---
abstract: 'We construct examples of countable linear groups $\Gamma < {{\operatorname{SL}}}_n({{\mathbf{R}}})$ with no non-trivial normal abelian subgroup that admit a faithful, sharply $2$-transitive action on a set. The stabilizer of a point in the this action does not contain an involution.'
author:
- 'Yair Glasner and Dennis D. Gulko'
bibliography:
- 'yair.bib'
title: |
    Non-split linear sharply 2-transitive groups,\
    after Rips-Segev-Tent
---

Introduction
============

A sharply $2$-transitive group is, by definition, a permutation group $\Gamma \curvearrowright X$ which acts transitively and freely on ordered pairs of distinct points. Such a group is called [*[split]{}*]{} if it admits a non-trivial normal abelian subgroup. The following question, that has attracted the attention of algebraists for many years, was recently answered negatively, by Rips Segev and Tent in [@RST:s2t]

\[q:main\] Is every sharply $2$-transitive group split?

In Theorem \[thm:s2t\] we show that the answer remains negative even in the setting of countable linear groups. This contrasts nicely with the prior results of [@GlGu; @GMS:s2t] that show that the answer to the same question is positive for linear groups when the permutational characteristic of $\Gamma$ is not $2$; or in other words under the additional assumption that involutions in $\Gamma$ fix a point.

Splitting implies a tame, algebraic, structure theory. In particular with every split sharply $2$-transitive group $\Gamma$ one can associate a near field $N$, which is by definition a division ring that is distributive only from the right[^1]. In this case $\Gamma$ indeed [*[splits]{}*]{} as a semi-direct product $\Gamma= N^* \ltimes N$ of the multiplicative and additive groups of the near field. Moreover the given sharply $2$-transitive action is the unique faithful primitive permutation representation of this group and it is isomorphic to the natural action $N^* \ltimes N \curvearrowright N$ by affine transformation $x^{(a,b)} = x\cdot a+b$.

When $\Gamma \curvearrowright X$ is sharply $2$-transitive there exists an element flipping any two points in $X$, whose square must be trivial. This gives rise to a large set of involutions ${{\operatorname{Inv}}}(\Gamma) \subset \Gamma$. Such a pair of points determines the involution and since $\Gamma$ is transitive on pairs of points all involutions are conjugate. Any nontrivial element, and in particular any involution, can have either $0$ or $1$ fixed points. If an involution does fix a point then the map ${{\operatorname{Inv}}}(\Gamma) {\rightarrow}X$ taking an involution to its fixed point is a $\Gamma$ invariant bijection. Consequently the $\Gamma$ action on ${{\operatorname{Inv}}}(\Gamma)$ by conjugation is $2$-transitive and in particular, the order of the product of two different involutions is independent of the choice of the specific involutions. This gives rise to the following definition:

Let $\Gamma$ be a sharply $2$-transitive permutation group on $X$. If the stabilizer of a point contains an involution let $p = {{\operatorname{Ord}}}(\sigma \tau)$ be the order of the product of two distinct involutions. We define the [*[permutational characteristic of $\Gamma$]{}*]{} to be $${{{\it {p}}\operatorname{-char}}}(\Gamma)=\left\{ \begin{array}{ c l }
    2\hspace{8pt} & {\text{Involutions do not fix a point}}\\
    p \hspace{8pt} & p < \infty \\
    0\hspace{8pt} & p =\infty
\end{array}\right.$$

It is not difficult to verify that ${{{\it {p}}\operatorname{-char}}}(\Gamma)$ is either $0$ or prime, and that it coincides with the characteristic of the near field whenever $\Gamma$ splits. We refer the readers to [@Kerby:mst], [@GlGu] for more details.

In two papers [@Z1:finite_s2t; @Z2:finite_s2t] from 1936 H. Zassenhaus completed a full classification of finite sharply $2$-transitive groups. He started by showing that every such group splits, and then gave a complete classification of finite near fields. Contrary to the situation for skew fields, the latter classification involves non-trivial examples of finite near fields. In [@T1:cont_s2t; @T2:cont_s2t] Tits proved that every locally compact connected sharply $2$-transitive group splits. The near fields here are just ${{\mathbf{R}}},{{\mathbf{C}}}$ or $\mathbf{H}$. In [@Turk:s2t_char3] Türkelli proved that every sharply $2$-transitive group $\Gamma$ with ${{{\it {p}}\operatorname{-char}}}(\Gamma)=3$ splits. In [@GlGu; @GMS:s2t] it was shown that every linear sharply $2$-transitive group $\Gamma < {{\operatorname{GL}}}_n(k)$ with ${{{\it {p}}\operatorname{-char}}}(\Gamma) \ne 2$ splits. This was shown by us under the additional assumption that $\operatorname{char}(k) \ne 2$, an assumption that was deemed unnecessary by [@GMS:s2t] who also relaxed the linearity assumption.

Let $\Delta \curvearrowright Y$, $\Gamma \curvearrowright X$ be two actions of groups on sets. An [*[embedding]{}*]{} of such actions $(\phi,\iota): (\Delta \curvearrowright Y) {\hookrightarrow}(\Gamma \curvearrowright X)$ is a monomorphism $\phi: \Delta {\rightarrow}\Gamma$ and a $\phi$-equivariant injective map $\iota: Y {\rightarrow}X$.

