I was wondering if there are some "algebraic" fixed point theorems, in group theory.

More precisely, given a group $G$ and a group morphism $f : G \to G$, what conditions on $G$ and $f$ should we demand, so that $f$ has a non-trivial fixed point (i.e. $\exists x \neq 1_G, f(x)=x$) ?

Here are my thoughts :

  1. This « non-trivial fixed point condition » is sometimes a strong condition. For instance, if $G = \mathbb Z$, then the only $f \in \text{Hom}(G,G)$ to have a non-trivial fixed point is the identity.

  2. The set of fixed point $\{y \in G \mid f(y)=y\}$ is a subgroup of $G$.

  3. Let $G = \mathbb Z / n\mathbb Z$. Assume that $n=ab$ with $a,b>1$. If $f([1]_n) = [a+1]_n$, then $f$ has a non trivial fixed point, namely $x=[b]_n$.
  4. This question may be «artificial» ; I don't know if a morphism with a non trivial fixed point can be useful in other contexts...
  5. I don't see a natural way to turn this problem into a « group action » problem (to get some results about fixed points). I tried $G \curvearrowright \text{Im}(f)$ by defining $g \bullet f(x) := f(g)f(x) = f(gx)$, but this doesn't seem to help...

Thank you in advance !

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    $\begingroup$ For starters, a finite group which admits a fixed-point free automorphism of prime order is nilpotent. (moreover, the special case of order 2 in fact implies that the group is abelian) $\endgroup$ – M.U. Dec 27 '15 at 10:30
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    $\begingroup$ related: mathoverflow.net/questions/56464/… $\endgroup$ – M.U. Dec 27 '15 at 10:46
  • $\begingroup$ Apparently, the result quoted below was the subject of a talk here. $\endgroup$ – Watson Jul 13 '18 at 15:27

The only fixed point theorem involving finite groups I know is the following:

$p$-group fixed point theorem: Let $G$ be a finite $p$-group acting on a finite set $X$. Then $|X^G| \cong |X| \bmod p$. In particular, if $|X| \not \equiv 0 \bmod p$, then $G$ has a fixed point.

For example, applied to the conjugacy action of a finite $p$-group on itself, we conclude that such a group has nontrivial center. We can get a statement of your form by asking that $G$ is a $p$-group and $f$ has order a power of $p$.

Other applications are given here.

  • $\begingroup$ Thank you for your answer. It may be easy, but I don't see what is the action you are considering if $G$ is a $p$-group and $f$ has order a power of $p$ ? $\endgroup$ – Watson Jan 3 '16 at 17:18
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    $\begingroup$ @Watson: the action of the cyclic subgroup of $\text{Aut}(G)$ generated by $f$ on $G$. $\endgroup$ – Qiaochu Yuan Jan 3 '16 at 18:08
  • $\begingroup$ Thank you, that's clear. I think you meant "the action of the cyclic subgroup of $\text{Aut}(G)$ generated by $f$ on $G \setminus \{1\}$.", isn't it ? $\endgroup$ – Watson Jan 3 '16 at 18:47
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    $\begingroup$ @Watson: yes, for applying the theorem. $\endgroup$ – Qiaochu Yuan Jan 3 '16 at 19:26

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