- #1
Silversonic
- 130
- 1
Homework Statement
Determine the index [G], where H is a subgroup of the group G and;
[itex] G = GL(2,Z_p) [/itex]
[itex] H = SL(2,Z_p) [/itex]
p is a prime
Where [itex]GL(2,Z_p)[/itex] is the general linear group of 2x2 invertible matrices with entries in [itex]Z_p[/itex], [itex]SL(2,Z_p)[/itex] is the general linear group of 2x2 invertible matrices with entries in [itex]Z_p[/itex] but with determinant 1.
Homework Equations
[itex]|GL(2,Z_p)| = (p^2 -1)(p^2 -p) [/itex]
Lagrange's theorem
[itex]|GL(2,Z_p)| = [G]|SL(2,Z_p)| [/itex]
The Attempt at a Solution
To be able to do this, I need to intuitively work out the number of distinct cosets, or work out the order of [itex]SL(2,Z_p)[/itex] in terms of p.
I know how to derive the order of [itex]GL(n,Z_p)[/itex] (note: not necessarily a 2x2 matrix), but doing the same for the special linear group doesn't seem to be as straightforward.
I've looked at it this way, the determinant of [itex]SL(2,Z_p)[/itex] must be one, so
[itex] ac-bd = 1 [/itex] - the determinant
Thus if we let
[itex] ac = m [/itex]
Then
[itex] bd = m - 1 [/itex]
[itex] 1 \leq m \leq p^2 [/itex]
So, as [itex] a [/itex] or [itex] c [/itex] cannot be zero, [itex] a [/itex] and [itex] c [/itex] can have values ranging from [itex] 1 [/itex] to [itex] p-1 [/itex]. i.e. [itex] a [/itex] can have [itex] (p-1) [/itex] different values and so can [itex] c [/itex]. So the number of matrices in [itex]SL(2,Z_p)[/itex] is
(Number of combinations of a) times (Number of combinations of c) times (Number of combinations of b) times (Number of combinations of d)
=
[itex] (p-1) [/itex] times [itex] (p-1) [/itex] times (Number of combinations of b) times (Number of combinations of d)
=
[itex] (p-1)^2 [/itex] times (Number of combinations of b) times (Number of combinations of d)
How would I work out the number of possible combinations of [itex] b [/itex] and [itex] d [/itex] from this? Once I work that out, I should times [itex] (p-1)^2 [/itex] by those to get the order of [itex]SL(2,Z_p)[/itex].
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