Simple groups of Morley rank $3$ are algebraic

Abstract

There exists no bad group (in the sense of Gregory Cherlin); namely, any simple group of Morley rank 3 is isomorphic to ${\mathrm{PSL}}_2(K)$ for an algebraically closed field $K$.

1. Introduction

Model theory is a branch of mathematical logic concerned with the study of classes of mathematical structures by considering first-order sentences and formulas. There are numerous interactions between model theory and other areas of mathematics, such as number theory, sometimes these interactions are spectacular, such as Pila’s work on the André–Oort Conjecture 15 or Hrushovski’s proof of the function field Mordell–Lang conjecture 9. In particular model theory of abelian groups is used in the latter, and a result of Wagner on abelian groups of finite Morley rank is used in a recent paper concerning the Mordell–Lang conjecture 3.

Morley rank is a model-theoretical notion of dimension. It generalizes the dimension of an algebraic variety (when the ground field is algebraically closed). In this paper, we are concerned with groups of finite Morley rank. The main example of such a group is an algebraic group defined over an algebraically closed field in the field language (Zilber 20). Independently, in the late 1970s Gregory Cherlin 6, §6 and Boris Zilber 20 formulated the following algebraicity conjecture.

Conjecture 1.1 (Cherlin–Zilber conjecture or algebraicity conjecture).

An infinite simple group of finite Morley rank is algebraic over an algebraically closed field.

This is the main conjecture on groups of finite Morley rank, and it is still open. Most studies on groups of finite Morley rank focus on this conjecture.

In the 1980s, Borovik proposed to attack the algebraicity conjecture by transferring methods from the classification of the finite simple groups, and to analyze a minimal counterexample from its involutions. Borovik’s program has been very effective for several important classes of groups of finite Morley rank, including locally finite groups 18. Its main success is the main theorem of 1 which ensures that any simple group of finite Morley rank with an infinite abelian subgroup of exponent 2 satisfies the Cherlin–Zilber conjecture.

However, despite numerous papers on the subject, the Cherlin–Zilber conjecture is still open. In this paper we show that any simple group of Morley rank 3 is algebraic over an algebraically closed field. Due to the absence of any internal group theoretic structure allowing local analysis, one resorts to a more geometrical analysis. We note that such an analysis is also encountered in the Borovik program, but it is often associated there with the geometry of involutions or other aspects of local analysis.

As a matter of fact, in 6 the algebraicity conjecture was formulated as a result of local analysis of simple groups of Morley rank 3. The main result of 6 can be summarized as follows, where a bad group is a nonsolvable group of Morley rank 3 containing no definable subgroup of Morley rank 2.

Fact 1.2 (Cherlin 6).

Let $G$ be an infinite simple group of Morley rank at most $3$. Then $G$ has Morley rank $3$, and one of the following two assertions is satisfied:

•

there is an algebraically closed field $K$ such that $G\simeq {\mathrm{PSL}}_2(K)$,

•

$G$ is a bad group.

Thus bad groups have become a major obstacle to the Cherlin–Zilber conjecture. These groups have been studied in 6, 13, and 14, whose results are summarized in Facts 2.3 and 2.4, respectively. Later, it was shown that no bad group is existentially closed 11 or linear 12. However, these groups appeared very resistant, and only sparse supplementary information was known on them.

Furthermore, Nesin has shown in 14 that a bad group acts on a natural geometry, which is not very far from being a non-Desarguesian projective plane of Morley rank 2. However, Baldwin 2 discovered non-Desarguesian projective planes of Morley rank 2. Thus, the question of the existence, or not, of a bad group was still fully open. In this paper, we show that bad groups do not exist.

Main Theorem 1.3.

There is no bad group (in the sense of Cherlin).

Note that other more general notions of bad groups have been introduced independently by Corredor 7 and by Borovik and Poizat 4, where a bad group is defined to be a nonsolvable connected group of finite Morley rank all of whose proper connected definable subgroups are nilpotent. Such a bad group has similar properties to original bad groups. Moreover, Jaligot later introduced a more general notion of bad groups 10 and obtained similar results. However, we recall that, in this paper, a bad group is defined as a nonsolvable group of Morley rank 3 containing no definable subgroup of Morley rank 2.

Our proof of Main Theorem 1.3 goes as follows. First we note that it is sufficient to study simple bad groups since for any bad group $G$, the quotient group $G/Z(G)$ is a simple bad group by 13, §4, Introduction.

Then we fix a simple bad group $G$, and we introduce a notion of line as a coset of a Borel subgroup of $G$ (Definition 3.1). In §3 we study their behavior, mainly in regards with conjugacy classes of elements of $G$.

In §4 we propose a definition of a plane (Definition 4.1). This section is dedicated to proving that $G$ contains a plane (Theorem 4.14). This result is the key point of our demonstration. Roughly speaking, we show that for each nontrivial element $g$ of $G$ such that $g=[u,v]$ for $(u,v)\in G\times G$, the union of the preimages of $g$, by maps of the form ${\mathrm{ad}}_v:G\to G$ defined by ${\mathrm{ad}}_v(x)=[x,v]$, is almost a plane, and from this, we obtain a plane.

In last section, §5, we try to show that our notions of lines and planes provide a structure of projective space over the group $G$. Indeed, such a structure would provide a division ring (see 8, p. 124, Theorem 7.15), and probably it would be easy to conclude. However, a contradiction occurs along the way, and achieves our proof.

The other simple groups of dimension 3

•

If $G$ is a nonbad simple group of Morley rank 3, then $G$ is isomorphic to ${\mathrm{PSL}}_2(K)$ for an algebraically closed field $K$ (Fact 1.2). As in §3 we may define a line in $G$ to be a coset of a connected subgroup of dimension 1, and we may define a plane as in §4. It is possible to show that two sorts of planes occur: the cosets of Borel subgroups; and the subsets of the form $aJ$ where $J$ is defined to be

–

the set of involutions when the characteristic $c$ of $K$ is not 2,

–

the set of involutions and the identity element when $c=2$.

