Show that $\left|PGL(2,q)\right| = (q-1) q (q+1)$

Question

I have thought about quite a while, I figured that if $q = 2$ or $q = 3$, then $(q + 1)! = (q + 1)(q)(q − 1)$ because $0! = 1$ and $1! = 1$ respectively. I'm sure that I can use this somehow but I can't figure out how, maybe someone knows and elegant way to show that $\left|PGL(2,q)\right|$ is indeed equal to $(q-1) q (q+1)$

Help is appreciated :)

Answer 1

$GL(n,q)=(q^n-1)(q^n-q)(q^n-q^2) \ldots (q^n-q^{n-1})$

$|PGL(n,q)|=|SL(n,q)|=\frac{|GL(n,q)|}{|Z(GL(n,q))|}$

$|PGL(2,q)| = (q^2-1)(q^2-q)/(q-1)$

$=(q+1)q(q-1)$

Answer 2

Let us count all the regular (=invertible) matrices, i.e. the elements of $\;GL(2,q)\;$ : look at the vector space of ordered pairs $\;V:=(\Bbb F_q)^2\;$ . From this $\;q^2\;$ vectors, there are $\;q^2-1\;$ which can be a regular matrix's first column, and then for the second column we can choose all the vectors in $\;V\;$ but those which are a multiple scalar of the first column, so we're left with $\;q^2-1\;$ choices, and all in all\

$$(q^2-1)(q^2-q)=q(q-1)(q^2-1)=q(q-1)^2(q+1)$$

To get $\;|PGL(2,q)|=|SL(2,q)|\;$ now just observe that this is the kernel of the homomorphism $\;\det: GL(2,q)\to\Bbb F_q^*\;$ , so its image's order is $\;|\Bbb F_q^*|=q-1\;$ .