MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $C$.

share|cite|improve this answer

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

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.