Are there conditions on the finite groups $A$, $B$, $C$ or the morphisms $A \leftarrow C\rightarrow B$ that restrict the number of infinite conjugacy classes in the amalgamated product $A \overset{C}{*}B$?
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1$\begingroup$ No, unless your group is either finite or infinite dihedral. $\endgroup$– MishaCommented Sep 25, 2012 at 12:56
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3$\begingroup$ @Misha, not quite. Consider $A$ and $B$ of order 4 and $C$ of order 2. $\endgroup$– Anton KlyachkoCommented Sep 25, 2012 at 14:34
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1 Answer
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If your morphisms $A \leftarrow C \rightarrow B$ are injective (which I'm assuming you want, since you say amalgamated product, not pushout), then the group $A\ast_C B$ is a virtually free group. In particular, there will be infinitely many conjugacy classes of infinite elements if the group is not finite or virtually cyclic.
The group will be finite if one of the morphisms is surjective. Otherwise, the index of both morphisms is $>1$. If both indices are $2$, then there is a surjection to $\mathbb{Z}/2 \leftarrow 1 \rightarrow \mathbb{Z}/2$, and the group is virtually cyclic. In this case, each conjugacy class has finitely many elements.
Otherwise, the group will be virtually free (non-cyclic), and there will be infinitely many conjugacy classes which are infinite. If you'd like more information on this result, have a look at Serre's book Trees.
answered Sep 25, 2012 at 15:02