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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.

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    $\begingroup$ could you recall the meaning of almost malnormal and weakly malnormal? ($H\le G$ is malnormal if $gHg^{-1}\cap H=\{1\}$ for every $g\in G\smallsetminus H$) $\endgroup$
    – YCor
    Commented Sep 23, 2013 at 17:45
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    $\begingroup$ At least you'll need to replace in the conclusion "the centralizer of $g\in G$ is virtually cyclic" instead of "cyclic": indeed if $H$ has a finite subgroup $F$ that is normal in both $A$ and $B$ then many elements not conjugate to $A\cup B$ will centralize $F$. (Also: in the question: do you mean that $g$ is in a conjugate of a factor, or its centralizer?) $\endgroup$
    – YCor
    Commented Sep 23, 2013 at 18:23
  • $\begingroup$ Thanks for the answer. The conclusion of the Karrass-Solitar result is for the centralizer of $g$ in $G$. $\endgroup$
    – bemihai
    Commented Sep 23, 2013 at 18:39

3 Answers 3

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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.

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  • $\begingroup$ Thanks for the nice argument. I understand that the same result doesn't hold for an almost malnormal amalgam, but can one prove a similar type result, eventually repalcing cyclic by virtually cyclic, as Yves suggested ? More precisely, what can we say about the centralizer of a nontrivial element in an amalgamated free product with almost malnormal amalgam ? $\endgroup$
    – bemihai
    Commented Sep 23, 2013 at 20:22
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    $\begingroup$ @m.b.: What is true (similarly to the 'easy exercise' in my answer) is that if $H$ is almost malnormal on one side then the Bass--Serre tree is 'almost 2-acylindrical', and it follows that the centralizer of any hyperbolic element is virtually cyclic. On the other hand, there may be elliptic elements with very large centralizers, as Yves' answer shows. Another nice counterexample is $SL_2(\mathbb{Z})\cong \mathbb{Z}/4*_{\mathbb{Z}/2}\mathbb{Z}/6$, which of course has a non-trivial centre. $\endgroup$
    – HJRW
    Commented Sep 23, 2013 at 20:30
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    $\begingroup$ @m.b.: On further thought, I think we can perhaps rescue the theorem. In all the counterexamples we've seen, $g$ has finite order. On the other hand, if $g$ is of infinite order, then $\mathrm{Fix}(g)$ is at worst a tree of diameter 2, as in the above answer, and I think the same argument applies. That is: if $H$ is almost malnormal in $A$ and $g$ is of infinite order, then it is indeed the case that $Z(g)$ is either virtually cyclic or conjugate into $A$ or $B$. $\endgroup$
    – HJRW
    Commented Sep 23, 2013 at 21:00
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    $\begingroup$ @ HJRW: Are these results well-known in geometric group theory ? (if yes, can you, please, indicate a reference ?) Thanks again. $\endgroup$
    – bemihai
    Commented Sep 24, 2013 at 8:43
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    $\begingroup$ @m.b.: I don't know a reference, but these kinds of argument are completely standard. This exact statement may or may not appear in the literature. You could try checking the papers that cite the Karrass--Solitar paper. $\endgroup$
    – HJRW
    Commented Sep 24, 2013 at 10:34
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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 $A\cup B$.

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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 :-)

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