Amalgamated free products and Stallings' theorem on ends of groups Let $G=A\ast_C B$ be a non-trivial free product with amalgamation. Then, if $C$ has index greater than two in $A$ or $B$, $G$ has infinitely many ends if it is infinite and the amalgamating subgroup is finite, by Stallings' Theorem on Ends of Groups. My question is, > What does it mean for the amalgamating subgroup to be finite? Do you have to take two finite subgroups of $A$ and $B$, say $H$ and $K$ respectively, and amalgamate them (i.e. they are finite _before_ the amalgamation), or can $H$ and $K$ be infinite but they map to a finite subgroup of $G$ (so they are finite _after_ the amalgamation). For example, $G=F_2 \ast_{C_n} C_n=\langle a, b, c; b=c, c^n\rangle$ and here $H=\langle b\rangle\cong\mathbb{Z}$ with $K=\langle c\rangle\cong C_n$. (I'm pretty sure this is a free product with amalgamation - but I'm never quite sure about this stuff...)

The amalgamation only identifies two isomorphic subgroups, it doesn't perform any further quotienting. So the group you are amalgamating over has to be isomorphic to a subgroup of both $A$ and $B$. Thus your first statement is correct.

You cannot form an amalgamated free product of $F_2$ and a finite group with non-trivial amalgamation, because $F_2$ has no non-trivial finite subgroups.