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It is well known that a finite group admitting an automorphism of order 2 that fixes only the identity is abelian and has odd order. Moreover, the automorphism is inversion.

Is anything known about finite groups admitting an automorphism of order 2 that fixes only the identity and one other element?

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    $\begingroup$ I don't have the book to hand, but if I recall correctly, Khukhro's book on nilpotent groups and their automorphisms includes several results on automorphisms with few fixed points. $\endgroup$
    – Colin Reid
    Commented Dec 2, 2012 at 13:12
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    $\begingroup$ its not well known to me, is the proof accessible? $\endgroup$
    – Mozibur Ullah
    Commented Dec 2, 2012 at 17:10
  • $\begingroup$ The group does not have to be nilpotent, consider $S_3$. $\endgroup$
    – Steve D
    Commented Dec 2, 2012 at 23:23
  • $\begingroup$ Hi Steve, I am interested in automorphisms of order 2. $\endgroup$
    – Michael Giudici
    Commented Dec 3, 2012 at 2:21
  • $\begingroup$ Hi Michael, yes I think my second comment was wrong, so I deleted it. My comment on $S_3$ is correct though. $\endgroup$
    – Steve D
    Commented Dec 3, 2012 at 6:04

1 Answer 1

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MacKay [On the structure of a special class of $p$-groups, Quart. J. Math. Oxford Ser (2) 38, 489-502] and, indipendently, Kiming [Structure and derived length of finite $p$-groups possessing an automorphism of $p$-power order having exactly $p$ fixed points, Math. Scand. 62, 153-172] showed that if a finite $p$-group $G$ admits an automorphism of order $p^n$ with exactly $p$ fixed points, then $G$ contains a subgroup $H$ of index bounded by a function of $p$ and $n$ which is nilpotent of class at most 2 (and $H$ is abelian if $p=2$).

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  • $\begingroup$ In the context of the original question, this tells us that $G$ has a $2$-Sylow subgroup that is abelian-by-(bounded size), since any automorphism of order $p$ must leave a $p$-Sylow subgroup invariant. $\endgroup$
    – Colin Reid
    Commented Dec 3, 2012 at 13:08

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