Determining Finiteness of (ZxZ)/H for a,b,c,d

[e12514]

Quick question:


Fix integers a,b,c,d.
Let H be the subgroup of ZxZ generated by (a,b) and (c,d).
When (in terms of a,b,c,d) is the factor group (ZxZ)/H finite?



I figured that if ad is not equal to bc then the factor group (ZxZ)/H is of order ad-bc, and if ad is equal to bc then the factor group (ZxZ)/H is infinite. But I can only figure out how to prove it by drawing diagrams and showing it geometrically. Are there any rigorous ways to prove that?

 

[matt grime]

What is non-rigorous about what you did, and how can you go about correcting that?

 

[e12514]

Non-rigorous: drawing diagrams and describing what's happening instead of precise lines of algebraic proof.

Rigorous: perhaps some proof not via a geometric point of view, and purely by considering the group ZxZ and the subgroup H and the factor groups algebraicly? (If that's possible)

 

[matt grime]

Firstly, it is not necessarily true that your first argument is not rigorous. Write out you 'non-rigorous' descriptions, and see if you can make them satisfy your notion of rigorous. If a proof is logically correct without any leaps of faith then it doesn't matter whether it is diagrammatic or not.

Anyway, the point is that your subgroup H is the set of all points (ax+cy, bx+dy), isn't it? Sums of multiples of (a,b) and (c,d). Now doesn't that look a lot like what happens when you apply a matrix to (x,y)? A matrix involving a,b,c,d, and isn't ad-bc very suggestive when you think about matrices?

 

[e12514]

Well yes. What I have is that the set of points in H are elements of the form (ma+nc, mb+nd) which is pictured geometrically by the set of points on ZxZ that lies on any multiple of the vector (a,b) or (c,d) or a combination of both. The number of elements in (ZxZ)/H is then the number of points contained inside inside one //gram plus the point on the bottom left vertex, which should be equal to its area, which is equal to the determinant of the matrix whose first column is a,b and second column is c,d , equalling (the absolute value of) ad-bc. With a diagram showing the elements of H and joining them up with parallel lines etc.

The only problem is I think that was just a description rather than a proper proof, and I probably need to go into more detail for proving certain parts, perhaps that the number of elements in H is equal to the area which is equal to the determinant... etc.