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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.

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  • $\begingroup$ It is assumed $p$ is prime. The obvious automorphisms are those coming field automorphisms of $F_q$, they form a cyclic group of order $n$, mapping injectively into the Out. The outer automorphism group of $SL_2(K)$ for an arbitrary field is actually reduced to $\mathrm{Aut}(K)$. (I think this is done in Dieudonné's 1951 Mem. AMS paper). Btw related: math.stackexchange.com/questions/1499574 $\endgroup$
    – YCor
    Dec 16, 2019 at 7:52
  • $\begingroup$ @YCor, I'm pretty sure that this is not true, since there are also the automorphisms coming from conjugation in $\operatorname{PGL}(2, \mathbb F_q)$. $\endgroup$
    – LSpice
    Dec 16, 2019 at 14:12
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    $\begingroup$ @LSpice Oh indeed I misread myself from my comment to the linked question. The assertion was that $\mathrm{Aut}(\mathrm{SL}_2(K))=\mathrm{Aut}(K)\ltimes\mathrm{PGL}_2(K)$. Hence $\mathrm{Out}(\mathrm{SL}_2(K))=\mathrm{Aut}(K)\ltimes\mathrm{PGL}_2(K)/\mathrm{PSL}_2(K)=\mathrm{Aut}(K)\ltimes(K^*/{K^*}^2)$. Hence for $K=F_{p^n}$ this yields $C_n\times C_2$ for $p$ odd and $C_n$ for $p=2$. $\endgroup$
    – YCor
    Dec 16, 2019 at 14:16

1 Answer 1

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"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, 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 and @YCor 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 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 $\langle\operatorname{Fr}\rangle \cong \operatorname C_n$.

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    $\begingroup$ For $G=SL(2,F_q)$ and $PSL(2,F_q)$, the outer automorphism groups are isomorphic to $Z_{(2,q-1)}\times Z_n$. Here the first factor is generated by the image (mod $Inn(G)$) of a diagonal automorphism and the second factor is generated by the image of the Frobenius automorphism of $F_q$, acting on matrix entries. This is a consequence of Steinberg's theorem quoted by @LSpice. $\endgroup$ Dec 16, 2019 at 12:11
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    $\begingroup$ Here is another reference for Steinberg's theorem: Theorem 30 in Robert Steinberg, "Lectures on Chevalley Groups," Amer. Math. Soc. University Lecture Series 66 (2016), page 93. There are no graph automorphisms in this case since the Dynkin diagram of type $A_1$ has only trivial symmetries. $\endgroup$ Dec 16, 2019 at 12:30
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    $\begingroup$ I've checked the 1951 Dieudonné Memoirs AMS with appendix by Hua. Actually Dieudonné determines the automorphism group of $\mathrm{SL}_n(K)$ (and its cousins) for arbitrary field $K$ when $n\ge 3$, but leaves open the case $n=2$. In the appendix, Hua solves the case of $\mathrm{(P)GL}_2(K)$, $\mathrm{SL}^\pm_2(K)$, but explicitly leaves the case of $\mathrm{SL}_2(K)$ open. $\endgroup$
    – YCor
    Dec 16, 2019 at 14:37
  • $\begingroup$ @YCor Have you found info about the outer automorphism group of $\text{SL}_2(K)$? Specifically, I am interested in $K = \mathbb{R}$. Thanks for any help and any references! $\endgroup$
    – XYSquared
    Nov 29, 2021 at 7:57
  • $\begingroup$ @XYSquared I'd check in Tits' papers from the 60s. I guess this was eventually solved (and that $\mathrm{SL}_2(\mathbf{R})$ has trivial outer automorphism group, as abstract group). $\endgroup$
    – YCor
    Nov 29, 2021 at 8:24

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