Let M2() be the ring of 2x2 matrices with integer entries, and define GL2() to be the subring of matrices invertiblePlanetmathPlanetmathPlanetmath over . Thus for MM2(),


Let Aut() be the ring of automorphismsPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of as a -module. Then GL2()Aut() as rings, under the obvious operationsMathworldPlanetmath.

To see this, we demonstrate a natural correspondence between endomorphisms of and M2() and show that invertible endomorphisms correspond to invertible matrices. Let φ: be any ring homomorphismMathworldPlanetmath. It is clear that φ is determined by its action on (1,0) and (0,1), since


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


Now, φ is surjectivePlanetmathPlanetmath if both (1,0) and (0,1) are in its image. But (1,0)imφ 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=±1. Similarly, (0,1)imφ if and only if y(ad-bc)=±1. Thus φ is surjective precisely when ad-bc=±1, i.e. precisely when the matrix representationPlanetmathPlanetmath of φ has determininant ±1. This then gives a map from Aut() to GL2() that is obviously a ring isomorphism. This concludes the proof.

A=Aut() has a simple and well-known set of generatorsPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath as a group:

r =(x,y)(y,x)
s =(x,y)(x,x+y)

Note that sm=(x,y)(x,mx+y) for any integer m. We now prove this fact.

Define the subgroupMathworldPlanetmathPlanetmath AA by A=<r,s>, the subgroup of A generated by r and s. If φ1,φ2A, define φ1φ2 if φ1 and φ2 are in the same A-coset.

Our objective is to show that A=A, which we can do by showing that each φe, where e is the identity transformation of A. This demonstration is essentially an application of the Euclidean algorithmMathworldPlanetmath. For suppose


Assume, by applying r if necessary, that ab, and choose m such that b=am+q,0q<a. Then rs-mφ(x,y)=r(ax+cy,(b-am)x+(d-cm)y)=r(ax+cy,qx+dy)=(qx+dy,ax+cy), so that


Continuing this process, we eventually see that


But ad-bc=±1, so we have bc=±1. Applying either sd or s-d as appropriate, we get


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

(x,y)(x,-y) =s-1rsrs-1r
(x,y)(-x,y) =rs-1rsrs-1r

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

This result is often phrased by saying that the matrices


generate GL2() as a multiplicative groupMathworldPlanetmath.

Title GL2()
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