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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 $C'$.

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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 $A*C$ splits non-trivially as an amalgamated free product $A'*{C'} B'$ then either $A$ splits over $C'$ or $C$ is conjugate into $C'$.

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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 $A*C$ splits non-trivially as an amalgamated free product $A'*{C'} B'$ then either $A$ splits over $C'$ or $C$ is conjugate into $C'$.