Such an embedding is obtained, in particular, whenever $\Delta < \Gamma$ is a subgroup and there is a $\Delta$-orbit in $Y$ that is permutation isomorphic to the given action of $\Delta$ on $X$. At the group theoretic level if the two actions are identified with coset actions $Y = A {\backslash}\Delta, X = B {\backslash}\Gamma$ then we are just looking at an embedding $\phi: \Delta < \Gamma$ such that $A = \Delta \cap B$.

Recently the first examples of non-split sharply $2$-transitive groups were given by Rips-Segev-Tent in [@RST:s2t]. Their examples are very general in the sense that they show that every transitive group action on a set $G \curvearrowright X$ with the properties that the stabilizer of a point does not contain an involution and the stabilizer of every pair of distinct points is trivial can be embedded into a non-split sharply $2$-transitive action of permutational characteristic $2$. Clearly these two conditions are also necessary for such an embedding.

Our main theorem comes to show that such non-split sharply $2$-transitive groups of permutational characteristic $2$ can be constructed even within the realm of linear groups. Our main theorem is:

\[thm:s2t\] (Linear s-2-t groups) Let $H < {{\operatorname{SL}}}_n({{\mathbf{R}}})$ be a countable group that contains neither involutions nor nontrivial scalar matrices. Assume that $H \curvearrowright X$ is a transitive permutation action with the property that the stabilizer of every pair of distinct points is trivial. Then there exists a larger countable group $H < H_1 < {{\operatorname{SL}}}_n({{\mathbf{R}}})$ which admits a sharply $2$-transitive non-split permutation representation $H_1 \curvearrowright X_1$ of permutational characteristic $2$ and an embedding: $$(H \curvearrowright X) {\hookrightarrow}(H_1 \curvearrowright X_1).$$

In some examples the resulting group $H_1$ admits uncountably many non-isomorphic faithful primitive permutation representations $H_1 \stackrel{\rho_{\alpha}}{\curvearrowright} X_{\alpha}$.

The last statement of the theorem should be contrasted with the well known fact that a split sharply $2$-transitive group admits a unique (up to isomorphism of permutations representations) faithful primitive permutation representation. Thus the nonsplit examples of [@RST:s2t] are the first natural candidates for sharply $2$-transitive groups that admit multiple faithful primitive actions. In order to actually construct such examples linearity comes in handy. We appeal to the results of [@GG:AOS] which ensure that any Zariski dense countable subgroup of ${{\operatorname{SL}}}_n({{\mathbf{R}}})$ with trivial center admits uncountably many non-isomorphic faithful primitive actions. But linearity is not essential here. Using the methods of [@RST:s2t], it is possible to construct groups that admit many faithful non-isomorphic sharply $2$-transitive actions! This will be shown in subsequent paper.

Reductions
==========

Given an involution $t \in {{\operatorname{SL}}}_n({{\mathbf{R}}})$ let ${{\mathbf{R}}}^n = W^+(t) \oplus W^-(t)$ denote the decomposition of ${{\mathbf{R}}}^n$ into its $\pm 1$ eigenspaces and ${\pi}_t = \frac{1 + t}{2} :{{\mathbf{R}}}^n {\rightarrow}W^{+}(t)$ the projection on $W^{+}(t)$ along $W^{-}(t)$. When possible, we will omit $t$ from the notation writing $W^{\pm}$ instead of $W^{\pm}(t)$. Our main technical theorem is the following, linear version of [@RST:s2t Theorem 1.1]:

\[thm:tec\] Let $G < {{\operatorname{SL}}}_n({{\mathbf{R}}})$ be a countable group, $t \in G$ an involution and $A < G$ a malnormal subgroup containing no involutions. Assume further that all the involutions in $G$ are conjugate (in $G$), that $\dim(W^{+}(t)) \ge \dim(W^{-}(t))+2$, and whenever $W^{+}(t)$ is contained in an eigenspace of ${\pi}_t \circ g$ for some $g \in G$ then either $g =1$ or $g =t$.

Then for any element $v \in G \setminus A$ there exists a countable extension $G \le G_1 < {{\operatorname{SL}}}_n({{\mathbf{R}}})$ with a malnormal subgroup $A_1 < G_1$ containing no involutions, such that $A_1 \cap G = A$ and an element $f \in A_1$ such that $A_1 t f = A_1 v$. Moreover all the involutions in $G_1$ are conjugate and the only elements of $G_1$ such that $W^{+}(t)$ is contained in an eigenspace of ${\pi}_t \circ g$ are $g =1$ and $g =t$.

The significant difference between this theorem and [@RST:s2t Theorem 1.1] is the linearity requirement $G,G_1 < {{\operatorname{SL}}}_n({{\mathbf{R}}})$. As we plan to use this theorem infinitely many times within an induction argument, it is particularly important for us that $G_1$ is realized as a linear group within the same ambient matrix group as the original group $G$. A few “auxiliary requirements” become necessary in order to prove the theorem in the linear setting: That $G$ be countable, with conjugate involutions and that if $g \not \in \{1,t\}$ then $W^+(t)$ is not contained in an eigenspace of ${\pi}_t \circ g$.

