Elements of specific order in $SL(2,q)$ Let $G = SL(2,q)$, the group of all invertible $2 \times 2$ matrices of determinant $1$ over $\mathbb{F}_q$, where $q$ is odd. 
(a) Determine the number of elements of order $2$ in $G$.  
(b) Find an element of order $4$ in $G$. 
Here are my thoughts so far: 
(a) If an element $A$ of $G$ has order $2$, then $A^2 = I \Rightarrow A^2 - I = 0 \Rightarrow f(x) = x^2-1 = (x-1)(x+1)$ is an annihilating polynomial for $A$. Since the minimal polynomial of $A$ must divide this annihilating polynomial, we have three possibilities for the minimal polynomial of $A$ : $m_A(x) = x-1, m_A(x) = x+1, m_A(x) = (x-1)(x+1)$. The first case is not possible; since a matrix satisfies its minimal polynomial, $m_A(x) = x-1$ would give that $A$ is the identity matrix in $SL(2,q)$, which has order $1$. In the second case, $A$ is the negative identity matrix, which does have order $2$ in $SL(2,q)$. The third case is not possible; in this case, the matrix $A$ has eigenvalues $1$ and $-1$, and since the determinant is the product of eigenvalues, this would not give a determinant $1$ matrix over $F_q$. 
Thus, I've got one element of order $2$ in $G$ so far, which is the negative identity matrix. The above process tells me that this is the only element of order $2$ in $G$. Is this correct ? I think the only other possibility is a scalar multiple $a \in \mathbb{F}_q$ of this, $-aI$, such that $a^2 \equiv 1$ (mod $q$) -- but I'm not sure how to count the number of these for general $q$.  
(b) If an element $A$ of $G$ has order $4$, we get that $f(x) = x^4-1 = (x-1)(1 + x + x^2 + x^3)$ is an annihilating polynomial for $A$. The minimal polynomial cannot involve the second factor there, since we only have a $2 \times 2$ matrix -- thus, I believe the minimal polynomial must be $m_A(x) = x-1$. But, then this means $A$ is the identity matrix, which has order $1$. How can I find an element of order $4$ in $G$ ? My methods haven't brought any fruition in this case. 
Thanks! 
 A: Edit: In part (b) we're asked to find an element of order $4$ in $G=SL_2(\mathbb{F}_q)$, where $q$ is odd. My original post contained a correct solution (i.e. I did find an element of order $4$). However, on the way to finding that element, I did say something that was incorrect. The mistake has been corrected below. Thanks go to the thoughtful reader who caught the mistake!
-- March 12th, 2022
Part (a): Your analysis for part (a) is correct. The minimal polynomial of $A$ must divide $x^2-1=(x-1)(x+1)$, and since $A$ is $2\times2$, we must have either $m_A(x)=x-1$, $m_A(x)=x=1$, or $m_A(x)=(x-1)(x+1)$.
If $m_A(x)=x-1$, then since $A$ satisfies $m_A(x)$, we would have that $A-I=0$, which would imply that $A=I$. But since $I$ has order $1$, we can't have $m_A(x)=x-1$.
If $m_A(x)=(x-1)(x+1)$, then $A$ would have $1$, $-1$ for eigenvalues, so it wouldn't be an element of $SL_2(\mathbb{F}_q)$. So it can't be that $m_A(x)=(x-1)(x+1)$.
If $m_A(x)=x+1$, then since $A$ satisfies $m_A(x)$, we would have that $A+I=0$, which implies that $A=-I$. Hence, there is exactly one element of order $2$ in $SL_2(\mathbb{F}_q)$. (Note: we are implicitly using the fact that $q$ is odd, since if $q$ were even, then $1\equiv-1$).
You had asked if we could get other elements of order $2$ of the form $\alpha I$, for some scalar $\alpha$. If $\alpha I$ had order $2$ then $(\alpha I)^2=I$, which would imply that $\alpha$ satisfies the equation $\alpha^2=1$ in $\mathbb{F}_q$. Since $\mathbb{F}_q$ is a field, the only elements that satisfy this equation are $\pm1$.
Part (b): Let $A\in SL_2(\mathbb{F}_q)$ be an element of order $4$. Then $A^4=I$, so $m_A(x)$ divides $x^4-1=(x-1)(x+1)(x^2+1)$.
Here's an idea for finding an element of order $4$: We've already shown that: (1) an element $A$ of $G=SL_2(\mathbb{F}_q)$ has order $1$ iff $m_A(x)=x-1$; (2) an element $A$ of $G$ has order $2$ iff $m_A(x)=x+1$; and (3) there is no element $A$ of $G=SL_2(\mathbb{F}_q)$ with $m_A(x)=(x-1)(x+1)$. So the idea is to see if we can find an element $A$ such that $m_A(x)$ divides $x^2+1$. In other words, let's try to find element $A$ with $A^2+I=0$.
Note: if an element $A$ of $G$ satisfied $A^2+I=0$, then it would also satisfy $A^4=I$. So it's order would have to be $1$, $2$, or $4$. On the other hand, an element $A$ with $A^2+I=0$ could not satisfy $A-I=0$ or $A+I=0$. So it's order could not be $1$ or $2$. So if $A^2+I=0$, then $A$ would have to be order $4$.
Finally, let's note that if we had an element $A$ of $G$ with $m_A(x)=x^2+1$, then from the rational canonical form, we would get that $A$ would be similar to $$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$ We can check that the above matrix does satisfy $A^2+I=0$, so it is, in fact, an element order $4$ in $SL_2(\mathbb{F}_q)$.
Note: we are again assuming that $q$ is odd since otherwise we'd have $1\equiv-1$. Since all we had to do was find one element of order $4$, we are finished with part (b).
