Math-1200 · mathematics · Group theory

Math-1200

Kourovka Notebook Problem 21.20

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Statement

Definitions and mathematical statement

All groups and actions are definable in a common sufficiently saturated first-order structure, and parameters are allowed. The Morley rank of a nonempty definable set is at least zero; rank at least $k+1$ means that there are infinitely many disjoint definable subsets of rank at least $k$. Finite Morley rank means this dimension is finite. A definable group is connected if it has no proper definable subgroup of finite index.

Let $M$ be a nontrivial simple group of finite Morley rank with no involutions. Here simple means no nontrivial proper normal subgroup, and an involution is an element of order exactly two. Suppose $M$ acts by automorphisms on a connected definable group $S$ of finite Morley rank. The action map $M\times S\to S$ must be definable. It is faithful when its kernel is $1$, and irreducible when there is no definable subgroup $1<H<S$ invariant under every element of $M$.

Prove that every such faithful irreducible action is equivalent to $M$ acting on itself by conjugation. With a left-action convention, this means that there is a group isomorphism $\eta:S\to M$ such that \[ \eta(m\cdot s)=m\eta(s)m^{-1} \qquad(m\in M,\ s\in S). \] Allowing an automorphism of the acting copy of $M$ gives the same notion of equivalence after adjusting $\eta$. No nilpotence condition on the proper subgroups of $M$ is assumed. Connectedness is required of the target, while its simplicity is part of the proposed conclusion. Irreducibility excludes even disconnected definable invariant subgroups, not only connected ones.

Short English statement

Is conjugation the only faithful irreducible definable action of an involution-free simple group of finite Morley rank on a connected group?

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