# Is there a highly transitive action of a finitely generated torsion simple group?

Is there a highly transitive action of a finitely generated torsion simple group $G$ on $\mathbb{Z}$ ?

Highly transitive means $k$-transitive for each $k \in \mathbb{N}$, that is: for every two $k$-tuples $(x_1, \dots x_k) , (y_1, \dots, y_k) \in \mathbb{Z}^k$ of pairwise distinct elements, there is some $g \in G$ such that $g(x_i) = y_i$ for each $1 \leq i \leq k$.

• Without "simple" the answer is yes: take any infinite f.g. torsion group and consider the subgroup of permutations of itself generated by left translations and finitely supported permutations: it's finitely generated, torsion, and highly transitive. But this is of course never simple. – YCor Jan 2 '15 at 12:21
• The answer is also yes if you remove "finitely generated", because you can let $G$ be the set of all even permutations of $\mathbb{N}$ that move only finitely many elements. – Jeremy Rouse Jan 2 '15 at 16:16
• @Pablo: yes, Thompson's group $V$ of the dyadic Cantor set is a f.p. simple group and admits such an action (on $\mathbf{Z}[1/2]/\mathbf{Z}$). – YCor Jan 2 '15 at 16:28
• $V$ is not a torsion group so it doesn't answer the main question. – YCor Jan 2 '15 at 17:37
• @YCor In fact I hope to be able to show that groups which admit some sort of interesting actions on sets will have nice properties like residual finiteness, and in particular will not be simple. What I really have is: Let $G$ be a finitely generated group acting faithfully and highly transitively on $\mathbb{Z}$ such that the orbit of each $z \in \mathbb{Z}$ under every finitely generated subgroup of infinite index in $G$ is finite. I want to be able to show that $G$ is at least residually finite, but I am not even able to show that it is not simple. – Pablo Jan 2 '15 at 23:08

## 2 Answers

Yes, such groups exist. Simple groups in my paper "Palindromic subshifts and simple periodic groups of intermediate growth" are like that.

• As mentioned in your paper, N. Matte Bon showed that the Grigorchuk group embeds in a full topological group, yielding many actions of it on $\mathbf{Z}$. Do you know if any of these actions is $\infty$-transitive? – YCor Nov 2 '16 at 20:56
• – YCor Nov 2 '16 at 23:06
• The Grigorchuk group can not be highly transitive, since it is a 2-group, so in particular, can not act as a cycle on a set of cardinality 3. – Volodymyr Nekrashevych Nov 3 '16 at 1:12

As you expressed interest in Thompson's group V in the comments -- the following is a highly transitive faithful permutation representation of V on $\mathbb{Z}$ (although as Ives de Cornulier has already mentioned this group is not a torsion group):

Let $r(m)$ denote the residue class $r+m\mathbb{Z}$, where $0 \leq r < m$. Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$, let the class transposition $\tau_{r_1(m_1),r_2(m_2)}$ be the permutation of $\mathbb{Z}$ which interchanges $r_1+km_1$ and $r_2+km_2$ for every $k \in \mathbb{Z}$ and which fixes everything else.

Then we have $V \cong \langle \kappa, \lambda, \mu, \nu \rangle$, where $\kappa := \tau_{0(2),1(2)}$, $\lambda := \tau_{1(2),2(4)}$, $\mu := \tau_{0(2),1(4)}$ and $\nu := \tau_{1(4),2(4)}$ satisfy the defining relations

• $\kappa^2 = \lambda^2 = \mu^2 = \nu^2 = 1$,

• $\lambda\kappa\mu\kappa\lambda\nu\kappa\nu\mu\kappa\lambda\kappa\mu = 1$,

• $\kappa\nu\lambda\kappa\mu\nu\kappa\lambda\nu\mu\nu\lambda\nu\mu = 1$,

• $(\lambda\kappa\mu\kappa\lambda\nu)^3 = (\mu\kappa\lambda\kappa\mu\nu)^3 = 1$,

• $(\lambda\nu\mu)^2\kappa(\mu\nu\lambda)^2\kappa = 1$,

• $(\lambda\nu\mu\nu)^5 = 1$,

• $(\lambda\kappa\nu\kappa\lambda\nu)^3\kappa\nu\kappa(\mu\kappa\nu\kappa\mu\nu)^3 \kappa\nu\kappa\nu = 1$,

• $((\lambda\kappa\mu\nu)^2(\mu\kappa\lambda\nu)^2)^3 = 1$,

• $(\lambda\nu\lambda\kappa\mu\kappa\mu\nu\lambda\nu\mu\kappa\mu\kappa)^4 = 1$,

• $(\mu\nu\mu\kappa\lambda\kappa\lambda\nu\mu\nu\lambda\kappa\lambda\kappa)^4 = 1$,

• $(\lambda\mu\kappa\lambda\kappa\mu\lambda\kappa\nu\kappa)^2 = 1$, and

• $(\mu\lambda\kappa\mu\kappa\lambda\mu\kappa\nu\kappa)^2 = 1$

given in:

Graham Higman, Finitely presented infinite simple groups, Notes on Pure Mathematics, Department of Pure Mathematics, Australian National University, Canberra, 1974. MR0376874.

• Can you think of a highly transitive action of a finitely generated simple group $G$ with the orbit of every $z \in \mathbb{Z}$ under the action of each $g \in G$ finite? If $G$ is torsion then the last condition is automatic. – Pablo Jan 2 '15 at 22:53