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Hello! Let $G=A\underset{C}\star B$ be an amalgamated Product. Let $a\in A$. If a is conjugated to an Element $b\in B$, then $a$ is conjugated to an Element $c\in C$. The Question is: Why is that true? It is clear, when $a\in A\cap C=C$. It seems to be very easy. But at the moment, i think, i make a fault while im calculating with the elements.

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up vote 3 down vote accepted

There is also a simple geometric proof using Bass-Serre theory, which does not require much in the way of calculation. Let $G$ act on the Bass-Serre tree $T$ of the amalgamated free product $A*_CB$. The stabilizer of every edge of $T$ is conjugate to $C$, and the vertices of the tree $T$ are partitioned into two subsets $V_A$, $V_B$ so that the stabilizer of every vertex in $V_A$ is conjugate to $A$ and the stabilizer of every vertex in $V_B$ is conjugate to $B$. By your assumption, the element $a$ fixes some vertex in $V_A$ and some vertex in $V_B$, so $a$ fixes the edge path between those two vertices, so $a$ fixes some edge along that edge path, so $a$ is conjugate to an element of $C$.

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You can get the answer by Theorem 4.6 in Combinatorial Group Theory by W. Magnus, A. Karrass, and D. Solitar.

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Thanks. Sorry. Wanted to ask this in math.stackexchange. –  Peter May 11 '12 at 15:23
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