In the proof we follow the strategy of [@RST:s2t]. For the convenience of the reader we quote here three of their propositions. The first proposition reduces the main theorem to two special cases.

\[prop:hyp\] It is enough to prove Theorem \[thm:tec\] under the additional assumption that $v,v^{-1} \not \in AtA$ and either $v^{-1} \not \in AvA$ or $v$ is an involution.

See [@RST:s2t Section 2], the exact same reduction works for us here.

The next two theorems treat these two cases respectively:

\[thm:RST\_free\] ([@RST:s2t Theorem 3.1]) Let $G$ be a group, $A < G$ a malnormal subgroup containing no involutions, $t,v \in G$ two elements with $t$ an involution and $v,v^{-1} \not \in AtA$ and $v^{-1} \not \in AvA$. If $A$ already contains an element $f$ such that $Atf=Av$ then take $G_1 =G, A_1 =A, f=f$. Otherwise set $$G_1 := G * \langle f \rangle, \qquad  A_1 = \langle A, f, tfv^{-1} \rangle.$$ Where $\langle f \rangle = {{\mathbf{Z}}}$.

Then $A_1$ is malnormal in $G_1$, $A_1tf=A_1v$ and $G \cap A_1 = A$.

\[thm:RST\_HNN\] [@RST:s2t Theorem 4.1] Let $G$ be a group, $A < G$ a malnormal subgroup containing no involutions, $t,v \in G$ two involutions such that $v,v^{-1} \not \in AtA$. If $A$ already contains an element $f$ such that $Atf=Av$ then take $G_1 =G, A_1 =A, f=f$. Otherwise set $$G_1= \langle G,f \ | \ f^{-1}tf=v \rangle, \qquad  A_1 := \langle A, f \rangle.$$

Then $A_1$ is malnormal in $G_1$, $A_1tf = A_1 v$ and $G \cap A_1 = A$.

The contribution of the current paper is in showing that in the setting of the above two theorems; when the group $G < {{\operatorname{SL}}}_n({{\mathbf{R}}})$ is countable linear, and subject to all of the auxiliary conditions described in Theorem \[thm:tec\], then $G_1$ is still linear. In fact $G_1$ is still contained in the same ambient matrix group ${{\operatorname{SL}}}_n({{\mathbf{R}}})$. This is achieved in Theorems \[thm:free\_prod\] and \[thm:HNN\] appearing in Section \[sec:linearity\]. In turn this observation is enough to carry out the full Rips-Segev-Tent construction inside the ambient matrix group.

Some projective dynamics
========================

All our notation is taken from [@BG:Dense_Free Section 3] and we refer the readers to that paper for more details.

We use extensively the action $SL_n({{\mathbf{R}}}) \curvearrowright {{\mathbb{P}}}:= {{\mathbf{R}}}^n/R^{*}$ on the projective space. If $0 \ne v \in {{\mathbf{R}}}^n, \langle 0 \rangle \ne W < {{\mathbf{R}}}^n$ are a nontrivial vector and subspace in ${{\mathbf{R}}}^n$ we denote the corresponding projective point, and subspace by $[v] \in {{\mathbb{P}}}, [W] < {{\mathbb{P}}}$. Fix a norm ${\left\Vert\cdot\right\Vert}$ on ${{\mathbf{R}}}^n$, this gives rise in to a norm on the exterior product ${{\mathbf{R}}}^n \wedge {{\mathbf{R}}}^n$ which is used in turn to endow the projective space ${{\mathbb{P}}}= {{\mathbf{R}}}^{n}/R^{*}$ with the metric $$d([v],[w]) = \frac{{\left\Vertv \wedge w\right\Vert}}{{\left\Vertv\right\Vert} \cdot {\left\Vertw\right\Vert}}$$ We will denote the $\epsilon$ neighborhood of a set in this metric by $(\Omega)_{\epsilon} = \{x \in {{\mathbb{P}}}\ | \ d(x,\Omega) < \epsilon\}$.

Any nontrivial matrix $0 \ne M \in M_n({{\mathbf{R}}})$ gives rise to a partially defined map $[M]: {{\mathbb{P}}}\setminus [\ker(M)] {\rightarrow}{{\mathbb{P}}}$. A projective point $[v] \in {{\mathbb{P}}}$ is moved by $M$ if and only if $v$ is not an eigenvector of $M$. Indeed $[M]$ is not defined on $[v]$ if and only if $v$ is an eigenvector with eigenvalue $0$ and $[v]$ is a fixed point of $[M]$ if and only if $v$ is an eigenvector with a nonzero eigenvalue. For example our auxiliary condition in Theorem \[thm:tec\] requires that whenever $g \not \in \{1,t\}$ then the matrix ${\pi}_t \circ g  = \frac{g + tg}{2}$ should move at least one projective point in $[W^{+}(t)]$. If $M \in {{\operatorname{GL}}}_n({{\mathbf{R}}})$ then the map $[M]$ is defined on the whole projective space and is in fact a bilipschitz homeomorphism with respect to the metric defined above.

\[lem:general\_pos\] Let $B \in {{\operatorname{PGL}}}_r({{\mathbf{R}}})$. If $v_0,v_1,\ldots, v_r < {{\mathbf{R}}}^r$ are $r+1$ vectors in general position that are all eigenvectors of $B$. then $B$ is a scalar matrix.

