# How many elements does GL(2,Z2) have?

## Homework Statement

If $Z$2={[0],[1]} with addition and multiplication modulo 2. How many elements does GL(2,$Z$2)={2 by 2 matrices with determinant 1 and entries from $Z$2} have?

## The Attempt at a Solution

A$\in$GL(2,$Z$2) then det(A)=[1]={...,-3,-1,1,3,5,...}
Suppose A=|a b:c d| then det(a)=ad-bc. If a,b,c,d$\in$$Z$2 then a.d can equal [1] or [0].
If a.d=[1] then b.c=[0] so that det(A)=1
If a.d=[0] then b.c=[1] so that det(A)=1

If a.d=[1] then a=[1] and d=[1] meaning that b.c=[0] so b=[0] or [1] and c=[0].
From this one can see that there are 3 possible matrices with determinant 1 and a.d=[1].
The same logic shows that if a.d=[0] there are another 3 matrices with determinant 1.

Therefore GL(2,$Z$2) has 6 elements.

I haven't been given a solution to this problem and I'm new to equivalence classes and $Z$n so I'm not sure if have made any stupid mistakes in solving this problem?

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Hurkyl
Staff Emeritus
Gold Member
By the way, there's a classic counting argument for counting the size of $GL(n, k)$ where k is a finite field of q elements. The basic idea is that elements of GL(n,k) are the same thing as ordered bases on the vector space kn.
P.S. GL(n,k) is not the nxn matrices over k of determinant 1: it's the invertible nxn matrices. The notions of being invertible and determinant 1 just happen to coincide over $\mathbb{F}_2$: this is the only field where that happens.