Are the non-free factors of Grushko decomposition of a finitely generated convex-cocompact (but not cocompact) subgroup of PSL$(2,\mathbb{R})$ finite? See Grushko decomposition theorem.
Are the non-free factors of Grushko decomposition of a finitely generated convex–cocompact (but not cocompact) subgroup of $\operatorname{PSL}(2,\mathbb{R})$ finite?
In the cocompact case it is not true, since the group is not a free group and cannot be split into a non-trivial free product.
For convex–cocompact but not cocompact I know of particular examples with affirmative answer. Is it always the case?
 A: For a discrete, noncocompact subgroup $\Gamma < \text{PSL}(2,\mathbb R)$, the quotient $\mathbb H^2 / \Gamma$ is a noncompact, 2-dimensional oriented orbifold, i.e. a noncompact surface with an orbifold locus consisting of cone singularities forming a closed, discrete subset.
Assuming in addition the finite type hypothesis, the underlying surface of the quotient orbifold is obtained from some closed oriented surface by removing a finite subset, and the orbifold locus is a finite set.
Such an orbifold has a spine (an orbifold deformation retract, thus having the same fundamental group) which is a finite graph of groups, whose vertices include all of the cone singularities and perhaps some other points with trivial vertex group, and whose edges all have trivial group.
The Grushko decomposition of the graph-of-groups fundamental group therefore has the form $A_1 * ... * A_K * F_n$ for some finite cyclic groups $A_1,...,A_K$ and some finite rank free group $F_n$.
So yes, the non-free factors of the Grushko decomposition are all finite.
