Free subgroup of an amalgamated free product I have the following question:
Let $G=A\underset{C}\star B$ be the free product of two groups $A$ and $B$ with amalgam $C$, such that $C\cap vCv^{-1}=1$ for all reduced words $v$ in G with length $\geq k$. Let $w\in G$ be a reduced word of length $\geq 2k+1$.
I want to show that $\langle A,wAw^{-1} \rangle\cong A\star A$. 
How can i show that $A\cap wAw^{-1}=1$?
Thanks for help!
 A: Think of this problem in the Bass-Serre tree $T$ for the amalgam. The group $G$ acts on $T$, and the tree $T$ has a vertex $V_A$ with stabilizer subgroup $A$, a vertex $V_B$ with stabilizer $B$, and an oriented edge $E = V_A \rightarrow V_B$ with stabilizer $C$, so that every vertex is in the $G$-orbit of either $V_A$ or $V_B$ and every edge is in the orbit of $E$. 
I am going to make some assumptions about your reduced word $w$ just to focus on a special case: $w=w_1 \ldots w_{2l+1}$ has odd length $2l+1$; and the first and last letters $w_1, w_{2l+1}$ are in $B$.  (A few other very similar special cases are needed for the complete proof.)
With these assumptions, the path in the tree $T$ from vertex $V_A$ to vertex $w \cdot V_A$ has length $2l+2$ and passes through the following sequence of vertices: 
$$V_A \rightarrow V_B \leftarrow w_1 \cdot V_A \rightarrow w_1 w_2 \cdot V_B \leftarrow w_1 w_2 w_3 \cdot V_A \rightarrow w_1 w_2 w_3 w_4 \cdot V_B \leftarrow \ldots 
$$
$$ \ldots \rightarrow w_1 w_2 \cdots w_{2L-1} w_{2L} \cdot V_B \leftarrow w_1 w_2 \cdots w_{2l-1} w_{2l} w_{2l+1} \cdot V_A = w \cdot V_A
$$
Furthermore, the directed edges represented by the arrows in this sequence are
$$E, w_1 \cdot E, w_1 w_2 \cdot E, w_1 w_2 w_3 \cdot E, \ldots, w_1 w_2 \cdots w_{2L-1} w_{2l} E, w_1 w_2 \cdots w_{2l-1} w_{2l} w_{2l+1} E
$$
Suppose now that $A \cap w A w^{-1}$ contains a nontrivial element $g \in G$. The stabilizer of $w \cdot V_A$ equals $w A w^{-1}$, so $g$ fixes the two vertices $V_A$, $w \cdot V_A$, which are the endpoints of the path. Since $T$ is a tree, $g$ must fix every point along that path, in particular every vertex and every oriented edge along that path. So $g$ fixes both of the edges $E$ and $w_1 w_2 \cdots w_k \cdot E$. Set $v=w_1w_2\cdots w_k$. The stabilizer of $vE$ is $vCv^{-1}$, so $g \in C \cap vCv^{-1}$, contradiction.
By the way, it looks to me like this proof does not use the full power of your hypothesized inequality $2l+1 \ge 2k+1$, it only uses $2l+1 \ge k$. I presume you need the full power of the inequality elsewhere in the proof.
