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

Continuing this process, we eventually see that

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

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

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

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

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

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

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

This result is often phrased by saying that the matrices

generate $G{L}_2(\mathbb{Z})$ as a multiplicative group.

Title	$G{L}_2(\mathbb{Z})$
Canonical name	GL2mathbbZ
Date of creation	2013-03-22 16:31:38
Last modified on	2013-03-22 16:31:38
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	7
Author	rm50 (10146)
Entry type	Application
Classification	msc 20G15