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In article <1159544026.292114.112580@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<dince24@xxxxxxxxx> wrote:
>I need to prove that the order of GL ( n , Fp) where p is prime is :
>p^4 - p^3 - p^2 + p
Surely this is wrong, as larger n will give you larger groups. Do you
mean GL(2,Fp)?
>The hint in the book says to look at 2x2 matrices from the field and
>the invertible ones, but I don't even know how to start.
>Thanks for any help.
Let us consider a 2 x 2 matrix with coefficients in the field. The
order of the full ring of 2 x 2 matrices with coefficients in F_p is
p^4 (p choices for each of the 4 entries).
So the order of GL(n,Fp) is p^4 - #{A in M_{2x2}(Fp) : A is not invertible}
Now: when is a 2 x 2 matrix invertible? If you think of its rows as
vectors in Fp^2, then the matrix is invertible if and only if the two
vectors are linearly independent; so a matrix in M_{2x2}(Fp) is not
invertible if and only if the two rows are linearly independent, which
happens if and only if one of the rows is a multiple of the other.
If the first row is zero, then the matrix is not invertible.
If the first row is not zero, then the matrix is linearly
independent if and only if the second row is a multiple of the
first. Start counting.
--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org
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