$\DeclareMathOperator\SL{SL}$Let $q = p^r$ be a prime power. Let $H$ denote the subgroup of $\SL(2,\overline{\mathbb{F}}_q)$ consisting of matrices of the form $\begin{pmatrix}a & b\\ b^q & a^q\end{pmatrix}$, where $a,b\in\mathbb{F}_{q^2}$. Some notes I'm reading claim that $H$ is conjugate to $\SL(2,\mathbb{F}_q)$.

This is a very naive question — How does one classify the conjugates of $\SL(2,\mathbb{F}_q)$ inside $\SL(2,\overline{\mathbb{F}}_q)$? How do you generally classify the conjugates of $G(\mathbb{F}_q)$ inside $G(\overline{\mathbb{F}}_q)$ when $G$ is say a split reductive algebraic group over $\mathbb{F}_p$?

Also, since I expect this is easy to answer for the experts here, what is the explicit matrix that conjugates $H$ to $\SL(2,\mathbb{F}_q)$?

References would be appreciated!

  • 2
    $\begingroup$ What do you mean by "classify"? A somewhat unhelpful answer is that they are classified by elements of $G( \overline{\mathbb F}_q)$ modulo the centralizer of $G(\mathbb F_q)$ inside $G( \overline{\mathbb F}_q)$, and this centralizer is almost always the center of $G(\overline{\mathbb F}_q)$. The "classification" is the bijection that sends an element $g$ to $g G(\mathbb F_q) g^{-1}$. $\endgroup$
    – Will Sawin
    Oct 9, 2021 at 23:43
  • $\begingroup$ The main question might be why this given subgroup can be conjugate in such a way. The answer might follow from a general principle, but most likely not from a classification of conjugates in the suggested way. $\endgroup$
    – YCor
    Oct 10, 2021 at 6:51
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    $\begingroup$ But I don't see why the set of such matrices (assuming they have determinant 1) should form a subgroup. Actually, it seems it is not: if $a\in\mathbf{F}_4-\mathbf{F}_2$, then $\begin{pmatrix}1 & a\\1&a+1\end{pmatrix}\begin{pmatrix}1 & a+1\\1&a\end{pmatrix}=\begin{pmatrix}1+a&0\\a&a\end{pmatrix}$. $\endgroup$
    – YCor
    Oct 10, 2021 at 7:01
  • $\begingroup$ @WillSawin You're right. This is a badly phrased question. I will likely delete this question in a few minutes and perhaps ask a better question later. $\endgroup$ Oct 10, 2021 at 19:37
  • $\begingroup$ @YCor I actually made a typo. The bottom row should be $b^q,a^q$. Anyway, I've realized this is a bad question. I will delete this in a few minutes and perhaps ask a better question later. $\endgroup$ Oct 10, 2021 at 19:37

1 Answer 1


If $\theta$ lies in $\mathbb F_{q^2} \setminus \mathbb F_q$, then $\operatorname{Int}\begin{pmatrix} 1 & 1 \\ \theta & \overline\theta \end{pmatrix}\begin{pmatrix} a & b \\ \overline b & \overline a \end{pmatrix}$, where $\overline\cdot$ is the non-trivial Galois conjugation, lies in $\operatorname{SL}_2(\mathbb F_q)$. We can easily bring $\begin{pmatrix} 1 & 1 \\ \theta & \overline\theta \end{pmatrix}$ into $\operatorname{SL}(2, \overline{\mathbb F_q})$ by multiplying it by $\delta^{-1}$, where $\delta^2 = \overline\theta - \theta$.

$\DeclareMathOperator\tr{tr}$If $q$ is an odd prime power, then we may choose $\theta$ so that $\theta^2$ lies in $\mathbb F_q$. Then $\operatorname{Int}\begin{pmatrix} 1 & 1 \\ \theta & -\theta \end{pmatrix}\begin{pmatrix} a & b \\ \overline b & \overline a \end{pmatrix}$ equals $\dfrac1 2\begin{pmatrix} \tr(a + b) & \tr(\theta^{-1}(a - b)) \\ \tr(\theta(a + b)) & \tr(a - b) \end{pmatrix}$ for all $a, b \in \mathbb F_{q^2}$.


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