TITLE: Centralizers in amalgamated free products
QUESTION [11 upvotes]: An old result of Karrass and Solitar from 1970 says that if $g$ is a nontrivial element in an amalgamated free product $G=A*_HB$, with $H$ malnormal in $G$, then the centralizer of $g$ in $G$ is either infinite cyclic or it is in a conjugate of a factor. 
Is it sufficient to assume that $H$ is malnormal in $A$ or in $B$ ? Does the same result hold if we assume that $H$ is almost or weakly malnormal in $G$ ?  
$H<G$ is almost malnormal if $gHg^{-1}\cap H$ is finite, for all $g\in G\setminus H$ and $H<G$ is weakly malnormal if there exist $g_1,...,g_n\in G$ such that $\bigcap_{k=1}^{n}{g_k H g_k^{-1}}$ is finite.

REPLY [3 votes]: I wanted to answer this question to give a solid journal citation (as requested in the comments to HJRW's answer), but also I feel there is some historical interest here.
It seems that the result you are looking for is in the next paper of Karrass and Solitar. To paraphrase the question: an old result of Karrass and Solitar from 1970 1971 says that if $g$ is a nontrivial element in an amalgamated free product $G=A*_HB$, with $H$ malnormal in $A$ or $B$, then the centralizer of $g$ in $G$ is either infinite cyclic or it is in a conjugate of a factor.
This follows from Theorem 1 of the paper Karrass, Abraham, and Donald Solitar. "The free product of two groups with a malnormal amalgamated subgroup." Canadian Journal of Mathematics 23.5 (1971): 933-959 (DOI)
Karrass and Solitar actually prove this result in the more general case when the subgroup $C$ is "$r$-step malnormal in $G$". The question here corresponds to being $2$-step malnormal, and indeed being $r$-step malnormal corresponds to being $r$-acylindrical, as in HJRW's answer. So, as they say, there is nothing new under the sun :-)