The plane $J$ is normalized by $G$, and there is no such a plane in a bad group (Lemma 5.12). Another important difference between $G$ and a bad group is the presence of a Weyl group. Indeed, the first lemma of this paper is not verified in $G$ (Lemma 3.2), because we have $jT=Tj$ for any torus $T$ and any involution $j\in N_G(T)\setminus T$.

•

The group ${\mathrm{SO}}_3(\mathbb{R})$ is not of finite Morley rank and is not even stable 13. However, our definitions of lines and planes naturally extend to ${\mathrm{SO}}_3(\mathbb{R})$. Then, as above, the set $J$ of involutions in ${\mathrm{SO}}_3(\mathbb{R})$ forms a plane, and the presence of a Weyl group is again a major difference between ${\mathrm{SO}}_3(\mathbb{R})$ and bad groups. Moreover, we note that the plane $J$ has a structure of projective plane, whereas this is false in ${\mathrm{PSL}}_2(K)$ 5, Fact 8.15.

Note

In very recent preprints 1719, by analyzing the present paper, Poizat and Wagner generalize our main result to other groups, and they eliminate other groups of finite Morley rank.

2. Background material

A thorough analysis of groups of finite Morley rank can be found in 5 and 1. In this section we recall some definitions and known results.

2.1. Borovik–Poizat axioms

Let $(G,\,\cdot \,,^{-1},1,\ldots )$ be a group equipped with additional structure. This group $G$ is said to be ranked if there is a function “rk” which assigns to each nonempty definable set $S$ an integer, its “dimension” ${\mathrm{rk\,}}(S)$, and which satisfies the following axioms for every definable sets $A$ and $B$.

Definition

For any integer $n$, ${\mathrm{rk\,}}(A)>n$ if and only if $A$ contains an infinite family of disjoint definable subsets $A_i$ of rank $n$.

Definability

For any uniformly definable family $\{A_b\nobreakspace :\nobreakspace b\in B\}$ of definable sets and for any $n\in \mathbb{N}$, the set $\{b\in B\nobreakspace :\nobreakspace {\mathrm{rk\,}}(A_b)=n\}$ is also definable.

Finite Bounds

For any uniformly definable family $\mathfrak{F}$ of finite subsets of $A$, the sizes of the sets in $\mathfrak{F}$ are bounded.

These axioms were introduced in 16, where it is shown that the groups as above satisfy a fourth axiom, namely the additivity axiom, and they are precisely the groups of finite Morley rank. Moreover, the function ${\mathrm{rk\,}}$ assigns to each definable set its Morley rank. In this paper, as in 5 and 1, the Morley rank will be denoted by ${\mathrm{rk\,}}$.

2.2. Morley degree

A nonempty definable set $A$ is said to have Morley degree $1$ if for any definable subset $B$ of $A$, either ${\mathrm{rk\,}}B<{\mathrm{rk\,}}A$ or ${\mathrm{rk\,}}(A\setminus B)<{\mathrm{rk\,}}A$. The set $A$ is said to have Morley degree $d$ if $A$ is the disjoint union of $d$ definable sets of Morley degree 1 and Morley rank ${\mathrm{rk\,}}A$.

Fact 2.1.

•

Every nonempty definable set has a unique degree 5, Lemmas 4.12 and 4.14.

•

Let $X$ and $Y$ be definable subsets of Morley degree $d$ and $d',$ respectively. Then $X\times Y$ has Morley degree $dd'$ 5, Proposition 4.2.

•

A group of finite Morley rank has Morley degree $1$ if and only if it is connected, namely it has no proper definable subgroup of finite index 6, §2.2.

Moreover, the following elementary result will be useful for us.

Fact 2.2.

Let $f:E\to F$ be a definable map. If the set $E$ has Morley degree $1$ and $r={\mathrm{rk\,}}f^{-1}(y)$ is constant for $y\in F$, then the Morley degree of $F$ is $1$.

Proof.

Let $B$ be a definable subset of $F$ of Morley rank ${\mathrm{rk\,}}F$. We show that ${\mathrm{rk\,}}(F\setminus B)<{\mathrm{rk\,}}F$. By the additivity axiom, we have ${\mathrm{rk\,}}E=r+{\mathrm{rk\,}}F$ and

$${\mathrm{rk\,}}f^{-1}(B)=r+{\mathrm{rk\,}}B=r+{\mathrm{rk\,}}F={\mathrm{rk\,}}E.$$

Since $E$ has Morley degree 1, we obtain ${\mathrm{rk\,}}f^{-1}(F\setminus B)={\mathrm{rk\,}}(E\setminus f^{-1}(B))<{\mathrm{rk\,}}E$, and by the additivity axiom again

$${\mathrm{rk\,}}(F\setminus B)={\mathrm{rk\,}}f^{-1}(F\setminus B)-r<{\mathrm{rk\,}}E-r={\mathrm{rk\,}}F,$$

so $F$ has Morley degree 1.

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2.3. Bad groups

The main properties of bad groups are summarized in the following facts, where a Borel subgroup of a bad group $G$ is defined to be an infinite definable proper subgroup of $G$.

Fact 2.3 (6, §5.2 and 13).

Let $G$ be a simple bad group, and let $B$ be a Borel subgroup of $G$.

(1)

$B=C_G(b)$ for any nontrivial element $b$ of $B$.

(2)

$B$ is connected, abelian, self-normalizing, and of Morley rank 1.

(3)

$C_G(x)$ is a Borel subgroup for each nontrivial element $x$ of $G$.

