Centralizers in amalgamated free products 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.
 A: The answer [edit: of the second question about weakly/almost malnormal in $G$] is "no" in both cases and you need to reformulate the question as I suggested in the comments. If $F$ is a nontrivial finite group, $U,V$ are nontrivial groups and $G$ is the amalgam of $A=U\times F$ and $B=V\times F$ over $H=F$, then $F$ being finite, it is obviously almost and weakly malnormal. However, if $u$ and $v$ are nontrivial elements of $U$ and $V$, then the centralizer of $uv$ contains $F$, so is not infinite cyclic, and is not conjugate inside $A\cup B$.  
A: 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 :-)
A: As Yves points out, the answer is 'no' if you replace 'malnormal' by 'almost malnormal'.  However, the answer to your first question is 'yes': it is the case that the Karrass--Solitar result remains true if you only assume that $H$ is malnormal in $A$.
The action of a group $G$ on a tree $T$ is called $k$-acylindrical if, for every $g\neq 1$, the diameter of the fixed-point set of $g$ is at most $k$.  If $H$ is malnormal in both $A$ and $B$ then the action on the Bass--Serre tree is 1-acylindrical; if it's just malnormal in $A$, the action is 2-acylindrical.
Easy exercise: If $T$ is $k$-acylindrical for some $k$ and $g$ acts hyperbolically, then the centralizer of $g$ is infinite cyclic.
So, after conjugating, we're left with the case in which $g$ is in $A$ or $B$.   Note that the centralizer $Z(g)$ maps $\mathrm{Fix}(g)$ to itself.  If $g\in A$ and is not conjugate into $H$ then $\mathrm{Fix}(g)$ is a single vertex stabilized by $A$, so $Z(g)\subseteq A$ as required.  On the other hand, if $g\in B$ then $\mathrm{Fix}(g)$ is contained in a tree of diameter at most 2, with the central vertex stabilized by $B$.  Every automorphism of $\mathrm{Fix}(g)$ with no edge inversions fixes the central vertex, and it follows that $Z(g)$ is indeed contained in $B$, as required.
