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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...)

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You are misquoting Stallings's theorem. Consider the infinite dihedral group. You need $C$ to have index at least $3$ in one of $A$ or $B$. –  user641 Apr 30 '12 at 18:39
    
I don't want to be "that guy", but you still do not understand Stallings's theorem. The group $C_4\ast_{C_2} C_4$ fits your question, but is one-ended. –  user641 May 3 '12 at 5:48
    
@SteveD: What, you don't want to be helpful!? You are right, I do not understand Stallings' theorem. As I said in the comments to the answer, below, I prefer to ignore this stuff and just pretend it doesn't exist... –  user1729 May 3 '12 at 8:59
    
@SteveD: I have to admit I didn't even read the bit about Stallings' Theorem, since my attention was drawn immediately to the fact that user1729 didn't understand amalgamated products, and so I just concentrated on that. The question itself could have been asked without any reference to Stallings' Theorem. –  Tara B May 3 '12 at 9:37
    
@TaraB: Yes, but my motivation was to try and understand Stallings' theorem. I just got bogged down on the way... –  user1729 May 3 '12 at 9:53

1 Answer 1

up vote 5 down vote accepted

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.

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Ah, okay. One day I will understand these things... –  user1729 Apr 30 '12 at 15:21
    
Are you still not quite clear on the definition of the amalgamated product? I can add some more detail if there's something you don't understand. –  Tara B Apr 30 '12 at 17:30
    
No, I just keep on forgetting the definition. I had a picture in my head, but the arrows went the wrong way... –  user1729 May 1 '12 at 8:31
    
Oh, I see. Okay. Are you familiar with Bass-Serre theory? If so, it helps to keep in mind the picture of an amalgamated product as the fundamental group of a graph of groups consisting of a single edge (the vertex groups are $A$ and $B$ and the edge group is $C$). –  Tara B May 1 '12 at 8:35
    
I am familiar with Bass-Serre theory, and I know the picture you are talking about...but...I've never really found it useful, and so I tend to just ignore amalgamated free products and pretend they don't exist. I should probably have another crack at "Trees", and actually do some examples this time... –  user1729 May 1 '12 at 8:40

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