Let $F$ a free group acting freely on a graph $X$. Does this imply that $X$ is a tree?

If $F=\mathbb Z$, the answer is no. For example $\mathbb Z$ acts on $\mathbb Z^2$ by translations.

But what about the other free groups?

As $F_2$ contains $F_k$ as a subgroup $\forall k$, it would suffice to show that the claim is also falso for $F=F_2$.

However I could not think of any graph which is not a tree on which $F_2$ acts freely. In particular, letting one generetor ($b$) act trivially does not give a free action, because then $1\neq aba^-1$ acts as the identity.

  • 4
    $\begingroup$ Copying the example of $F = {\mathbb Z}$, $F_2$ acts freely on the Cayley graph of $F_2 \times H$ for any group $H$. $\endgroup$ – Derek Holt Jul 16 '18 at 13:14
  • $\begingroup$ I think I got it. If I understood it correctly, it works for any grpah of the form $X \times A$ wtih $A$ an arbitrary graph and $X$ the Cayley graph of $F_2$ $\endgroup$ – stacky Jul 16 '18 at 14:37

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