By definition, $r+1$ vectors in $R^n$ are in [*[general position]{}*]{} if any $i$ of them span an $i$-dimensional subspace as long as $i \le r$. By counting considerations there must be at least one eigenspace $V< {{\mathbf{R}}}^r$ of dimension $l$ containing at least $l+1$ of the eigenvectors $\{v_0,v_1,\ldots, v_r\}$. Since the vectors are in general position then $l=r$, so $B$ has an eigenvalue with $r$-linearly independent eigenvectors and is hence a scalar.

Linearity proofs {#sec:linearity}
================

\[prop:free\_prod\] Let $n \ge 3$ and $G,H < {{\operatorname{SL}}}_n({{\mathbf{R}}})$ be two countable groups that contain no nontrivial scalar matrices. Let ${\pi}: {{\mathbf{R}}}^n {\rightarrow}W$ be a linear projection with $\dim(\Im({\pi})) - \dim(\ker({\pi})) \ge 2$. For $f \in {{\operatorname{SL}}}_n({{\mathbf{R}}})$ let $\Phi_f:G*H {\rightarrow}{{\operatorname{SL}}}_n({{\mathbf{R}}})$ be defined by $\Phi_f(g)=g, \forall g \in G$ and $\phi_f(h) = fhf^{-1}, \ \forall h \in H$.

Then for a Baire generic choice of $f \in {{\operatorname{SL}}}_n({{\mathbf{R}}})$ the map $\Phi_f$ is injective and its image contains no nontrivial scalar matrices. Moreover whenever $\go \not \in G$ then $W$ is not contained in an eigenspace of ${\pi}\circ \Phi_f(\go)$.

Let $\go \in G*H$ and write it as a reduced word $\go = h_1 g_1 \cdot...\cdot h_k g_k,$ with $g_i, \in G, h_i \in H$, all of them nontrivial except possibly $g_k$ and $h_1$. We may (and shall) also assume that $\go \not \in G$ since in this case there is nothing to prove. Set $${{\mathcal{U}}}(\go) ={\left\{f\in{{\operatorname{SL}}}_n({{\mathbf{R}}}): {\pi}\Phi_f(\go) {\text{ is scalar on }} W\right\}}, \qquad \qquad {{\mathcal{U}}}=  \bigcap_{\go \in G*H \setminus G} {{\mathcal{U}}}(\go).$$ By Lemma \[lem:general\_pos\] ${{\mathcal{U}}}(\go)$ is Zariski open in ${{\operatorname{SL}}}_n$. Indeed if $\dim(W^{+})=r$ and $\{w_0,w_1,\ldots, w_r \}$ are $r+1$ vectors in general position in $W^{+}$, the complement is characterized by the equations $\{{\pi}\Phi_f (\go) w_i \wedge  w_i = 0\}_{0 \le i \le r}$. Since ${{\operatorname{SL}}}_n({{\mathbf{R}}})$ is connected, if we show that ${{\mathcal{U}}}(\go)$ is nonempty it will follow from the Baire category theorem that ${{\mathcal{U}}}$ is nonempty and the theorem will be proved.

Choose a basis $\{v_1,v_2,\ldots, v_n\}$ for ${{\mathbf{R}}}^n$ a vector $v \in W$ and a number $L >> 0$. Let $a^{+} = [v_1]$, $a^{-} = [v_2]$, $x = [v]$, $H^{+} = [{{\operatorname{Span}}}\{v_2,v_3,\ldots, v_n\}]$ and $H^{-} = [{{\operatorname{Span}}}\{v_1,v_3,\ldots, v_n\}]$ three projective points and two projective hyperplanes corresponding to these vectors. Define $f = f(L) \in {{\operatorname{SL}}}_n({{\mathbf{Z}}})$ by the requirements: $$f v_1 = L v_1, \qquad f v_2 = \frac{1}{L} v_2, \qquad f v_i = v_i , \ \ \ \forall 3 \le i \le n.$$ The dynamics of $f$ on ${{\mathbb{P}}}$ is proximal in the sense that $\lim_{L {\rightarrow}\infty} f(L)^{\pm 1}(x) = a^{\pm}, \forall x \not \in H^{\pm}$ respectively. Note that $${\pi}\Phi_f(\go)(x) = \pi f h_1 f^{-1} g_1 f h_2 f^{-1} \ldots f h_k f^{-1} g_k (x).$$ Consider the sequence of points $f^{-1} g_k(x)$, $f h_k f^{-1} g_k(x)$, up to $f h_1 f^{-1} g_1 f \ldots f h_k f^{-1} g_k (x)$. Our strategy is to choose all of the above data in such a way that these points alternate, coming arbitrarily close to $a^{-}$ and $a^{+}$ respectively when $L {\rightarrow}\infty$. If we insist also that $x \ne {\pi}(a^{+})$ (or that $x \ne {\pi}g_1 (a^{+})$ when $h_1 = 1$) it will follow that when $L$ is large enough ${\pi}\Phi_f(\go)$ does not fix the projective point $x \in W$ and the claim will follow. To carry this out we require the following:

