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...
 A: 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.
A: 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.
