Math-1211 · mathematics · Group theory

Math-1211

Kourovka Notebook Problem 21.39

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Statement

Definitions and mathematical statement

A group $G$ is locally finite if every subgroup generated by finitely many elements is finite. A subgroup $C\le G$ is characteristic if $\alpha(C)=C$ for every automorphism $\alpha$ of $G$. The group is characteristically simple if it is nontrivial and its only characteristic subgroups are $1$ and $G$; proper normal subgroups are not excluded unless they are characteristic.

The automorphism group acts on the underlying set of $G$ by evaluation. Having finitely many automorphism orbits means that there exist $g_1,\ldots,g_t\in G$ such that every $g\in G$ has the form $\alpha(g_i)$ for some $i$ and some $\alpha\in\operatorname{Aut}(G)$. These are orbits of all abstract automorphisms, not only conjugation orbits.

A group is residually finite if each nonidentity $g$ survives in some finite quotient: there is a homomorphism $\pi:G\to Q$ to a finite group with $\pi(g)\ne1$. The question asks whether there exists a locally finite, characteristically simple group with finitely many automorphism orbits which is not residually finite.

For a proposed witness the last condition means that some fixed $g\ne1$ is killed by every homomorphism into every finite group. This must coexist with finiteness of each finitely generated subgroup; local finiteness alone does not specify quotient separation. No countability, finite generation, or bound on the sizes of the finite subgroups is assumed. The requested example must be infinite, since any finite group separates its elements through its identity map.

Short English statement

Can a locally finite characteristically simple group have finitely many automorphism orbits while failing residual finiteness?

Sources

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