(4)

If $A$ is another Borel subgroup of $G$, then $A$ is conjugate with $B$, and either $A=B$ or $A\cap B=\{1\}$.

(5)

$G=\bigcup _{g\in G}B^g$.

(6)

$G$ has no involution.

Fact 2.4 (14, Lemma 18).

Let $A$ and $B$ be two distinct Borel subgroups of a simple bad group $G$. Then ${\mathrm{rk\,}}(ABA)=3$, ${\mathrm{rk\,}}(AB)=2$, and $AB$ has Morley degree 1.

The following result, due to Delahan and Nesin, was proved for a more general notion of bad groups and is used in our final argument.

Fact 2.5 (5, Proposition 13.4).

A simple bad group $G$ cannot have an involutive definable automorphism.

3. Lines

In this paper $G$ denotes a fixed simple bad group. We fix a Borel subgroup $B$ of $G$, and we denote by $\mathscr{B}$ the set of Borel subgroups of $G$.

In this section, we define a line of $G$, and we provide their basic properties. We note that, by conjugation of Borel subgroups (Fact 2.3(4)), any Borel subgroup is a line in the following sense.

Definition 3.1.

A line of $G$ is a subset of the form $uBv$ for two elements $u$ and $v$ of $G$.

We denote by $\Lambda$ the set of lines of $G$.

We note that, by Fact 2.3(2), each line has Morley rank 1 and Morley degree 1.

Lemma 3.2.

Let $uBv$ and $rBs$ be two lines. Then $uBv=rBs$ if and only if $uB=rB$ and $Bv=Bs$.

Proof.

We may assume that $uBv=rBs$. Then we have

$$B=u^{-1}rBsv^{-1}=u^{-1}rsv^{-1}B^{sv^{-1}},$$

so $u^{-1}rsv^{-1}\in B^{sv^{-1}}$ and $B=B^{sv^{-1}}$. Now $sv^{-1}$ belongs to $B$ since $B$ is self-normalizing by Fact 2.3. Hence we obtain $Bv=Bs$, and the equality $uB=rB$ follows from $uBv=rBs$.

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By the above lemma, the set $\Lambda$ identifies with $(G/B)_l\times (G/B)_r,$ where $(G/B)_l$ (resp. $(G/B)_r$) denotes the set of left cosets (resp. right cosets) of $B$ in $G$. Then $\Lambda$ is a definable set. Moreover, since $G$ is connected of Morley rank 3 and $B$ has Morley rank 1, the Morley rank of $\Lambda$ is 4 and its Morley degree is 1. In particular, $\Lambda$ is a uniformly definable family.

Lemma 3.3.

Two distinct elements $x$ and $y$ of $G$ lie in one and only one line $l(x,y)$. Moreover, the map $l:\{(x,y)\in G\times G\nobreakspace |\nobreakspace x\neq y\}\to \Lambda$ is definable.

Proof.

By Fact 2.3(5), there exists $v\in G$ such that $y^{-1}x$ belongs to $B^v$. Then $x$ and $y$ lie in $uBv$ for $u=yv^{-1}$.

Now, if $rBs$ is a line containing $x$ and $y$, then we find two elements $b_1$ and $b_2$ of $B$ such that $x=rb_1s$ and $y=rb_2s$. Thus $y^{-1}x=s^{-1}b_2^{-1}b_1s$ is a nontrivial element of $B^s$. But $y^{-1}x$ belongs to $B^v$ by the choice of $v$, hence we have $B^s=B^v$ (Fact 2.3(4)). Since $B$ is self-normalizing, $sv^{-1}$ belongs to $B,$ and we obtain $Bs=Bv$, so there exists $b\in B$ such that $s=bv$. This implies that $u=yv^{-1}=(rb_2s)(s^{-1}b)=rb_2b$ belongs to $rB$, and $rBs=uBv$ is the unique line containing $x$ and $y$.

Moreover, since $\Lambda$ is a uniformly definable family, the set $\{(x,y)\in G\times G\nobreakspace | x\neq y\}\times \Lambda$ is definable, and

$$\Gamma =\{((x,y),uBv)\in (G\times G)\times \Lambda\nobreakspace |\nobreakspace x\neq y,\nobreakspace x\in uBv,\nobreakspace y\in uBv\}$$

is a definable subset of it. But $\Gamma$ is precisely the graph of the map $l$, hence $l$ is definable.

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Lemma 3.4.

If $uBv=(uBv)^g$ for $uBv\in \Lambda \setminus \mathscr{B}$ and $g\in G$, then $g=1$.

Proof.

We have $uBv=g^{-1}uBvg$, so $uB=g^{-1}uB$ and $Bv=Bvg$ by Lemma 3.2, and $g$ belongs to the Borel subgroups $B^{u^{-1}}$ and $B^v$. If $g$ is nontrivial, then $B^{u^{-1}}=B^v$ (Fact 2.3(4)), and $vu$ belongs to $N_G(B)=B$. Consequently, $u$ belongs to $v^{-1}B$, and we obtain $uBv=B^v$, contradicting $uBv\not \in \mathscr{B}$. Thus $g=1$.

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Definition 3.5.

For each $g\in G$ and each definable subset $X$ of $G$, we consider the following subsets of $\Lambda$:

$$\mathscr{L}(g,X)=\{l(g,x)\in \Lambda\nobreakspace |\nobreakspace x\in X\setminus \{g\}\},$$

$$\Lambda _X=\{\lambda \in \Lambda\nobreakspace |\nobreakspace \lambda \cap X\ {\mathrm{is\ infinite}}\}.$$

Since the map $l$ is definable (Lemma 3.3), the set $\mathscr{L}(g,X)$ is definable for each $g\in G$ and each definable subset $X$ of $G$. Moreover, by the Definablity axiom, the set $\Lambda _X=\{\lambda \in \Lambda\nobreakspace |\nobreakspace {\mathrm{rk\,}}(\lambda \cap X)=1\}$ is definable too.

