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This might help.

Lemma If $A$ does not split freely and $C$ is a non-trivial subgroup of $A$ then the HNN extension $G=A*_C$ does not split freely.

The proof uses Bass--Serre theory---see Serre's book Trees.

Proof. Let $T$ be the Bass--Serre tree of a free splitting of $G$. Because $A$ does not split freely, $A$ stabilizes some unique vertex $v$. But $C$ is non-trivial, so $C$ also stabilizes a unique vertex, which must be $v$. Therefore, $G$ stabilizes $v$, which means the free splitting was trivial. QED

A similar argument shows the following.

Lemma If $A*C$ A *_C $splits non-trivially as an amalgamated free product$ A'*{C'} A' *_{C'} B'$then either$A$splits over$C'$or$C$is conjugate into$C'$. 2 Tried to fix broken LaTeX. This might help. Lemma If$A$does not split freely and$C$is a non-trivial subgroup of$A$then the HNN extension$G=A*_C$does not split freely. The proof uses Bass--Serre theory---see Serre's book Trees. Proof. Let$T$be the Bass--Serre tree of a free splitting of$G$. Because$A$does not split freely,$A$stabilizes some unique vertex$v$. But$C$is non-trivial, so$C$also stabilizes a unique vertex, which must be$v$. Therefore,$G$stabilizes$v$, which means the free splitting was trivial. QED A similar argument shows the following. Lemma: If$A*C$splits non-trivially as an amalgamated free product$A'*{C'} B'$then either$A$splits over$C'$or$C$is conjugate into$C'$. 1 This might help. Lemma If$A$does not split freely and$C$is a non-trivial subgroup of$A$then the HNN extension$G=A*_C$does not split freely. The proof uses Bass--Serre theory---see Serre's book Trees. Proof. Let$T$be the Bass--Serre tree of a free splitting of$G$. Because$A$does not split freely,$A$stabilizes some unique vertex$v$. But$C$is non-trivial, so$C$also stabilizes a unique vertex, which must be$v$. Therefore,$G$stabilizes$v$, which means the free splitting was trivial. QED A similar argument shows the following. Lemma: If$A*C$splits non-trivially as an amalgamated free product$A'*{C'} B'$then either$A$splits over$C'$or$C$is conjugate into$C'\$.