There's a particular property of the elements of $\mathrm{GL}(2,q)$, the general linear group of $n\times n$ matrices over a finite field of order $q$, that I don't understand.
I know that the order of $\mathrm{GL}(2,q)$ is $(q^2-1)(q^2-q)=q(q+1)(q-1)^2$, since there are $q^2-1$ possible vectors for the first column, excluding the $0$ vector, and $q^2-q$ possible vector for the second column, excluding all multiples of the first.
So the order of any element must divide $q(q+1)(q-1)^2$ by Langrange. However, there is a further detail that
any element of $\mathrm{GL}(2,q)$ must have order dividing $q(q-1)$ or $(q-1)(q+1)$.
Is there a reason why one can narrow down the order to divide one of those smaller factors of $q(q+1)(q-1)^2$? Thanks!
Say $q=p^n$, with $p$ prime, $n\gt 0$.
If the matrix is diagonalizable, then you can find elements $a$ and $b$ of $\mathbb{F}_q$ such that $$A\approx \left(\begin{array}{cc} a&0\\ 0&b \end{array}\right).$$ Since $ab\neq 0$, then since $a^{q-1}=b^{q-1} = 1$, it follows that $A^{q-1}=I$, so the order divides $q-1$. Selecting $a$ to be a primitive root shows that we cannot get away with anything smaller.
If $A$ is not diagonalizable but has repeated eigenvalue, then it is similar to the matrix $$\left(\begin{array}{cc} a & 1 \\ 0 & a \end{array}\right).$$ The $n$th power of this matrix is $$\left(\begin{array}{cc} a & 1\\ 0 & a\end{array}\right)^n = \left(\begin{array}{cc} a^n & na^{n-1}\\ 0 & a^n \end{array}\right).$$ For this to be the identity, we need $n$ to be a multiple of the order of $a$ in the multiplicative group of units, and for $n$ to be a multiple of the characteristic. The order of $a$ is coprime to the characteristic, so the order is a multiple of $p$ that divides $p(q-1)$; selecting $a$ to be a primitive root shows we can get away with nothing smaller.
If the matrix has no eigenvalues, then it is diagonalizable over the field of $q^2$ elements, and so by the first case above has order dividing $q^2-1$; selecting a primitive polynomial for $\mathbf{F}_{q^2}$ over $\mathbf{F}_q$ will produce a matrix of order exactly $q^2-1$.
So in any case, the order of an element either divides $p(q-1)$ (if it has eigenvalues), or divides $q^2-1$ (if it has no eigenvalues).