Order of SL(2,Zp), p is a prime.

[Silversonic]

Homework Statement

Determine the index [G], where H is a subgroup of the group G and;

$G = GL(2,Z_p)$

$H = SL(2,Z_p)$

p is a prime

Where $GL(2,Z_p)$ is the general linear group of 2x2 invertible matrices with entries in $Z_p$, $SL(2,Z_p)$ is the general linear group of 2x2 invertible matrices with entries in $Z_p$ but with determinant 1.

Homework Equations

$|GL(2,Z_p)| = (p^2 -1)(p^2 -p)$

Lagrange's theorem

$|GL(2,Z_p)| = [G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />]|SL(2,Z_p)|$

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 $SL(2,Z_p)$ in terms of p.

I know how to derive the order of $GL(n,Z_p)$ (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 $SL(2,Z_p)$ must be one, so

$ac-bd = 1$ - the determinant

Thus if we let

$ac = m$

Then

$bd = m - 1$

$1 \leq m \leq p^2$

So, as $a$ or $c$ cannot be zero, $a$ and $c$ can have values ranging from $1$ to $p-1$. i.e. $a$ can have $(p-1)$ different values and so can $c$. So the number of matrices in $SL(2,Z_p)$ is

(Number of combinations of a) times (Number of combinations of c) times (Number of combinations of b) times (Number of combinations of d)

=

$(p-1)$ times $(p-1)$ times (Number of combinations of b) times (Number of combinations of d)

=

$(p-1)^2$ times (Number of combinations of b) times (Number of combinations of d)

How would I work out the number of possible combinations of $b$ and $d$ from this? Once I work that out, I should times $(p-1)^2$ by those to get the order of $SL(2,Z_p)$.

 

[micromass]

Hmmm, what if you apply the isomorphism theorem to the map

$$GL(2,\mathbb{Z}_p)\rightarrow \mathbb{Z}_p:A\rightarrow det(A)$$

??