Highly transitive actions of \(\mathrm{Out}(F_n)\). (Q355381)

Let \(G\) be a group and \(F_n=\langle x_1,x_2,\ldots,x_n\rangle\) be the free group on the free generators \(x_1,x_2,\ldots,x_n\). There is a natural identification of \(G^n\) with \(\Hom(F_n,G)\) and the set of epimorphisms \(\mathrm{Epi}(F_n,G)\) with the set \(V_n(G)=\{\underline g\in G^n:\langle\underline g\rangle=G\}\), of all generating \(n\)-tuples.NEWLINENEWLINE The group \(\Aut(G)\) acts on \(\Hom(F_n,G)\) from the left and \(\Aut(F_n)\) acts on \(\Hom(F_n,G)\) from the right.NEWLINENEWLINE Let \(\overline V_n(G)\) be the set of all \(\Aut(G)\)-orbits on \(V_n(G)\). Since the \(\Aut(F_n)\) action preserves \(V_n(G)\) and commutes with the \(\Aut(G)\) action, an action of \(\Gamma=\mathrm{Out}(F_n)=\Aut(F_n)/\mathrm{Inn}(F_n)\) on \(\overline V_n(G)\) is induced.NEWLINENEWLINE An action of a group on a set is called \(k\)-transitive if it is transitive on ordered \(k\)-tuples of distinct points. It is called highly transitive if it is \(k\)-transitive for every \(k\in\mathbb N\).NEWLINENEWLINE A Tarski monster group is a noncyclic group \(G\) all of whose subgroups are cyclic. The existence of Tarski monster groups that do not admit a law was established by \textit{A. Yu. Ol'shanskiĭ} et al., [see Geom. Topol. 13, No. 4, 2051-2140 (2009; Zbl 1243.20056)].NEWLINENEWLINE Now we can state the authors' main results in this paper.NEWLINENEWLINE Theorem. Let \(G\) be a Tarski Monster and \(n\geq 4\) then the action of \(\Gamma=\mathrm{Out}(F_n)\) on \(\overline V_n(G)\) is highly transitive. Moreover, this action is faithful if and only if \(G\) does not satisfy a group law.NEWLINENEWLINE To prove that the action is highly transitive, the authors argue by induction on \(k\) and for the basic step they prove theNEWLINENEWLINE Proposition. For every \(n\geq 3\), \(\Aut(F_n)\) acts transitively on \(V_n(G)\), where \(G\) is a Tarski monster group.NEWLINENEWLINE Since the action \(\Aut(F_n)\curvearrowright V_n(G)\) is transitive, so is the quotient action \(\mathrm{Out}(F_n)\curvearrowright\overline V_n(G)\).NEWLINENEWLINE For the faithfulness of the action they proveNEWLINENEWLINE Lemma. For every \(n\geq 3\), \(\mathrm{Out}(F_n)\) acts faithfully on \(\overline R_n(F_{n-1})\).NEWLINENEWLINE Where, for a group \(G\) with \(n>d(G)\) (\(d(G)\) denotes the minimal number of generators of \(G\)), \(R_n(G)=\{\varphi\in V_n(G)\mid\langle\varphi(x_1),\varphi(x_2),\ldots,\varphi(x_{n-1})\rangle=G\) for some basis \(\{x_1,x_2,\ldots,x_n\}\) of \(F_n\}\) and \(\overline R_n(G)<\overline V_n(G)\) is the image of this (invariant) set, modulo \(\Aut(G)\).NEWLINENEWLINE Proposition. Let \(G\) be a Tarski Monster and \(n\geq 3\). Then the action of \(\mathrm{Out}(F_n)\) on \(\overline V_n(G)\) is faithful if and only if \(G\) satisfies no group law.NEWLINENEWLINE The authors notify that the direction \textit{\dots the action is faithful implies that the group satisfies no group law} is valid in general. From this they rise theNEWLINENEWLINE Theorem. For any general group \(G\), the following are equivalent:NEWLINENEWLINE 1. The action of \(\mathrm{Out}(F_n)\) on the \(\Aut(G)\)-classes of \(\Hom(F_n,G)\) is faithful for all large enough \(n\).NEWLINENEWLINE 2. The group \(G\) does not satisfy a group law.NEWLINENEWLINE The paper concludes with some remarks and some questions. For example. What about \(\mathrm{Out}(F_2)\) and \(\mathrm{Out}(F_3)\)? Do they admit a highly transitive action on a set?