Order of Group of 2 x 2 Matrices

[e(ho0n3]

Homework Statement
Let G be the group of all 2 x 2 matrices
<br /> \begin{pmatrix}<br /> a &amp; b \\ c &amp; d<br /> \end{pmatrix}<br />
where a, b, c, d are integers modulo p, p a prime number, such that ad - bc ≠ 0. G forms a group relative to matrix multiplication. What is the order of G?

The attempt at a solution
Let's consider the case p = 3. According to the book, G has 48 elements. I don't see how this is possible. The possible values for a, b, c, d are 0, 1, 2, so there are 3 * 3 * 3 * 3 = 81 possible matrices. The ones we don't want are those that satisfy ad = bc and there are 3 + (3 * 2) * 2 = 15 of these. There should be 81 - 15 = 66 matrices in G. Right?

 

[Dick]

I don't know how you are counting the ad=bc cases, but there are 33, not 15. Be really careful counting the cases with zero entries. It helps to split the cases by the number of zero entries.

 

[e(ho0n3]

You're right. I messed up counting the cases with 0 entries. I will heed your advice.

 

[e(ho0n3]

In order to solve this problem I need to know how many ways one can write a nonnegative integer k < p as the product of two nonnegative integers x,y < p. I haven't been able to figure this one out. Any tips?

 

[Dick]

In order to solve this problem I need to know how many ways one can write a nonnegative integer k < p as the product of two nonnegative integers x,y < p. I haven't been able to figure this one out. Any tips?

Uh, have you forgotten you are working modulo p? Every nonzero element has a multiplicative inverse. There are p-1 ways.

 

[e(ho0n3]

The problem doesn't explicitly say that ad - bc ≠ 0 (mod p), although I don't think it matters.

But you're right that any nonzero element will have a multiplicative inverse, since p is prime. But how does that help? For example, say p = 31 and I want to find how many ways there are of writing 12 as a product of two positive integers. How would I go about it?

 

[Dick]

If you want to solve mn=q modulo p, you can pick ANY m and then set n=q*m^(-1). It has just as many solutions as there are nonzero numbers mod p.

 

[e(ho0n3]

You're right. I didn't think of that for some reason. Thanks.