Here is the proof of a theorem I am working through in Geometric Group Theory by Clara Loh:
The first paragraph shows that if $F$ is free, then it admits a free action on a (non-empty) tree. This paragraph gives me no trouble. On the other hand, the second paragraph, which proves that $F$ admitting a free action on a (non-empty) tree implies it is free, is giving trouble. First, why is $S$ a free generating set? Isn't that tantamount to assuming that $F$ is free, which is what we are trying to prove? Second, why does the left action of $F$ on $Cay(F,S)$ imply that $F$ is free?
Here is proposition 4.1.10:
Proposition 4.1.10 (Free actions on Cayley graphs) Let $G$ be a group and let $S$ be a generating set of $G$. Then the left translation action on the Cayley graph $Cay(G,S)$ is free if and only if $S$ does not contain any elements of order $2$
Theorem 3.3.3 might be relevant, but I'm not sure...
Theorem 3.3.3 (Cayley trees and free groups) Let $G$ be a group, let $S \subseteq G$ be a generating set satisfying $s \cdot t \neq e$ for all $s,t \in S$. If the Cayley graph $Cay(G,S)$ is a tree, then $S$ is a free generating set of $G$.