1
$\begingroup$

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...)

  • 0
    @YvesCornulier: Yes that seems to be a typo on my part. I don't remember exactly what the question was like when I left that response, but I believe I was more focused on finding groups of that type that had finitely many ends.2012-08-23

1 Answers 1

5

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.

  • 0
    Doing some exercises about free products with amalgamation would probably be very helpful in understanding them. I think there are some good ones in one of the books called "Combinatorial Group Theory", but I can't remember whether it was Magnus-Karrass-Solitar or Lyndon and Schupp.2012-05-01