# Order of Group of 2 x 2 Matrices

1. Sep 11, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Let G be the group of all 2 x 2 matrices
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
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?

2. Sep 11, 2008

### 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.

3. Sep 11, 2008

### e(ho0n3

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

4. Oct 2, 2008

### 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?

5. Oct 2, 2008

### Dick

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

6. Oct 2, 2008

### 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?

7. Oct 2, 2008

### 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.

8. Oct 2, 2008

### e(ho0n3

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