Lemma 3.6.

Let $\lambda _1,\ldots ,\lambda _n$ be $n$ lines. Then $\lambda _1\cup \cdots \cup \lambda _n$ is a definable set of Morley rank $1$ and Morley degree $n$.

Proof.

For each $i$, the set $A_i=\lambda _i\cap (\bigcup _{j\neq i}\lambda _j)$ has at most $n-1$ elements by Lemma 3.3, and $\lambda _1\cup \cdots \cup \lambda _n$ is the disjoint union of $\lambda _1\setminus A_1,\ldots ,\lambda _n\setminus A_n, \bigcup _{i=1}^nA_i$. Since each line $\lambda _i$ has Morley rank 1 and Morley degree 1 (Fact 2.3(2)), the result follows.

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Lemma 3.7.

If $\Lambda _0$ is a definable subset of $\Lambda$, then $\bigcup \Lambda _0$ is a definable subset of $G$. Moreover, if $\Lambda _0$ is infinite, then $\bigcup \Lambda _0$ has Morley rank at least $2$.

Proof.

Since $\Lambda _0$ is a definable subset of the uniformly definable family $\Lambda$, the set $\bigcup \Lambda _0=\{x\in G\nobreakspace |\nobreakspace \exists \lambda \in \Lambda _0,\nobreakspace x\in \lambda \}$ is definable. Moreover, if $\Lambda _0$ is infinite, then $\bigcup \Lambda _0$ has Morley rank at least 2 by Lemma 3.6.

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Corollary 3.8.

The subset $\bigcup \Lambda _X$ of $G$ is definable for each definable subset $X$ of $G$.

4. Planes

Our original aim, before we reached a contradiction, was to find a definable structure of projective space on our bad group $G$. In this section, we introduce a notion of planes, and we show that $G$ has such a plane (Theorem 4.14). We fix a definable subset $X$ of $G$, of Morley rank 2.

Definition 4.1.

The definable subset $X$ of $G$ is said to be a plane if it satisfies $\widetilde{X}=X$ where

$$\widetilde{X}=\{g\in G\nobreakspace |\nobreakspace {{\mathrm{rk\,}}}(\mathscr{L}(g,X))=1\}.$$

Lemma 4.2.

The set $\widetilde{X}$ is a definable subset of $\bigcup \Lambda _X$.

Proof.

If $g\in G$ does not belong to $\bigcup \Lambda _X$, then $l(g,x)\cap X$ is finite for each $x\in X$, and since $X$ has Morley rank 2, the set $\mathscr{L}(g,X)$ has Morley rank 2, so $g\not \in \widetilde{X}$. Thus $\widetilde{X}$ is contained in $\bigcup \Lambda _X$.

We show that $\widetilde{X}$ is definable. We consider the set

$$A=\{(g,\lambda )\in G\times \Lambda\nobreakspace |\nobreakspace g\in \lambda ,\nobreakspace \exists x\in X\setminus \{g\},\nobreakspace x\in \lambda \}$$

and the map $f:A\to G$ defined by $f(g,\lambda )=g$. We note that, since $\Lambda$ is a uniformly definable family, $A$ is definable, and $f$ is definable too. Moreover, the preimage by $f$ of each $g\in G$ is $f^{-1}(g)=\{g\}\times \mathscr{L}(g,X)$, and we have ${\mathrm{rk\,}}(f^{-1}(g))={\mathrm{rk\,}}\mathscr{L}(g,X)$. Consequently, we obtain $\widetilde{X}=\{g\in G\nobreakspace |\nobreakspace {\mathrm{rk\,}}(f^{-1}(g))=1\}$, and $\widetilde{X}$ is definable.

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Lemma 4.3.

The Morley ranks of $\Lambda _X$ and of $\bigcup \Lambda _X$ are at most $2$. Moreover, $\Lambda _X$ is infinite if and only if ${\mathrm{rk\,}}\bigcup \Lambda _X=2$.

Proof.

We consider the surjective definable map

$$l_0:(X\times X)\cap l^{-1}(\Lambda _X)\to \Lambda _X$$

defined by $l_0(x,y)=l(x,y)$. For each $\lambda \in \Lambda _X$, we have $l_0^{-1}(\lambda )=\{(x,y)\in (\lambda \cap X)\times (\lambda \cap X)\nobreakspace |\nobreakspace x\neq y\}$, and since ${\mathrm{rk\,}}\lambda =1$, we obtain ${\mathrm{rk\,}}(\lambda \cap X)=1$ and ${\mathrm{rk\,}}l_0^{-1}(\lambda )=2$. But we have

$${\mathrm{rk\,}}((X\times X)\cap l^{-1}(\Lambda _X))\leq {\mathrm{rk\,}}(X\times X)=2{\mathrm{rk\,}}X=4,$$

hence ${\mathrm{rk\,}}\Lambda _X$ is at most $4-2=2$.

We show that ${\mathrm{rk\,}}\bigcup \Lambda _X\leq 2$. We consider the definable set

$$A=\{(x,\lambda )\in G\times \Lambda _X\nobreakspace |\nobreakspace x\in \lambda \setminus X\}$$

and the definable map $l_1:A\to \Lambda _X$ defined by $l_1(x,\lambda )=\lambda$. For each $\lambda \in \Lambda _X$, we have ${\mathrm{rk\,}}\lambda =1={\mathrm{rk\,}}(\lambda \cap X)$, so, since each line has Morley degree 1, the preimage $l_1^{-1}(\lambda )$ is finite. Consequently, we obtain ${\mathrm{rk\,}}A\leq {\mathrm{rk\,}}\Lambda _X\leq 2$. But the definable map $l_2:A\to (\bigcup \Lambda _X)\setminus X$, defined by $l_2(x,\lambda )=x$, is surjective, hence the Morley rank of $(\bigcup \Lambda _X)\setminus X$ is at most ${\mathrm{rk\,}}A\leq 2$. Since $X$ has Morley rank 2, we obtain ${\mathrm{rk\,}}\bigcup \Lambda _X\leq 2$.

