Order of Group of 2*2 matrix

Question

Let G be the group of 2*2 matrices [ a b ; c d] where a,b,c,d are integers modulo p, p is prime number, such that ad-bc≠0. G forms group under relative to matrix multiplication. What is o(G)? Let H be the subgroup of the G defined by H= { (a b; c d) €G | ad-bc=1} . What is order of H?

I got the direct formula for computation, but How can we get that relation or is there any simple way to do this??

Answer 1

For any element $A\in H$, you can multiply the top row by any non-zero number to get an element of $G$. So $|G|=(p-1)|H|$.
Now to count $H$.

• If $a\neq0$, then $d=(1+bc)/a$, so there are how many matrices? How many choices for $a,b,c$?
• If $a=0$, then $c=-b^{-1}$. That gives how many other matrices? How many choices for $b,d$?

Answer 2

As an alternative to Michael's answer (which does give the right hints), you can count the tuples $(a,b,c,d)$ for which $ad-bc=0$:

• If $a=0$ (one case) and $d$ arbitrary ($p$ cases) and $b=0$ (one case), then $c$ is arbitrary ($p$ cases), so this yields $p\cdot p=p^2$ tuples.
• If $a=0$ (one case) and $d$ arbitrary ($p$ cases) and $b\neq 0$ ($p-1$ cases), then $c$ must be $=0$, so this case yields $p\cdot(p-1)=p^2-p$ tuples.
• If $a\neq 0$ ($p-1$ cases) and $d=0$ (one case) and $b=0$ (one case), then $c$ is arbitrary ($p$ cases), so this yields $(p-1)\cdot p=p^2-p$ tuples.
• If $a\neq 0$ ($p-1$ cases) and $d=0$ (one case) and $b\neq 0$ ($p-1$ cases), then $c$ must be $=0$, so this yields $(p-1)\cdot(p-1)=p^2-2p+1$ tuples.
• If $a\neq 0$ ($p-1$ cases) and $d\neq0$ ($p-1$ cases), then $b$ can be arbitrary $\neq 0$ ($p-1$ cases), and $c$ is determined by $a$, $d$ and $b$, so this yields $(p-1)\cdot(p-1)\cdot(p-1)=p^3-3p^2+3p-1$ tuples.

In total, these five cases add up to $$p^2+p^2-p+p^2-p+p^2-2p+1+p^3-3p^2+3p-1=p^3+p^2-p$$ "disallowed" tuples, so the number of elements of $G$ is $$p^4-(p^3+p^2-p)=p^4-p^3-p^2+p=(p^2-1)\cdot(p^2-p)$$ (the last term being the standard way of expressing the group's order).