Let $G$ be some discrete finitely generated group acting cocompactly on a leafless tree $T$. Is it true that for any natural number $n$ there is a finite graph $\Gamma$ such that: $T$ is the universal cover of $\Gamma$ and the girth of $\Gamma$ is at least $n$?
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$\begingroup$ I don't see how the group $G$ enters the picture here. $\endgroup$– Michal KotowskiCommented Dec 10, 2013 at 20:32
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1$\begingroup$ When a graph allows a cocompact group action on it this implies some regularity of this graph. $\endgroup$– Tomek OdrzygozdzCommented Dec 10, 2013 at 20:42
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1 Answer
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Let $T$ be a tree admitting a cocompact proper action of a discrete group $G$. Then $G$ has a finite index subgroup $F$ free of finite rank. Let $(F_n)$ be a sequence of finite index subgroups such that $\limsup F_n=\{1\}$ (i.e. for every $g\neq 1$, eventually $g\notin F_n$); for instance $(F_n)$ is decreasing and $\bigcap F_n=\{1\}$; these exist because free groups are residually finite. Then the quotients $F_n\backslash T$ is a family of graphs covered by $T$ of girth tending to $\infty$.
On the other hand, for a leafless tree $T$ of bounded valency, to have a cocompact automorphism group is not enough: precisely a necessary and sufficient condition for $T$ to cover a finite graph is that the automorphism group of $T$ is cocompact and unimodular (necessity is due to Bass and Kulkarni 1990).