Now it follows from Lemmas 3.6 and 3.7 that $\Lambda _X$ is infinite if and only if ${\mathrm{rk\,}}\bigcup \Lambda _X=2$.

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Proposition 4.4.

For each $g\in \widetilde{X}$, we have ${{\mathrm{rk\,}}}(\mathscr{L}(g)\cap \Lambda _X)=1$.

Moreover, if $X$ has Morley degree $1$, then $\widetilde{X}=\{g\in G\nobreakspace |\nobreakspace {{\mathrm{rk\,}}}(\mathscr{L}(g)\cap \Lambda _X)=1\}$ and $G\setminus \widetilde{X}=\{g\in G\nobreakspace |\nobreakspace \mathscr{L}(g)\cap \Lambda _X\nobreakspace {\mathrm{is\ finite}}\}$.

Proof.

First we note that $\mathscr{L}(g)\cap \Lambda _X=\mathscr{L}(g,X)\cap \Lambda _X$ for any $g\in G$. For each $g\in G$, we consider the definable map $l_g:X\setminus \{g\}\to \mathscr{L}(g,X)$ defined by $l_g(x)=l(g,x)$. In particular, the preimage $l_g^{-1}(\lambda )$ of each $\lambda \in \mathscr{L}(g,X)$ is $(\lambda \cap X)\setminus \{g\}$.

We show that ${{\mathrm{rk\,}}}(\mathscr{L}(g,X)\cap \Lambda _X)\leq 1$ for each $g\in G$. We may assume that $\Lambda _X$ is infinite. Then, by Lemma 4.3, the set $\bigcup \Lambda _X$ has Morley rank 2. Let $g\in G$ and $u_g:\bigcup (\mathscr{L}(g)\cap \Lambda _X)\setminus \{g\}\to \mathscr{L}(g)\cap \Lambda _X$ be the map defined by $u_g(x)=l(g,x)$. Since each line has Morley rank 1, the preimage of each element of $\mathscr{L}(g)\cap \Lambda _X$ has Morley rank 1. Consequently, we have

$${\mathrm{rk\,}}(\mathscr{L}(g)\cap \Lambda _X)= {\mathrm{rk\,}}\bigcup (\mathscr{L}(g)\cap \Lambda _X)-1\leq {\mathrm{rk\,}}\bigcup \Lambda _X-1=1.$$

Let $g\in \widetilde{X}$. We show that ${{\mathrm{rk\,}}}(\mathscr{L}(g)\cap \Lambda _X)=1$. For each $\lambda \in \mathscr{L}(g,X)\setminus \Lambda _X$, the set $l_g^{-1}(\lambda )=(\lambda \cap X)\setminus \{g\}$ is finite, and since $g\in \widetilde{X}$, we have ${\mathrm{rk\,}}\mathscr{L}(g,X)=1$. Consequently, $l_g^{-1}( \mathscr{L}(g,X)\setminus \Lambda _X)$ has Morley rank at most 1, and $l_g^{-1}( \mathscr{L}(g,X)\cap \Lambda _X)$ has Morley rank ${\mathrm{rk\,}}X=2$. But the set $l_g^{-1}(\lambda )=(\lambda \cap X)\setminus \{g\}$ is infinite of Morley rank 1 for each $\lambda \in \mathscr{L}(g,X)\cap \Lambda _X$. Hence we obtain ${\mathrm{rk\,}}(\mathscr{L}(g,X)\cap \Lambda _X)=2-1=1$.

Now we assume that $X$ has Morley degree 1. Let $g\in G$ such that

$${{\mathrm{rk\,}}}(\mathscr{L}(g,X)\cap \Lambda _X)=1.$$

We show that $g\in \widetilde{X}$. Since the set $l_g^{-1}(\lambda )=(\lambda \cap X)\setminus \{g\}$ is infinite of Morley rank 1 for each $\lambda \in \mathscr{L}(g,X)\cap \Lambda _X$, the set $l_g^{-1}( \mathscr{L}(g,X)\cap \Lambda _X)$ has Morley rank

$$1+{{\mathrm{rk\,}}}(\mathscr{L}(g,X)\cap \Lambda _X)=2={\mathrm{rk\,}}X.$$

Then, since $X$ has Morley degree 1, the preimage of $\mathscr{L}(g,X)\setminus \Lambda _X$ has Morley rank at most 1. Moreover, for each $\lambda \in \mathscr{L}(g,X)\setminus \Lambda _X$, the preimage $l_g^{-1}(\lambda )=(\lambda \cap X)\setminus \{g\}$ is finite and nonempty, so we obtain

$${\mathrm{rk\,}}(\mathscr{L}(g,X)\setminus \Lambda _X)={\mathrm{rk\,}}l_g^{-1}(\mathscr{L}(g,X)\setminus \Lambda _X)\leq 1.$$

This shows that ${\mathrm{rk\,}}\mathscr{L}(g,X)=1$ and $g\in \widetilde{X}$.

Furthermore, since ${{\mathrm{rk\,}}}(\mathscr{L}(g,X)\cap \Lambda _X)\leq 1$ for each $g\in G$, we obtain $G\setminus \widetilde{X}=\{g\in G\nobreakspace |\nobreakspace \mathscr{L}(g)\cap \Lambda _X\nobreakspace {\mathrm{is\ finite}}\}$, as desired.

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Corollary 4.5.

