# $G{L}_{2}(\mathbb{Z})$

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

$$\phi (x,y)=(ax,bx)+(cy,dy)=(ax+cy,bx+dy)=\left(\begin{array}{cc}\hfill a\hfill & \hfill c\hfill \\ \hfill b\hfill & \hfill d\hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)$$ |

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

$$\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$$ |

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 |