I read the paper "Kurosh rank of intersections of subgroups of free products of orderable groups", which can be found here: <http://arxiv.org/abs/1109.0233>.
The proposition 2.3 states that: Let $H$ be a subgroup of $G$. Let $T'$ be a $H$-subtree of $T$. Then $κ\_{T}(H)$ = $κ\_{T′}(H)$.
In particular, $T\_{H}/H$ is ﬁnite (as a graph) if and only if $κ\_T(H)$ is ﬁnite.


They leave it as Bass-Serre exercise.
I am not very good in Bass-Serre theory, so I would appreciate if anyone can help me with this proof.



Bass- Serre theory exercise
First you need that every vertex of $T$ whose $H$-stabilizer is nontrivial is contained in the $H$-minimal subtree $T\_H \subset T$, and so each such vertex is contained in $T'$. This implies that $c\_T(H)=c\_{T'}(H)$ where $c$ is the number of orbits of vertices with nontrivial stabilizer. 


Next you need that the equivariant deformation retraction from $T$ to $T'$ descends to a deformation retraction from $T/H$ to $T'/H$, which again is a consequence of the fact that vertices not in $T\_H$ have trivial stabilizers. It follows that $rank(T/H)=rank(T'/H)$ where $rank$ means the rank of the first homology group of a graph. 


So $\kappa\_{T}(H)=c\_T(H)+rank(T/H)=c\_{T'}(H)+rank(T'/H)=\kappa\_{T'}(H)$.


