Is there a link between free actions and free groups?

Question

I am trying the prove the Hausdorff paradox. To do so I need to a define free groups and prove how free group: $F$ of rank 2 acting on a set $X$ means that $X$ is $F$-paradoxical. I understand what a free action on a set is, but would be interested to see if there is a link between free actions and free groups. Do free actions generate free groups in some way or are they different concepts entirely?

Answer 1

There is no particular relationship between free groups and free actions (and in particular, they are definitely not named similarly because of any special relationship). They are both "free" in the more general sense of abstract algebra (an object generated by a set with no relations), but in different categories: a free group is a free object in the category of groups, while a set with a free $G$-action is a free object in the category of $G$-sets.

Answer 2

Free actions arise naturally in topology, and are a fruitful way of connecting topology with group theory. In particular, we can understand basically all groups using free actions in a really nice way. So, to start with, some topology: suppose $S$ is a nice* topological space then it has a simply connected universal covering space $\widetilde{S}$ on which the fundamental group $\pi_1(S)$ acts freely on. Clearly, $\pi_1(S)$ need not be free here!

Now the connection to group theory: suppose $G$ is a group given by a finite set of generators and relators, so $g=\langle X\mid R\rangle$. The associated presentation complex $S$ consists of the "bouquet" of $|X|$-many loops (a graph with a single vertex and then loops labelled by the generators), and 2-cells corresponding to elements of $R$ (words correspond to closed paths on our bouquet so this makes sense). Then the universal cover $\widetilde{S}$ is the associated "Cayley complex"; it is the Cayley graph of $G$ with generators $X$, with some added $2$-cells (indeed, every loop can be tiled by $2$-cells).

If $G$ is free on $X$, so $G=\langle X\mid -\rangle$, then the associated presentation complex has no $2$-cells and hence $G$ acts freely on its Cayley graph (which is a tree). This gives you one direction of Lee Mosher's comment, that groups are free if and only if they act freely on a tree.

*nice=path connected, locally path connected and locally simply connected