1.  $g_k(x) \not \in H^{-}$ (in particular $x \not \in H^{-}$ if it so happens that $g_k=1$)

2.  $g_i (a^{+}) \not \in H^{-}, {\text{ for }} 1 \le i < k$

3.  $h_i (a^{-}) \not \in H^{+}, {\text{ for }} 1 < i \le k$

4.  ${\pi}(a^{+}) \ne x,$ if $h_1 \ne 1$ and ${\pi}g_1 (a^{+}) \ne x$ otherwise

Since $\{g_1,g_2,\ldots, g_{k-1}\}$ are by assumption non scalar matrices we can choose $v_1$ which is not an eigenvector of any of them. $v_2$ is chosen to be linearly independent of $v_1$ and not an eigenvector of any of the matrices $\{h_2,\ldots,h_k\}$. Now choose the rest of the basis $\{v_3,\ldots,v_n\}$ in such a way that $H^{\pm}$ stay away from the finitely many points as required in the second and third conditions. In addition we choose these hyperplanes in such a way that $g_k([W^{+}]) \not \subset H^{+}$ in order to make the first condition possible. Now choose $v \in W$ so as to actually satisfy the first and fourth conditions. The latter is always possible by our assumption that $\dim(W) > \dim(\ker({\pi}))+1 = \dim {\pi}^{-1}({{\operatorname{Span}}}\{v_1\}) = \dim {\pi}^{-1}({{\operatorname{Span}}}\{g_1 v_1\})$.

It is clear by all these choices, arguing by induction on $m$, that $$\lim_{L {\rightarrow}\infty} {\pi}\Phi_f(L)(\go)(x) =
\left \{
   \begin{array}{ll}
   {\pi}(a^{+}) \ne x, \ & {\text{if }} h_1 \ne 1 \\
   {\pi}g_1 (a^{+}), \ & {\text{if }} h_1 = 1 
   \end{array} \right \}
\ne x$$ Which concludes the proof.

Similar linearity statements for free products are well known and established for example in [@shalen:free]. However since we didn’t find the exact statement we needed we included the complete proof.

The previous theorem gives rise to our linear version of Theorem \[thm:RST\_free\]

(Free product) \[thm:free\_prod\] Under the assumptions of Theorem \[thm:tec\] set $G_1 = G*{{\mathbf{Z}}}$ where the group ${{\mathbf{Z}}}= \langle f \rangle$. Then there exists an element of infinite order $\ell \in {{\operatorname{SL}}}_n({{\mathbf{C}}})$ such that the natural map $\Phi_{\ell}: G_1 {\rightarrow}{{\operatorname{SL}}}_n({{\mathbf{R}}})$ defined by $g \mapsto g, \forall g \in G$ and $f \mapsto \ell$ is an isomorphism onto its image $\langle G, \ell \rangle < {{\operatorname{SL}}}_n({{\mathbf{R}}})$. Moreover all the involutions in $G_1$ are conjugate, and the only elements of $G_1$ such that $W^{+}(t)$ is contained in an eigenspace of ${\pi}_t \circ \Phi_{\ell}(g)$ are $g =1$ and $g =t$.

(Of Theorem \[thm:free\_prod\]) Let $H$ be any infinite cyclic group in ${{\operatorname{SL}}}_n({{\mathbf{R}}})$ and $G < {{\operatorname{SL}}}_n({{\mathbf{R}}})$ the given group. Applying Proposition \[prop:free\_prod\] to these groups $G,H$ yields everything we need leaving only the verification of the fact that all involutions in $G*H$ are conjugate. Let $\sigma \in G*{{\mathbf{Z}}}$ be any involution. Being an element of finite order it must stabilize a vertex in the Bass-Serre tree. Since ${{\mathbf{Z}}}$ contains no involutions $\sigma$ is conjugate into $G$ and by our assumption all involutions in $G$ are already conjugate.

And here is our linear version of Theorem \[thm:RST\_HNN\]

(HNN extension) \[thm:HNN\] Under the assumptions of Theorem \[thm:tec\] let $t \ne v \in G$ be an involution and set $G_1= \langle G,f \ | \ f^{-1}tf=v \rangle$ and $A_1 := \langle A, f \rangle.$ There exists an element of infinite order $\ell \in {{\operatorname{SL}}}_n({{\mathbf{C}}})$ such that $\ell^{-1} t \ell = v$ and such that the natural map $\Phi_{\ell}: G_1 {\rightarrow}{{\operatorname{SL}}}_n({{\mathbf{R}}})$ defined by $g \mapsto g, \forall g \in G$ and $f \mapsto \ell$ is an isomorphism onto its image. Moreover all the involutions in $G_1$ are conjugate and the only elements of $G_1$ such that $W^{+}(t)$ is contained in an eigenspace of ${\pi}_t \circ g$ are $g =1$ and $g =t$.

