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Thanks for any help.

Suppose $S$ is a simple group of Lie type of prime characteristic $p$. we know that every automorphism of $S$ is composite of inner, diagonal, field and graph automorphism of $S$. Suppose $g$ is an automorphism of $S$ of prime order $r$. Is it true that $g$ is exactly one of inner or diagonal or field or graph automorphism of $S$ not a composition. For $r=p$ is there any result for this question?

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    $\begingroup$ There's an automorphism of Spin(8), of order 3, that does not preserve any pinning. Its fixed-point subgroup is SL(3). I don't know whether it exists in characteristic 3. There is a nice treatment of all characteristic zero examples at www2.bc.edu/mark-reeder/Torsion.pdf $\endgroup$
    – David Treumann
    Dec 17, 2014 at 19:01
  • $\begingroup$ You can have a look at the "Atlas of finite groups" by J.Conway et al., which has a lot of info on $O_8^+(\mathbb{F}_3)$. $\endgroup$
    – Dima Pasechnik
    Dec 17, 2014 at 20:33
  • $\begingroup$ @Maryam: Note that your header is incomplete. Also, it might be helpful to indicate what your motivation for the question is. $\endgroup$
    – Jim Humphreys
    Dec 17, 2014 at 22:05

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An indicative example (when $p$ is odd) is : take an odd prime $r$ which divides $p(p^{2}-1)$, and let $S = {\rm PSL}(2,p^{r}).$ Let $\alpha$ be the automorphism of $S$ induced by the Frobenius automorphism of ${\rm GF}(p^{r}).$ Then $\alpha$ has order $r$ and has fixed point subgroup ${\rm PSL}(2,p)$ which contains an element $x$ of order $r.$ Then the automorphism $\beta$ which is the composition of $\alpha$ with the inner automorphism of $S$ induced by $x$ results in an automorphism of order $r.$

(Later edit: Note also that when $S = {\rm PSL}(n,q)$ and $q$ has the form $p^{2m}$ for some prime $p,$ positive integer $m,$ then there is a field automorphism inducing an automorphism of order $2$ on $S,$ and this commutes with the (automorphism induced on $S$ by) the transposed inverse automorphism which also has order $2.$ The composition of these two automorphisms still has order $2$).

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