For $g,h\in{\rm SL}(2,q)\setminus\{\pm I\}$ with ${\rm tr}(g)={\rm tr}(h)$, how does $h$, when ${\rm SL}(2,q)$-conjugated, relate to $g^{\pm 1}?$

Question

This question is less open-ended than you might think.

The Question:

For $g,h\in{\rm SL}(2,q)\setminus\{\pm I\}$ with ${\rm tr}(g)={\rm tr}(h)$, how does $h$, when ${\rm SL}(2,q)$-conjugated, relate to $g^{\pm 1}?$

Context:

According to a preprint:

[I]f $g,h\in \operatorname{SL}(2,\Bbb R)\setminus \{\pm I\}$ and $\operatorname{tr}(g)=\operatorname{tr}(h)$ then $h$ is $\operatorname{SL}(2,\Bbb R)$-conjugate to $g^{\pm 1}$, while if $g,h\in\operatorname{SL}(2,\Bbb C)\setminus\{\pm I\}$ and $\operatorname{tr}(g)=\operatorname{tr}(h)$ then $h$ is $\operatorname{SL}(2,\Bbb C)$-conjugate to $g$.

Thoughts:

My guess is there's some function $\varphi: \{(h,g)\}\to\operatorname{SL}(2,q)$ or maybe $\psi:\{(h,g)\}\to \Bbb Z$ such that

$$k_{h,g}hk_{h,g}^{-1}=\varphi(h,g)\tag{1}$$

or

$$l_{h,g}hl_{h,g}^{-1}=g^{\psi(h,g)},\tag{2}$$

where $k_{h,g}, l_{h,g}\in\operatorname{SL}(2,q)$.

I think $(1)$ says nothing, now that I think about it, but I'll leave it here to articulate what I mean exactly.

Something like $(2)$ seems likely.

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For copy-and-paste convenience for anyone who would like to answer (and to show more work/effort):

Let $A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ and $X=\begin{pmatrix} x & y \\ z & t\end{pmatrix}$ be in $\operatorname{SL}(2,q)$. Then

$$\begin{align} AXA^{-1}&=\begin{pmatrix} a & b \\ c & d\end{pmatrix}\begin{pmatrix} x & y \\ z & t\end{pmatrix}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}\\ &=\begin{pmatrix} dax+dbz-cay-cbt & -bax-b^2z+a^2y+abt \\ dcx+d^2z-c^2y-cdt & -bcx-bdz+acy+adt\end{pmatrix} \end{align}$$

Answer 1

There is no such relation as $(2)$ if $q=p^{2m}$ where $p$ is an odd prime number.

Note that any element of $GF(p)$ is a square in $GF(q)$. Indeed, if $x$ is a square in $GF(p)$, it is a square in $GF(q)$. If it is not a square in $GF(p),$ then $GF(p)(\sqrt{x})$ is nothing but $GF(p^2)$, which is a subfield of $GF(q)$, and $x$ is a square in $GF(q)$

Now, for $\alpha\in GF(q)^\times$, set $T(\alpha)=\begin{pmatrix}1 & \alpha \cr 0 & 1\end{pmatrix}$.

Direct computation show that if $T(\alpha)$ and $T(\beta)$ are conjugate, then $\beta\alpha^{-1}$ is a square on $GF(q).$ It is even an equivalence.

We also have $T(\alpha_1)T(\alpha_2)=T(\alpha_1+\alpha_2)$. In particular, $T(\alpha)^m=T(m \alpha)$ for all $m\in\mathbb{Z}$. Note that $GF(q)$ has characteristic $p$, so all the various powers of $T(\alpha)$ may be obtained by taking $m=0,\ldots, p-1$. For example, $T(\alpha)^{-1}=T((p-1)\alpha)$.

Take $g=T(1)$ and $h=T(\varepsilon)$, where $\varepsilon$ is a nonsquare element of $GF(q)$.

The previous considerations show that $g,g^2,\ldots, g^{p-1}$ are all conjugate to $g$ because $1,2,\ldots,p-1$ are square in $GF(q)$. However, $h$ is not conjugate to $g^0=I_2$ because $h\neq I_2$, and $h$ is not conjugate to $g$ either because $\varepsilon$ is not a square in $GF(q).$

All in all $h$ is not conjugate to any power of $g$, and $(2)$ cannot hold.

As a concrete example, you could take $p\equiv 3 \ [4]$ , $q=p^2$ and $\varepsilon=\sqrt{-1}$.