We have ${\mathrm{rk\,}}(\widetilde{X}\setminus X)\leq 1$.

Proof.

We remember that $\widetilde{X}$ is definable by Lemma 4.2, so the sets $Y=\widetilde{X}\setminus X$ and $A=\{(y,\lambda )\in Y\times \Lambda _X\nobreakspace |\nobreakspace y\in \lambda \}$ are definable too. Let $l_Y:A\to Y$ and $l_D:A\to \Lambda _X$ be the definable maps defined by $l_Y(y,\lambda )=y$ and $l_D(y,\lambda )=\lambda ,$ respectively. On the one hand, for each $\lambda \in \Lambda _X$, the set $\lambda \cap X$ is infinite, and since $\lambda$ has Morley rank 1 and Morley degree 1 (Fact 2.3(2)), the set $\lambda \cap Y$ is finite and $l_D^{-1}(\lambda )$ has Morley rank at most 0. This implies ${\mathrm{rk\,}}A\leq {\mathrm{rk\,}}\Lambda _X\leq 2$ (Lemma 4.3). On the other hand, for each $y\in Y$, we have ${{\mathrm{rk\,}}}(\mathscr{L}(y)\cap \Lambda _X)=1$ by Proposition 4.4, so $l_Y^{-1}(y)$ has Morley rank 1, and we obtain ${\mathrm{rk\,}}A=1+{\mathrm{rk\,}}Y$. Consequently, the Morley rank of $Y$ is at most 1.

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Lemma 4.6.

For each $g\in G$, the set $\mathscr{L}(g,X)$ is infinite.

Proof.

Indeed, $\bigcup \mathscr{L}(g,X)$ is definable (Lemma 3.7) and contains $X$. Since ${\mathrm{rk\,}}X=2$, we obtain ${\mathrm{rk\,}}(\bigcup \mathscr{L}(g,X))\geq 2$, and $\mathscr{L}(g,X)$ is infinite (Lemma 3.6).

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Corollary 4.7.

If the Morley degree of $X$ is not $1$, then ${\mathrm{rk\,}}\widetilde{X}<2$. In particular, any plane has Morley degree $1$.

Proof.

Let $n$ be the Morley degree of $X$, and let $X_1,\ldots ,X_n$ be $n$ definable subsets of $X$ of Morley rank 2 and Morley degree 1 such that $X$ is the disjoint union of $X_1,\ldots ,X_n$. For each $g\in \widetilde{X}$, we have ${\mathrm{rk\,}}\mathscr{L}(g,X)=1$, so we obtain ${\mathrm{rk\,}}\mathscr{L}(g,X_i)\leq 1$ for each $i$, and $g\in \widetilde{X_i}$ for each $i$ by Lemma 4.6. Thus $\widetilde{X}$ is contained in $\widetilde{X_1}\cap \widetilde{X_2}$. Since $X_1\cap X_2=\emptyset$, the set $\widetilde{X}$ is contained in $(X_1\cap Y_2)\cup (Y_1\cap X_2)\cup (Y_1\cap Y_2),$ where $Y_1=\widetilde{X_1}\setminus X_1$ and $Y_2=\widetilde{X_2}\setminus X_2$. Since $Y_1$ and $Y_2$ have Morley rank at most 1 by Corollary 4.5, we obtain ${\mathrm{rk\,}}\widetilde{X}<2$.

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Lemma 4.8.

We assume that $X$ has Morley degree $1$ and that $Y$ is another definable subset of $G$ of Morley rank $2$ and Morley degree $1$. If $X\cap Y$ has Morley rank $2$, then $\widetilde{X}=\widetilde{Y}$.

Proof.

Let $g\in G$. If $g$ belongs to $\widetilde{X\cap Y}$, then we have ${\mathrm{rk\,}}\mathscr{L}(g,X\cap Y)=1$. Since $X$ has Morley degree 1 and $X\cap Y$ has Morley rank 2, the set $X\setminus Y$ has Morley rank at most 1, and the set $\mathscr{L}(g,X\setminus Y)$ has Morley rank at most 1. Thus $\mathscr{L}(g,X)$ has Morley rank 1, and $g$ belongs to $\widetilde{X}$.

Conversely, if $g\in \widetilde{X}$, then $\mathscr{L}(g,X)$ has Morley rank 1, so $\mathscr{L}(g,X\cap Y)\subseteq \mathscr{L}(g,X)$ has Morley rank at most 1. Then Lemma 4.6 gives $g\in \widetilde{X\cap Y}$. This shows that $\widetilde{X\cap Y}=\widetilde{X}$. In the same way, we obtain $\widetilde{X\cap Y}=\widetilde{Y}$, so $\widetilde{X}=\widetilde{Y}$.

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For each $a\in G$, let $\mathscr{L}(a)=\mathscr{L}(a,G)$ be the (definable) set of lines containing $a$. Moreover, we note that $\mathscr{L}(1)=\mathscr{B}$.

Lemma 4.9.

Let $\Lambda _0$ be a definable subset of $\Lambda$. If ${\mathrm{rk\,}}\bigcup \Lambda _0=2$, then we have ${\mathrm{rk\,}}(\mathscr{L}(g)\cap \Lambda _0)\leq 1$ for each $g\in G$.

Moreover, if further ${\mathrm{rk\,}}\Lambda _0=2$, then the set $\{g\in G\nobreakspace |\nobreakspace {\mathrm{rk\,}}(\mathscr{L}(g)\cap \Lambda _0)=1\}$ has Morley rank $2$.

Proof.