(Of Theorem \[thm:HNN\]). Assume that we are in the setting of that theorem and $t,v \in G$ are the two given involutions. Conjugating $G$, if necessary, we may assume that $t = {{\operatorname{Diag}}}{\left(1,...,1,-1,...,-1\right)}$ is a diagonal matrix with $r$ ones and $n-r$ minus-ones along the diagonal. If $e_1,...,e_n$ denote the vectors of the standard basis then $W^{+}:=W^{+}(t) = \langle e_1, e_2, \ldots,e_r \rangle,$ and $W^{-}:=W^{-}(t) =\langle e_{r+1},e_{r+2}, \ldots, e_n \rangle$ are, respectively, the $\pm 1$ eigenspaces of $t$. By our assumption $r  > n-r$ which immediately implies also that $r\ge 2$.

Recall that $G_1 = \langle G,f \ | \ f^{-1}tf=v\rangle$, let $G_2 = \langle G,k \ | \ k^{-1}tk=t\rangle$. It follows from a standard argument involving the universal properties of these two HNN extensions that they are isomorphic. Indeed by assumption all the involutions in $G$ are already conjugate within $G$ so there is an element $h \in G$ such that $h^{-1}th = v$. Let $F: G_1 {\rightarrow}G_2$ be the homomorphism defined by the requirement that $F(g) = g, \forall g \in G, F(f) = k h$ and let $I: G_2 {\rightarrow}G_1$ be the homomorphism defined by the requirement $I(g)=g, \forall g \in G, I(k) = f h^{-1}$. Since $I \circ F$ fixes pointwise both $G$ and $f$ it must be the identity of $G_1$ and similarly $F \circ I$ is the identity of $G_2$. We will hence identify these two groups and in particular we will identify $G_1$ as the HNN extension $G_1 = \langle G,k \ | \ k^{-1}tk=t \rangle$ where $k = fh^{-1}$.

Let $Z = {{\operatorname{SL}}}(W^{+}) \times {{\operatorname{SL}}}(W^{-}) < C_{{{\operatorname{SL}}}_n({{\mathbf{R}}})}(t)$. This group is isomorphic to ${{\operatorname{SL}}}_{r}({{\mathbf{R}}}) \times {{\operatorname{SL}}}_{n-r}({{\mathbf{R}}})$ and in particular it is a connected closed subgroup of ${{\operatorname{SL}}}_n({{\mathbf{R}}})$. For any $u \in Z$ we obtain a homomorphism. $\Phi_u : G_1 {\rightarrow}{{\operatorname{SL}}}_n({{\mathbf{R}}})$ given by $g \mapsto g, \forall g \in G$ and $k \mapsto u$.

Our goal is to find some $u \in Z$ such that for every $w \in G_1, \ w \not \in \{1,t\}$ the element $\Phi_u(w)$ does not fix $[W^{+}]$ pointwise. This will show that $\Phi_u: G_1 {\rightarrow}\langle G, u \rangle$ is an isomorphism and at the same time it will establish the auxiliary requirement that the only elements of $G_1$ to fix $[W^{+}]$ pointwise be $1$ and $t$.

Let $w \in G_1 \setminus \{1,t\}$. Let $\{v_0,v_1,\ldots, v_r\}$ be any $r+1$ vectors in general position within $W^{+}$. By Lemma \[lem:general\_pos\] we see that $$\begin{aligned}
{{\mathcal{U}}}_{w} & := & \{u \in Z \ | \  W^{+} {\text{ is not contained in an eigenspace of }} {\pi}_t \circ \Phi_u(w) \} \\
& = & Z \setminus \left(\bigcap_{i=0}^r \{u \in Z \ | \ ({\pi}_t \circ g (v_i)) \wedge  v_i = 0 \} \right)\end{aligned}$$ is the complement of a closed subvariety of $Z$. If we can show that ${{\mathcal{U}}}_{w}$ is nonempty it will follow from the Baire category theorem that ${{\mathcal{U}}}= \bigcap_{w \in G_1 \setminus \{1,t\}} {{\mathcal{U}}}_{w}$ is a dense $G_{\delta}$ subset of $Z$. In particular we would have found some $u \in Z$ satisfying all of our requirements.

From here on we will fix $w \in G_1 \setminus \{1,t\}$ and write this element in a reduced canonical form as $w=g_1 k^{\delta_1} g_2 \ldots k^{\delta_m}g_{m+1}$ where $g_i \in G, ,\delta_i \in \{\pm 1\}$. That the word is reduced means that it is subject to the additional restrictions that $g_i \ne 1$, for $i \in \{2,\ldots,m\}$, and that $k^{\delta_{i}} g_{i+1} k^{\delta_{i+1}}$ is neither of the form $k^{-1}t k$ nor of the form $k t k^{-1}$ for any $i \in \{1 \ldots m\}$. Thus $$\Phi_u(w) =  g_1 u^{\delta_1} g_2 \ldots u^{\delta_m}g_{m+1}.$$ Let $S = \{g_1,g_1^{-1}, g_2,g_2^{-1}, \ldots, g_{m+1}^{-1}\}$ and set $S_0 = S \setminus \{1,t\}$.

