Finite groups whose non-trivial elements have no fixed points

Question

(I first asked this question on MathOverflow, but was recommended to ask here at Mathstackexchange instead.)

I am interested in finite groups $G$ acting on a finite set $X$ with the following property:

(*) fix(g)=$\emptyset$ for all $g\in G\setminus\{1\}$,

where

fix(g):=$\{x\in X|gx=x\}$

denotes the set of fixed points, i. e. (*) means: no non-trivial group element has a fixed point on $X$. (Equivalently: All non-trivial elements are derangements on $X$.)

As a starting point: The cyclic group $<(1,2,3,\dots,n)>\leq S_n$ acting naturally on $\{1,...,n\}$ serves as a simple example. Another example is subgroups of $GL_n(\mathbb{F}_q)$ whose elements all lack eigenvalue 1 (except of the identity matrix, of course) acting on $\mathbb{F}_q^{n}\setminus\{0\}$ (via left multiplication). However, there don't seem to be many of such matrix groups.

I'd like to see more (interesting) examples / classes of examples of such group actions. How rare are they?

Answer 1

There is a name for those actions, these are "free" group actions. There are many examples in the litterature. To name but a few :

A group $G$ acts on itself by left translation $g.h:=gh$. This action is (somehow) important because it allows you to identify any group with a subgroup of bijection on some set (namely the set $G$). In particular it leads to the Cayley Isomorphism theorem, every finite group is a subgroup of the permutation group of a finite set.

The second example appears sometimes :

Another example is when a topological group $G$ acts freely and properly (see Wikipedia for the definition) on a manifold $M$. Then $M/G$ admits naturally a manifold structure itself.

Answer 2

Exercise:

1) let $X$ be a set and let $G$ act on $G\times X$ as $g.(h,x)=(gh,x)$. Then the action is free.

2) Conversely, let $G$ act freely on some set $Y$. Then prove that there exists $X$ such that $Y$ is isomorphic to $Y'=G\times X$ as a $G$-set (in the sense that there exists a bijection $f:Y\to Y'$ such that $f(gy)=gf(y)$ for all $y\in Y$ and $g\in G$).