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?

  • 1
    $\begingroup$ 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 Dec 2 '12 at 13:12
  • 1
    $\begingroup$ its not well known to me, is the proof accessible? $\endgroup$ – Mozibur Ullah Dec 2 '12 at 17:10
  • $\begingroup$ The group does not have to be nilpotent, consider $S_3$. $\endgroup$ – Steve D Dec 2 '12 at 23:23
  • $\begingroup$ Hi Steve, I am interested in automorphisms of order 2. $\endgroup$ – Michael Giudici Dec 3 '12 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 Dec 3 '12 at 6:04

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 $p=2$).

| cite | improve this answer | |
  • $\begingroup$ 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 Dec 3 '12 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.