Help with Pojective Linear Groups

  • #1
Phillips101
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Hi,

F is a finite field. The problem is set up as follows: Let V be a 2-dimensional vector space over F. Let Omega=set of all 1-dimensional subspaces of V.

I've constructed PGL(2,F) by taking the quotient of GL(2,F) and the kernel of the action of GL(2,F) on Omega. Similarly for PSL(2,F). However, it now tells me to do something seemingly completely unrelated: Show that Omega can be identified with the set F union {infinity} in such a way that GL(2,F) acts on Omega as the set of Mobius transformations.

Thoughts on this bit so far: Clearly the matrix group and the mobius group are analagous... The identity is a=d and b=c=0 for instance in standard notation. How would I prove this rigorously?

It then goes on to say that I need to show PSL(2,F) under this action is all the transformations with determinants of a square in F. Huh?! I thought PSL(2,F) consisted of matrices with determinant 1?

Any help on either part would be much appreciated :)

Thanks
 

Answers and Replies

  • #2
fresh_42
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The basic underlying principle is the following:
In projective coordinates, we have points ##(a_1:a_2:\ldots :a_n)## which are equal to ##(\lambda a_1: \lambda a_2:\ldots : \lambda a_n)##, i.e. common factors do not count. This is the equivalence of ##PGL## being the factor group of ##GL## by its center.
 

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