We show that ${\mathrm{rk\,}}(\mathscr{L}(g)\cap \Lambda _0)\leq 1$ for each $g\in G$. Let $g\in G$ and $l_g:\bigcup (\mathscr{L}(g)\cap \Lambda _0)\setminus \{g\}\to \mathscr{L}(g)\cap \Lambda _0$ be the map defined by $l_g(x)=l(g,x)$. Since each line has Morley rank 1, the preimage of each element of $\mathscr{L}(g)\cap \Lambda _0$ has Morley rank 1. Consequently, we have

$${\mathrm{rk\,}}(\mathscr{L}(g)\cap \Lambda _0)= {\mathrm{rk\,}}\bigcup (\mathscr{L}(g)\cap \Lambda _0)-1\leq {\mathrm{rk\,}}\bigcup \Lambda _0-1=1,$$

as desired.

We suppose further that ${\mathrm{rk\,}}\Lambda _0=2$, and we show that $\{g\in G\nobreakspace |\nobreakspace {\mathrm{rk\,}}(\mathscr{L}(g)\cap \Lambda _0)=1\}$ has Morley rank 2. Let $U=\bigcup \Lambda _0$, $A=\{(u,\lambda )\in U\times \Lambda _0\nobreakspace |\nobreakspace u\in \lambda \}$, and let $f:A\to \Lambda _0$ be the map defined by $f(u,\lambda )=\lambda$. Then $A$ and $f$ are definable, and the preimage $f^{-1}(\lambda )$ of each $\lambda \in \Lambda _0$ has Morley rank ${\mathrm{rk\,}}\lambda =1$, so ${\mathrm{rk\,}}A=1+{\mathrm{rk\,}}\Lambda _0=3$. Now let $h:A\to U$ be the map defined by $h(u,\lambda )=u$. It is a definable map, and the preimage $h^{-1}(u)$ of each $u\in U$ has Morley rank of either 0 or 1 by the previous paragraph.

But the preimage of $U_0=\{u\in U\nobreakspace |\nobreakspace {\mathrm{rk\,}}h^{-1}(u)=0\}$ has Morley rank

$${\mathrm{rk\,}}h^{-1}(U_0)={\mathrm{rk\,}}U_0\leq {\mathrm{rk\,}}U=2<{\mathrm{rk\,}}A,$$

so the preimage of $U_1=\{u\in U\nobreakspace |\nobreakspace {\mathrm{rk\,}}h^{-1}(u)=1\}$ has Morley rank 3. Hence we obtain ${\mathrm{rk\,}}U_1=3-1=2$. Moreover, we note that

$$U_1=\{u\in U\nobreakspace |\nobreakspace {\mathrm{rk\,}}(\mathscr{L}(u)\cap \Lambda _0)=1\}=\{g\in G\nobreakspace |\nobreakspace {\mathrm{rk\,}}(\mathscr{L}(g)\cap \Lambda _0)=1\},$$

so $\{g\in G\nobreakspace |\nobreakspace {\mathrm{rk\,}}(\mathscr{L}(g)\cap \Lambda _0)=1\}$ has Morley rank 2.

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Proposition 4.10.

Let $X$ be a definable subset of $G$ of Morley rank $2$ and Morley degree $1$ . Then ${\mathrm{rk\,}}\widetilde{X}=2$ if and only if $\Lambda _X$ has Morley rank $2$.

In this case, $\Lambda _X$ and $\widetilde{X}$ have Morley degree $1,$ and $\widetilde{X}$ contains a generic definable subset of $X$.

Proof.

We consider the definable set $A=\{(x,\lambda )\in \widetilde{X}\times \Lambda _X\nobreakspace |\nobreakspace x\in \lambda \}$ and the definable maps $l_1:A\to \widetilde{X}$ and $l_2:A\to \Lambda _X$ defined by $l_1(x,\lambda )=x$ and $l_2(x,\lambda )=\lambda ,$ respectively. By Proposition 4.4, the preimage $l_1^{-1}(g)$ of each element $g$ of $\widetilde{X}$ has Morley rank 1, so ${\mathrm{rk\,}}A=1+{\mathrm{rk\,}}\widetilde{X}$. Moreover, the preimage $l_2^{-1}(\lambda )$ of each $\lambda \in \Lambda _X$ has Morley rank at most 1, so ${\mathrm{rk\,}}A\leq 1+{\mathrm{rk\,}}\Lambda _X$. Then we obtain ${\mathrm{rk\,}}\widetilde{X}\leq {\mathrm{rk\,}}\Lambda _X$. In particular, it follows from Lemma 4.3 that if ${\mathrm{rk\,}}\widetilde{X}=2$, then ${\mathrm{rk\,}}\Lambda _X=2$. Hence we may assume that ${\mathrm{rk\,}}\Lambda _X=2$.

At this stage, Lemma 4.3 gives ${\mathrm{rk\,}}\bigcup \Lambda _X=2$, and by Lemma 4.9 and Proposition 4.4, we obtain ${\mathrm{rk\,}}\widetilde{X}=2$. Moreover, it follows from Corollary 4.5 that $\widetilde{X}$ has Morley degree 1 and that $X\cap \widetilde{X}$ is a generic definable subset of $X$ contained in $\widetilde{X}$.

We show that the Morley degree of $\Lambda _X$ is 1. Let $l_0:\{(x,y)\in X\times X\nobreakspace |\nobreakspace x\neq y\}\to \Lambda$ be the definable map defined by $l_0(x,y)=l(x,y)$. Since the Morley degree of $X$ is 1, that of $\{(x,y)\in X\times X\nobreakspace |\nobreakspace x\neq y\}$ is 1 too. For each $\lambda \in \Lambda _X$, we have ${\mathrm{rk\,}}l_0^{-1}(\lambda )={\mathrm{rk\,}}((\lambda \cap X)\times (\lambda \cap X))=2$. Since ${\mathrm{rk\,}}\Lambda _X=2$, we obtain

$${\mathrm{rk\,}}l_0^{-1}(\Lambda _X)=2+{\mathrm{rk\,}}\Lambda _X=4={\mathrm{rk\,}}\{(x,y)\in X\times X\nobreakspace |\nobreakspace x\neq y\},$$

and since the Morley degree of $\{(x,y)\in X\times X\nobreakspace |\nobreakspace x\neq y\}$ is 1, the Morley degree of $l_0^{-1}(\Lambda _X)$ is 1 too. Now the Morley degree of $\Lambda _X$ is 1 by Fact 2.2.

