I'm looking for a reference for a description of the outer automorphism groups of $\operatorname{SL}(2,\mathbb{F}\_q)$ for $q = p^n$.


I'm sure such a thing must exist somewhere, but I'm having trouble locating a reference.



What is the outer automorphism group of $\operatorname{SL}(2,\mathbb{F}_q)$?
"Each automorphism $\sigma$ of $G$ can be written $\sigma = g f d i$, with $i$, $d$, $f$, and $g$ being inner, diagonal, field, and graph automorphisms, respectively" (Steinberg - [Automorphisms of finite linear groups](https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/automorphisms-of-finite-linear-groups/16023F257E0F21D57873B1450E9F15E4), 3.2). Here, as best as I can tell, $G$ is $\operatorname{PSL}(2, \mathbb F\_q)$ (not $\operatorname{PGL}(2, \mathbb F\_q)$); the definition of $G$ relies on a set that looks like $\mathfrak B$, whose definition I cannot find. A diagonal automorphism is one that arises by conjugation in the diagonal subgroup of $\operatorname{PGL}(2, \mathbb F\_q)$, not just of $\operatorname{PSL}(2, \mathbb F\_q)$; the only non-inner automorphism that we get this way is conjugation by $\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}$, where $\epsilon$ is a non-square in $\mathbb F\_q^\times$.


If an automorphism $\sigma$ of $\operatorname{SL}(2, \mathbb F\_q)$ induces the identity on $\operatorname{PSL}(2, \mathbb F\_q)$, then $g \mapsto g^{-1}\sigma(g)$ maps $\operatorname{SL}(2, \mathbb F\_q) \to \{\pm1\}$. However, $\operatorname{SL}(2, \mathbb F\_q)$ is generated by its unipotent elements, hence admits no non-trivial homomorphisms to a 2-torsion group (assuming $p \ne 2$). (EDIT: If $p = 2$, then $\operatorname{SL}(2, \mathbb F\_q) = \operatorname{PSL}(2, \mathbb F\_q)$, so there is no ambiguity.)


EDIT: I somehow misread the question as asking about the full automorphism group of $\operatorname{SL}(2, \mathbb F\_q)$, not just the outer automorphism group; and it's always a good time to break out a result of Steinberg. As [@RichardLyons](https://mathoverflow.net/questions/348440/what-is-the-outer-automorphism-group-of-operatornamesl2-mathbbf-q/348441#comment872757_348441) and [@YCor](https://mathoverflow.net/questions/348440/what-is-the-outer-automorphism-group-of-operatornamesl2-mathbbf-q/348441#comment872771_348440) point out, since there are no diagram automorphisms in type $\mathsf A\_1$, the outer automorphism group (of $\operatorname{SL}(2, \mathbb F\_q)$, which, we argued above, is the same as that of $\operatorname{PSL}(2, \mathbb F\_q)$) is $\langle\operatorname{Fr}\rangle \times \langle\operatorname{Int}\begin{pmatrix}
\epsilon & 0 \\
0 & 1
\end{pmatrix}\rangle \cong \operatorname C\_n \times \operatorname C\_2$, generated by the Frobenius and an appropriate conjugation in $\operatorname{PGL}(2, \mathbb F\_q)$, when $p \ne 2$.


If $p = 2$, then, again as [@YCor](https://mathoverflow.net/questions/348440/what-is-the-outer-automorphism-group-of-operatornamesl2-mathbbf-q/348441#comment872771_348440) points out, there is no extra conjugation coming from $\operatorname{PGL}(2, \mathbb F\_q)$ (since every element of $\mathbb F\_q$ is a square), so we get that the outer automorphism group is just $\langle\operatorname{Fr}\rangle \cong \operatorname C\_n$.


