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.



Centralizers in amalgamated free products
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.


