Order of matrices in $SL_2({\mathbb{F}_q})$

Question

Could you tell me how to prove that in $SL_2({\mathbb{F}_q})$ the only element of even order is $-I$ ($ \ I$ - identity matrix)?

I would really appreciate a thorough explanation, because I cannot find anything on my own.

Here, page 29, is why I'm asking.

Answer 1

Not true. Take $q = 5$, for instance, then $$ \begin{bmatrix} 2 & 0\\ 0 & 3 \end{bmatrix} $$ has order $4$.

More generally, your group has order $$ \frac{(q^2 -1) (q^2 - q)}{q-1} = (q-1) q (q+1), $$ so for $q$ odd you see a Sylow $2$-subgroup has order at least $8$.

(I'm only thinking of elements of order a power of $2$, then see user53845's answer.)

Addendum

The real question appears to be that $-I$ is the unique element of order $2$. Such an element must have two eigenvalues equal to $-1$, and then $$ \begin{bmatrix} -1 & 1\\ 0 & -1 \end{bmatrix}^{2} = \begin{bmatrix} 1 & -2\\ 0 & 1 \end{bmatrix} \ne I. $$ ($q$ odd assumed throughout, see the post of Marc van Leeuwen.)

Answer 2

It is false. For every element $g$ of odd order, $(-I)\cdot g$ must have even order.

Answer 3

The text only says that $-I$ is the unique element of order $2$, not the only element of even order. Even this is not quite true, but it is true in characteristic different from $2$, since there all involutions (elements of order$~2$) in $GL(n)$ are diagonalisable (being annihilated by the polynomial $X^2-1=(X+1)(X-1)$ that is split with simple roots), and for $n=2$ this can only mean there is one eigenvalue $1$ and one eigenvalue $-1$, so the determinant is $-1$ one one is not in $SL(2)$.

However in characteristic $2$ any element of the form $\begin{pmatrix}1&x\\0&1\end{pmatrix}$ has order $2$ and determinant$~1$, and is not a multiple of the identity if $x\neq0$.