# Finite generation in amalgamated free products

Let $G = A *_C B$ be an amalgamated free product of groups.

My question is: suppose $C$ and $G$ are finitely generated, can we prove that so is $A$?

I've been trying to prove it by contradiction. Any suggestion?

Thanks...

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Do you assume that $C \to A, C \to B$ are injective? (some authors do this) – Martin Brandenburg Oct 2 '10 at 20:32
I believe many people (including myself) assume $C\to A$ and $C\to B$ are injective when speaking about an amalgamated free product of groups; otherwise they (and I) say G is a pushout. – Lilly Oct 2 '10 at 21:46

## 2 Answers

Let $\{g_i\}$ be a set of generators for $G$ and $\{c_j\}$ a set of generators for $C$. For each $i$, write

$g_i= a_{i,1}b_{i,1}\ldots a_{i,n}b_{i,n}$

as usual, where $a_k\in A$, $b_k\in B$. Clearly, the finite set $\{a_{i,k},b_{i,k}\}$ generates $G$. Now, we shall see that $\{a_{i,k}\}\cup\{c_j\}$ generates $A$.

Any $a\in A$ can be written as a product

$a=\alpha_1\beta_1\ldots\alpha_m\beta_m$ (1)

where $\alpha_l$ is a product of $a_{i,k}^{\pm1}$'s, hence is in $A$, and $\beta_l$ is a product of $b_{i,k}^{\pm1}$'s, hence is in $B$.

An expression like the right hand side of (1) is called reduced if every $\alpha_i\notin C$ for $i>1$ and every $\beta_i\notin C$ for $i< m$. In other words, there's no 'obvious' way to reduce $m$ by combining one of the terms with its neighbouring terms.

The Normal Form Theorem for amalgamated products (as Lilly says below, this is Theorem 1 is Serre's Trees---this is where we use that the maps $C\to A$ and $C\to B$ are injective) asserts that every element has a unique reduced expression (up to the obvious ambiguity that we can insert elements of $C$ in the $A$-terms and cancel them in neighbouring $B$-terms). But (1) gives us two expressions for $a$, and the left hand side is clearly reduced. It follows that the right hand side is not reduced, unless $m=1$ and $\beta_1\in C$.

So if $m>1$ we know that the right hand side is not reduced, which means that some $\alpha_l\in C$ for $l>1$ or some $\beta_l\in C$ for $l< m$. Let's assume that $\beta_l\in C$---the case of $\alpha_l\in C$ is identical. Now we can write $\beta_l$ as a product of $c_j^{\pm1}$'s. We now know that

$\alpha'_l=\alpha_l\beta_l\alpha_{l+1}$ (2)

is a product of $a_{i,k}^{\pm1}$'s and $c_j^{\pm1}$'s. Use (2) to simplify the right hand side of (1), so it becomes

$a=\alpha_1\beta_1\ldots\beta_{l-1}\alpha'_l \beta_{l+1} \ldots \beta_m$ . (3)

The product on the right hand side of (3) has fewer terms than the product on the right hand side of (1), so we can continue by induction until the right hand side is reduced, at which point we have an expression of the form

$a=\alpha_1\beta_1$ (4)

with $\beta_1\in C$ and $\alpha_1$ a product of $a_{i,k}^{\pm1}$'s and $c_j$'s. So we are done.

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First paragraph is clear: G is generated by A and B (actually by their images in G). In particular you write $a$ as an alternating product in G. Since $\alpha_1$ may be 1 and some inverse of $a_{i,k}$ may appear in that expression of $a$, you actually mean $\alpha_l$ is in the group generated by all $a_{i,k}$'s, right? I'm pretty sure you cited Trees' Theorem 1 (p. 3). But now I don't follow you since your expression for $a$ is not in the normal form; when we speak about the normal form we need to choose coset representatives - right? Could you please elaborate more on your last two sentences? – Lilly Oct 4 '10 at 20:22
Regarding coset representatives: that's how Serre phrases it, but you don't have to phrase it like that---instead, you can just say that something's in reduced form if none of the `internal' factors are in $C$. If so, then you can always factor it into a product of coset reps just by pushing elements of $C$ to the left. Does that make sense? Let me know if not, and I'll try to edit the answer to make it clearer. – HJRW Oct 5 '10 at 4:57
Alternatively, there's another way of writing down the proof using van Kampen diagrams. Which would you prefer? – HJRW Oct 5 '10 at 4:58
I do know how to make an alternating product into the normal form. I don't understand why that alternating product of $a$ is reduced, and then I couldn't follow your inductive argument. I would appreciate if you could provide me some details... I don't know anything about van Kampen diagrams, so I'll leave them for latter. – Lilly Oct 6 '10 at 1:18
@Lilly: you can begin by choosing a set of coset representatives for $C$ in $A$, the $a_i$; then a set of coset representatives for $C$ in $B$, the $b_j$. Then each element of $G$ can be written uniquely as an alternating sum of $a_i$ and $b_j$ followed by an element of $c$. Do that for each of the finite set of $g_i$; only the coset reps that show up in these expressions are actually needed for the generation; then throw in the generating set for $C$ and you are done. – Arturo Magidin Oct 6 '10 at 4:10

Suppose f_1:C->A, f_2:C->B and C is generated by a finite set Q. Then

G = < ger(A) , ger(B) | rel(A) , rel(B) , {f_1(q)=f_2(q), q in Q} > .

In this presentation we can eliminate at most finitely many generators of A, namely those coming from C. Therefore if A is not finitely generated then G is not finitely generated.

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No, that just implies this particular presentation of $G$ isn't finite. – user641 Oct 7 '10 at 2:28
This seems a simpler argument but maybe I'm missing something... I understand you are applying Tietze transformations to remove generators, but I don't see that every presentation of G is not finitely generated - this is Steve's point. Does it follow from the standard presentation you gave? – Lilly Oct 7 '10 at 3:07
I'm sorry that doesn't prove anything! But I think I'll leave that. – Max Black Oct 9 '10 at 15:03
No, you have to use the normal form theorem somewhere. Otherwise, you're making the assertion for an arbitrary push-out. I'm pretty certain that's false (though I don't have an example to hand). – HJRW Oct 19 '10 at 15:50