# Possible orders of trace k elements in $SL_2(\mathbb F_q)$

As a continuation of this question I would like to ask about possible orders of trace $$k$$ elements in $$SL_2(q)$$. Here are examples which I know.

When trace is zero then we have $$x^2=-1$$ so it means that order of $$x$$ is $$2$$ in characteristic $$2$$ and it is $$4$$ in odd characteristic.

When trace is $$-1$$ then we have $$x^2=-x+1$$, so $$x^3=-x^2+x=-x+1+x=1$$. It means that order of $$x$$ is $$3$$.

When trace is $$t$$ which is order $$q-1$$ element in case $$q=2^n$$. Then order of $$x$$ is sometimes $$q+1$$ and $$x$$ generate $$\mathbb F_{q^2}$$ subalgebra. This is just guess, I have only checked this for $$q=2,4,8,16$$. Here is small test in GAP showing order of trace $$t$$ elements for $$q=2^n$$ for $$n=1..10$$, where $$t$$ is generator of the field multiplicative group:

gap> List([1..10],k->Order([[Z(2^k),1],[1,0]]*Z(2^k)^0));
[ 3, 5, 9, 17, 31, 65, 43, 51, 511, 25 ]


Let's call element imaginary when it is of trace $$0$$ in $$SL_2(q)$$. From above we know that order of such element is either $$2$$ or $$4$$. The next question we can ask is what order can have product of two imaginary elements. According to tests in GAP in characteristic two we obtain orders $$q-1$$, $$q+1$$ and divisors which are all orders in the group (tested only for few small q). In case of odd characteristic I do not have theory ready yet. Anyway set $$\{x^2=-1\}$$ seems to be interesting.

Here is some experimental data from GAP for answering this question. I do not have full picture yet. Let $$q=p^n$$, $$p$$ is prime number.
There are three types of subalgebras with one generated by one element: $$\mathbb F_q+\mathbb F_q$$, $$\mathbb F_q+\mathbb F_q\pmb i$$, $$\mathbb F_{q^2}$$ where $$\pmb i^p=1$$. The groups contained with invertible elements are $$C_{q-1}\times C_{q-1}$$, $$C_{2(q-1)}\times \underbrace{C_p\times...\times C_p}_{n-1}$$, $$C_{q^2-1}$$ with sizes $$(q-1)^2,q^2-q,q^2-1$$ respectively.
The three cases are distinguished by order of generator $$u$$ of the subalgebra. When it is divisor of $$q-1$$ then we are in case 1. When it is divisor of $$q$$ then we are in case 2. When it is divisor of $$q+1$$ then we are in case 3. Common divisor of $$q-1$$ and $$q+1$$ is $$2$$ and it happens in odd characteristic. In this case element of order $$2$$ generate case 1. In characteristic 2 zero divisor $$p^2=p$$ generate subalgebra of type 1. It is element of $$Q_{01}$$ (see this question for notation) with trace equal $$1$$.
Element of order $$p$$ in odd characteristic is belonging to $$Q_{12}$$ i.e. it is element of determinant $$1$$ and trace $$2$$ (and not belonging to $$\mathbb F_q1)$$.