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Say that a group is rich if it contains isomorphic copies of all finite groups.

It is easy to produce rich groups, and also rich locally finite groups, for instance the restricted direct product $A=\bigoplus_{n\ge 5}\mathrm{Alt}_n$ of alternating groups. Also the group of finitely supported permutations of an infinite set is an example.

This question is a follow-up to this question, which includes the observation that there is no rich group that embeds into every rich group. This question is an analogue, restricting to locally finite groups.

Main question. is there a rich group that embeds into every rich locally finite group?

An equivalent question is:

Does the above group $A$ embed into each rich locally finite group?

The equivalence between the two questions follows because every rich subgroup of $A$ contains a copy of $A$ (I can provide the simple argument upon request).

The specific family of alternating groups is not that important in this discussion, because if $(K_n)$ is a family of finite groups such that each finite group embeds in some $K_n$ and $K=\bigoplus K_n$, then $A$ and $K$ embed into each other.

A positive answer to the following would imply a positive answer to the main question:

Question 2. Let $G$ be a rich locally finite group and $F$ a finite subgroup. Is the centralizer of $F$ rich?

[Edit: Dave Benson proved that the answer is negative, see his answer. So this is not a helpful approach to Question 1.]

For instance, if we restrict to locally finite FC-groups (i.e., locally finite groups in which every finite subgroup has a centralizer of finite index), we obtain that every rich locally finite FC-group contains a copy of $A$.

Regarding Question 2, I could find in the literature a number of papers proving results of the form "if a locally finite group has some finite subgroup with a small centralizer, then it has a somewhat constrained structure (in particular, is not rich)". I don't know whether Question 2 has been addressed in the above form.

Just for context, $\mathrm{PSL}_n$ of an infinite locally finite field is an example of a locally finite group with a finite subgroup with trivial centralizer. But it is not rich.

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  • $\begingroup$ The paper "Subgroups of the Nottingham group" (Rachel Camina, core.ac.uk/download/pdf/82495463.pdf) is an interesting and nonobvious example of related things (in the context of pro-$p$-groups). $\endgroup$
    – Neil Strickland
    Commented Oct 28, 2022 at 11:33
  • $\begingroup$ When you define $A$, you mean it is a product of alternating groups, yes? $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Oct 28, 2022 at 11:50
  • $\begingroup$ @Carl-FredrikNybergBrodda typo fixed, thanks $\endgroup$
    – YCor
    Commented Oct 28, 2022 at 11:52
  • $\begingroup$ It is equivalent to say that for each rich locally finite group $G$, and each finite group $F$, there is an embedding $F \to G$ whose centralizer is rich? (This is sufficient by the same proof as Question 2 implies Question 1 and necessary since every finite subgroup of the group of finitely supported permutations has this property). It is not obvious to me if this is true for $F =\mathbb Z/2$ so maybe that could be a start. $\endgroup$
    – Will Sawin
    Commented Sep 20, 2023 at 18:09

1 Answer 1

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For your question 2, there exists a locally finite group that is rich in your sense, and has a finite subgroup whose centraliser is trivial. If you look at my recent paper "Centralisers of finite groups in locally finite simple groups" https://doi.org/10.1080/00927872.2023.2217255, it gives examples of finite simple groups acting as outer automorphisms of the locally finite version of $SL(\infty,\mathbb{F}_q)$ with trivial fixed points. If you take the semidirect product, this gives a negative answer for your question. As an explicit example, take the semidirect product $SL(\infty,\mathbb{F}_9)\rtimes A_6$, with the subgroup $A_6$, but there are many other examples in my paper.

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  • $\begingroup$ This is nice. A remark: for a group $G$, the existence of a finite subset whose centralizer is reduced to $Z(G)$ can be interpreted as: the closure $\mathrm{Inn}(G)$ is discrete (in $\mathrm{Aut}(G)$, endowed with the Onofri topology of pointwise convergence). For $G$ countable, $\mathrm{Aut}(G)$ is a Polish group. It is an open question whether there exists $G$ infinite countable locally finite such that $\mathrm{Aut}(G)$ is countable. A necessary condition for this is indeed that $\overline{\mathrm{Inn}(G)}$ is discrete, which is already a strong condition. $\endgroup$
    – YCor
    Commented Sep 20, 2023 at 15:25

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