By our assumption the set $$\Omega := \{v \in W^{+} \ | \ v {\text{ is not an eigenvector of }} {\pi}_t \circ g {\text{ for any }} g \in S_0 \}$$ is the complement of the union of finitely many proper linear subspaces of $W^{+}$. In particular $\Omega$ is a dense open subset of the vector space $W^{+}$. Let $v_1 \in \Omega$ be any vector and $v_2 \in \Omega$ a linearly independent vector that also avoids the finitely many one dimensional subspaces $\{ {{\operatorname{Span}}}\{{\pi}_t \circ g (v_1) \} \ | \ g \in S_0 \}$. Since $S_0$ is symmetric, these choices ensure that ${\pi}_t \circ g(v_i)$ and $v_j$ are linearly independent for every $i,j \in \{1,2\}$. Thus we may complete this pair of vectors to a basis $\{v_1,v_2, \ldots, v_r\}$ for $W^{+}$ in such a way that the two codimension one spaces ${{\operatorname{Span}}}\{v_1,v_3,v_4,\ldots, v_r\}$ and ${{\operatorname{Span}}}\{v_2,v_3,\ldots, v_r\}$ of $W^{+}$ do not contain any of the points $\{{\pi}_t \circ g (v_i) \ | \ g \in S_0, i \in \{1,2\}\}$. Finally let $\{v_{r+1}, v_{r+2},\ldots, v_n\}$ be a basis for $W^{-}$. So that $\{v_1,v_2,\ldots, v_n\}$ becomes a basis for the whole space which is compatible with the direct sum decomposition $W^{+} \oplus W^{-}$. Let $a^{+} = [v_1], a^{-} = [v_2]$, $H^{+} = [{{\operatorname{Span}}}\{v_2,v_3,\ldots, v_n\}]$ and $H^{-} = [{{\operatorname{Span}}}\{v_1,v_3,\ldots, v_n\}]$ be two projective points and two projective hyperplanes corresponding to these vectors. Our choices above imply that $$\label{eqn:prop_vi}
g(a^{\pm}) \not \in H^{-} \cup H^{+}, \qquad \forall g \in S_0.$$

Pick a number $L >> 0$ and define $u = u(L) \in Z$ by the requirements: $$u v_1 = L v_1, \qquad u v_2 = \frac{1}{L} v_2, \qquad u v_i = v_i , \ \ \ \forall 3 \le i \le n.$$ The dynamics of the element $u$ on the projective plane ${{\mathbb{P}}}$ is very proximal in the sense that $$\label{eqn:prox}
\lim_{L {\rightarrow}\infty} u(L)^{\pm 1}(x) = a^{\pm}, \qquad \forall x \not \in H^{\pm} {\text{ respectively}}.$$ We denote by $g(L) = \Phi_{u(L)}(w)$ in order to stress the dependence of this element on $L$.

If $m = 0$ then $g = g_1$ and as long as $g \not \in \{1,t\}$ any element $x \in \Omega$ will satisfy ${\pi}_t \circ g (x) \ne x$ as required. So let us assume that $m \ge 1$ and choose $x \in [W^{+}]$ to be any projective point such that $g_{m+1} (x) \not \in H^{+} \cup H^{-}$ and $g_1 (a^{\pm}) \not \in [{\pi}_t^{-1}(x)]$. Such a choice is possible, regardless of the value of $g_1$ in view of our assumption that $\dim W^{+} > \dim W^{-}$.

Now argue by induction on $m$ that $\lim_{L {\rightarrow}\infty} g_1^{-1} g(L) (x) = a^{\delta_1}$. If $m = 1$, $g_1^{-1}g(L) = u^{\delta_1} g_2$. Since by our choice of $x$, $g_2 \cdot x \not \in H^{+} \cup H^{-}$ the desired property follows directly from Equation (\[eqn:prox\]). Now applying the induction hypothesis to the word $\tilde{w} = g_2 k_2^{\delta_2} g_3 \ldots k_m^{\delta_m} g_{m+1}$ and $\tilde{g}(L) = \Phi_{u(L)}(\tilde{w})$ we obtain $$\lim_{L {\rightarrow}\infty} g_1^{-1} g(L) (x) =  \lim u(L)^{\delta_1} g_2 g_2^{-1} \tilde{g}(L) (x) 
= \lim_{L {\rightarrow}\infty} u(L)^{\delta_1} g_2 a^{\delta_2} = a^{\delta_1}$$ The equality before last, uses the induction hypothesis. The last equality is justified as follows. If $g_2 \ne t$ then by Equation (\[eqn:prop\_vi\]) $g_2 a^{\delta_2} \not \in H^{+} \cup H^{-}$, while if $g_2 = t$ then by the fact that the word $w$ is reduced we know that $\delta_1 = \delta_2$ and hence $a^{\delta_2} \not \in H^{\delta_1}$. In both cases the claim follows directly from Equation (\[eqn:prox\]). This completes the proof of the induction. By the Baire theoretic argument we have thus found $u \in Z$ such that $\Phi_u: G_1 {\rightarrow}{{\operatorname{SL}}}_n({{\mathbf{R}}})$ becomes an isomorphism and such that the induction hypothesis, according to which $W^{+}$ is not contained in an eigenspace of $\Phi_u(w)$ for any $w \in G_1 \setminus \{1,k\}$, is also satisfied.

