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

  • Thread starter gottfried
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Homework Statement


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


The Attempt at a Solution


A[itex]\in[/itex]GL(2,[itex]Z[/itex]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[itex]\in[/itex][itex]Z[/itex]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,[itex]Z[/itex]2) has 6 elements.

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

Answers and Replies

  • #2
Hurkyl
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Your answer looks right.

By the way, there's a classic counting argument for counting the size of [itex]GL(n, k)[/itex] 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.

So, you proceed by first counting the number of ways to pick the first basis element: qn - 1. Then, you count the number of ways to pick the second basis element: the only requirement is that it's not in the one-dimensional subspace spanned by our previous choice, so there are qn - q choices. Then....

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 [itex]\mathbb{F}_2[/itex]: this is the only field where that happens.
 
  • #3
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Cool thanks for the help.
 

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