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Let $G$ be a (countable) group, $H<G$ be a proper subgroup and $\theta:H\to G$ be an injective group homomorphism such that $\theta(H)\neq G$. Consider the HNN extension $\Gamma=<G,t \mid tHt^{-1}=\theta(H)>$

Similarly to link, if $H$ and $\theta(H)$ are malnormal in $\Gamma$, then the centralizer of a nontrivial element in $\Gamma$ is either infinite cyclic or it is in a conjugate of $G$.

Is it sufficient to assume that $H$ and $\theta(H)$ are malnormal in $G$ ? (eventually replacing cyclic by virtually cyclic ?)

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    $\begingroup$ One needs to be careful here. E.g., if $\theta:H \to H$ is the identity, then the $\Gamma$-centralizer of any $h \in H$ is the direct product $C_G(h) \times \langle t \rangle$. Also, if $\theta$ maps any $h \in H$ into $ghg^{-1}$ for some $g \in H$, then the centralizer of $t^{-1}g$ will not be cyclic either. The natural sufficient requirement for the desired property should be that $H^g \cap \theta(H)=\{1\}$ for all $g \in G$. $\endgroup$
    – Ashot Minasyan
    Commented Mar 4, 2014 at 12:44
  • $\begingroup$ Thank you very much for the comment. Actually, I realized that this is implicitly contained in the paper of Karrass and Solitar, namely if $H$ and $\theta(H)$ are malnormal in $G$ and $H^g\cap \theta(H)=1$, for all $g$, then the centralizer of a nontrivial element is as above. But is not clear what happens with the centralizer if a conjugate of $H$ touches $\theta(H)$. $\endgroup$
    – bemihai
    Commented Mar 6, 2014 at 13:52

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