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Let G be an infinite group wich is finitely generated.

Is that true that the size of all finite conjugacy classes is bounded?

What I know. If G is a finitely generated FC-group then it's true (follows from this). But if G isn't FC- or FZ-group, so it's center has an infinite index, then I don't see any problems to have conjugacy classes which sizes tends to infinity. But also I don't see any examples of it :-)

On the other hand Osin and Hull proved (see here) that there exists groups with finite number of conjugacy classes and that number of conjugacy classe may obey to almost any function. So it looks like an argument for existence a group which finite conjugacy classes are not bounded.

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2 Answers 2

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The answer is no: there exists a 2-generated group, having finite conjugacy classes of unbounded size.

Indeed B.H. Neumann (1937) produced a 2-generated group $G$ with normal subgroups $(H_n)_{n\ge 5}$ such that $H_n\simeq \mathrm{Alt}_n$. Since $H_n$ has a conjugacy class of size growing to infinity (say, the set of 3-cycles), so does $G$ (its conjugacy class in $G$ is contained in $H_n$, hence is finite too — actually it's unchanged in $G$, if one looks at the construction).

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  • $\begingroup$ I see the paper ''B. H. Neumann, Some remarks on infinite groups, Journ. London Math. Soc. 12 (1937) 120-127.'' but I don't see the theorem you are talking about. Can you please help with the reference? $\endgroup$
    – Andronick Arutyunov
    Commented Jun 18, 2022 at 14:38
  • $\begingroup$ @AndronickArutyunov this is the construction in the proof of Theorem 14 of this reference (well, to be precise, I should have said $(H_n)$ is indexed by odd numbers $\ge 5$ only, although it's not a real issue). $\endgroup$
    – YCor
    Commented Jun 18, 2022 at 14:45
  • $\begingroup$ got it! Thank you, I have to think about that construction. $\endgroup$
    – Andronick Arutyunov
    Commented Jun 18, 2022 at 14:54
  • $\begingroup$ @YCor. If I remember well, by another theorem of Neumann, if $G$ is a finitely generated FC-group then the center of $G$ has finite index. Therefore Schur's Theorem forces that the derived subgroup of $G$ is finite, and so the size of the conjugacy class of $G$ is bounded. Am I missing anything? $\endgroup$
    – Salvatore Siciliano
    Commented Jun 18, 2022 at 14:59
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    $\begingroup$ @SalvatoreSiciliano Yes, my example is not an FC-group and as already mentioned, by the (immediate) fact you're mentioning, no f.g. FC-group yields an example. There's no requirement in the OP's question that all conjugacy classes are finite. $\endgroup$
    – YCor
    Commented Jun 18, 2022 at 15:01
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The size of all conjugacy class of a group $G$ is bounded if and only if the derived subgroup $G^\prime$ of $G$ is finite.

This is a celebrated theorem of Neumann proved in the following paper:

B.H. Neumann: "Groups covered by permutable subsets", J. London Math. Soc. bf 29 (1954), 236–248.

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  • $\begingroup$ If G' isn't finite then there exist infinite conjugacy classes, sure. But I'm interested in a liitle bit other case. Group has infinite conjugacy classes. Is that true that the size of all finite (just finite!) classes is bounded? $\endgroup$
    – Andronick Arutyunov
    Commented Jun 18, 2022 at 14:19

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