It remains only to show that all the involutions in $G_1$ are conjugate. We argue on the Bass-Serre tree corresponding to the HNN extension $G_1 = \langle G, f \ | \ f^{-1}tf = v \rangle$. If $\sigma \in G_1$ is an involution then, being an element of finite order, it must fix a vertex in the Bass-Serre tree. Since the action of $G_1$ is transitive on the vertices all vertex stabilizers are conjugate to $G$. But by the induction hypothesis all involutions in $G$ itself are conjugate and the theorem is proved.

Conclusion
==========

(of Theorem \[thm:s2t\]) Let $H \curvearrowright X$ be the given action, fix any basepoint $x \in X$ and let $A = H_x$ be its stabilizer. The condition that the action on (ordered) pairs of distinct points is free is equivalent to the condition that $A$ be malnormal in H. We fix any involution $t \in SL_n({{\mathbf{R}}})$ and denote by $W^{\pm}$ its $\pm 1$ eigenspaces and by ${\pi}= {{\mathbf{R}}}^n {\rightarrow}W^+$ the projection on $W^{+}$ along $W^{-}$. We impose also the restriction $\dim(W^{+}) \ge \dim(W^{-})+2$. Applying Proposition \[prop:free\_prod\] to the groups $H$ and $\langle t \rangle$ and then replacing if necessary the given group $H$ by its conjugate obtained in that theorem we may assume that $G := \langle H, t \rangle \cong H * {{\mathbf{Z}}}/2{{\mathbf{Z}}}$ and that furthermore: (i) $A$ is malnormal in $G$, (ii) all involutions in $G$ are conjugate to $t$ and (iii) if $W^{+}$ is contained in an eigenspace of some $g \in G$ then $G\in \{1,t\}$. Indeed (i),(ii) are guaranteed by the properties of free products and (iii) follows from Proposition \[prop:free\_prod\].

The conditions (i),(ii),(iii) are exactly these needed in order to apply Theorem \[thm:tec\]. Let us enumerate the elements of the group $G$ as follows $G = \{v_1,v_2, \ldots\}$. Applying this theorem inductively to obtain a sequence of extensions $(A,G) < (A_1,G_1) < (A_2,G_2) < \ldots$, such that $G_i \cap A_{i+1} = A_i$. These come together with elements $f_i \in A_i$ such that $A_i t f_i = A_i v_i$. Let $G_{\omega} = \cup_{i=1}^{\infty} G_i$ and $A_{\omega} = \cup_{i=1}^{\infty} A_i$. These groups satisfy the properties (i),(ii),(iii). In addition $G \cap A_{\omega} = A$ and for every $v \in G$ there exists some $f \in A$ such that $Atf=Av$.

Applying the procedure of the previous paragraph to $(A_{\omega},G_{\omega})$ and continuing inductively we obtain a sequence $(A_{\omega}, G_{\omega}) < (A_{2\omega},G_{2\omega}), < \ldots$ and the union $H_1 = G_{\omega \cdot \omega} = \cup_{i=1}^{\infty} G_{i \omega}$ and $B_1 = A_{\omega \cdot \omega} = \cup_{i =1}^{\infty}A_{i\omega}$ satisfy the following conditions:

-   $B_1$ is malnormal in $H_1$

-   For every $v \in H_1$ there exits an element $f \in B_1$ such that $B_1 t f = B_1 v$

-   $H \cap B_1 = A$

The first two properties are respectively equivalent to the action $H_1 \curvearrowright B_1{\backslash}H_1$ being free and transitive on ordered pairs of distinct points. The last property implies that the action of $H$ on $B_1 \cdot H$ is isomorphic to the original given action $H \curvearrowright A {\backslash}H$.

We know that ${{{\it {p}}\operatorname{-char}}}(H_1)=2$ because the stabilizer of a point is $B_1$ and this group contains no involutions by construction. If $M \lhd H_1$ is a normal Abelian subgroup then $M \cap G_{\alpha}$ is a normal abelian subgroup of $G_{\alpha}$ for every ordinal $\alpha$ and by construction these groups clearly have non nontrivial normal abelian subgroups. Thus $M$ itself must be trivial. This proves that $H_1$ does not split. A similar argument shows that if $M,N \lhd H_1$ are two commuting normal subgroups then one of them must be trivial.

Finally we appeal to the main theorem of [@GG:AOS] (see also [@GG:Primitive]) to deduce that $H_1$ admits uncountably many non-isomorphic primitive permutation representations. We have to verify that $H_1$ is not of Affine or diagonal type. And this immediately follows from the fact that $H_1$ contains neither normal abelian subgroups, nor pairs of commuting normal subgroups respectively.

[Yair Glasner.]{} Department of Mathematics. Ben-Gurion University of the Negev. P.O.B. 653, Be’er Sheva 84105, Israel. [yairgl@math.bgu.ac.il]{}

[Dennis D. Gulko.]{} Department of Mathematics. Ben-Gurion University of the Negev. P.O.B. 653, Be’er Sheva 84105, Israel. [gulko@math.bgu.ac.il]{}

[^1]: Following [@RST:s2t] all the actions in this paper will be right actions. This choice is also responsible for the choice of right near fields here.