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Lemma 4.11.

Let $g$ be a nontrivial element such that $g=[u,v]$ for $(u,v)\in G\times G$. Then we have $\{x\in G\nobreakspace |\nobreakspace [x,v]=g\}=C_G(v)u$ and $\{y\in G\nobreakspace |\nobreakspace [u,y]=g\}=C_G(u)v$. In particular, they are two lines and have Morley rank $1$ and Morley degree $1$.

Proof.

The equalities are obvious. Moreover, by Fact 2.3 the sets $C_G(v)u$ and $C_G(u)v$ are two lines, and they have Morley rank 1 and Morley degree 1.

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Lemma 4.12.

For each $a\in G$, the set $a^G\cap B$ has exactly one element.

Proof.

We may assume $a\neq 1$. By Fact 2.3(5), there is $g\in G$ such that $a^g$ belongs to $B$. If $a^h\in B$ for $h\in G$, then $a$ is a nontrivial element of $B^{g^{-1}}\cap B^{h^{-1}}$. By Fact 2.3(4), we obtain $B^{g^{-1}}=B^{h^{-1}},$ and $h^{-1}g$ belongs to $N_G(B)=B$. But $B$ is abelian (Fact 2.3(2)), so $h^{-1}g$ centralizes $a^h$, and $a^h=(a^h)^{h^{-1}g}=a^g$. Hence $a^G\cap B=\{a^g\}$.

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The following result isolates a step of the proof of Theorem 4.14. Its proof and that of Theorem 4.14 were originally a lot more complicated, and Bruno Poizat provided a simplification.

For each $g\in G$, we consider the following definable subset of $G$:

$$X(g)=\{x\in G\nobreakspace |\nobreakspace \exists y\in G,\nobreakspace [x,y]=g\}.$$

Proposition 4.13.

For each nontrivial element $g$ of $G$, the set $X(g)$ has Morley rank at most $2$.

Proof.

We assume toward a contradiction that $X(g)$ has Morley rank 3. Then the Morley rank of $X(g^z)$ is 3 for each $z\in G$. We recall that, by Fact 2.3, the conjugacy class $g^G$ of $g$ has Morley rank ${\mathrm{rk\,}}g^G={\mathrm{rk\,}}G-{\mathrm{rk\,}}C_G(g)=2.$

We consider $V=\{(x,y)\in G\times G\nobreakspace |\nobreakspace [x,y]\in g^G\}$ and the definable surjective map $f:V\to g^G$ defined by $f(x,y)=[x,y]$. For each $z\in G$, we have

$$f^{-1}(g^z)=\{(x,y)\in G\times G\nobreakspace |\nobreakspace [x,y]=g^z\},$$

and by Lemma 4.11 this set has Morley rank ${\mathrm{rk\,}}f^{-1}(g^z)={\mathrm{rk\,}}X(g^z) +1=3 +1=4$, so ${\mathrm{rk\,}}V=4+{\mathrm{rk\,}}g^G=6$. Since $G\times G$ is a connected group of Morley rank 6, the set $V$ is a definable generic subset of $G\times G$, and there is $(x,y)\in V$ such that $(y,x)$ belongs to $V$. Thus $[x,y]\in g^G$ and its inverse $[y,x]\in g^G$ are conjugate, and they are equal by Lemma 4.12, contradicting that $G$ has no involution (Fact 2.3(6)).

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Theorem 4.14.

There is a plane in $G$.

Proof.

It is sufficient to show that there is a definable subset $X$ of $G$ satisfying the following properties:

(1)

its Morley rank is 2 and its Morley degree is 1;

(2)

$\Lambda _X$ has Morley rank 2.

Indeed, by Proposition 4.10, for such a subset $X$, the set $\widetilde{X}$ has Morley rank 2 and Morley degree 1, and it contains a generic definable subset of $X$. At this stage, Lemma 4.8 shows that $Y=\widetilde{X}$ is a plane.

We fix a nontrivial element $g$ such that $g=[u,v]$ for $(u,v)\in G\times G$.

1. For each $x\in X(g)$, there are infinitely many lines containing $x$ and contained in $X(g)$.

Since $x$ belongs to $X(g)$, there is $y\in G$ such that $[x,y]=g$. We note that, since $g$ is nontrivial, $x$ and $y$ are nontrivial, and we have $C_G(x)\neq C_G(y)$. In particular $C_G(x)y$ is a line, and it does not contain 1. Thus, for each $c\in C_G(x)$, the set $l_c=C_G(cy)x$ is a line, and by Lemma 4.11 we have $[r,cy]=[x,cy]=[x,y]=g$ for each $r\in l_c$. So $l_c$ is a line containing $x$ and contained in $X(g)$.

If $l_c=l_d$ for two elements $c$ and $d$ of $C_G(x)$, then we have $C_G(cy)=C_G(dy)$, and $C_G(cy)$ is a line containing $cy$ and $dy$. But $C_G(x)y$ is another line containing $cy$ and $dy$, and we have $C_G(x)y\neq C_G(cy)$ because $C_G(x)y$ does not contain $1$. Hence Lemma 3.3 gives $c=d$, and $\{l_c\in \Lambda\nobreakspace |\nobreakspace c\in C_G(x)\}$ is an infinite family of lines containing $x$ and contained in $X(g)$.

2. ${\mathrm{rk\,}}X(g)=2$.

By part 1 the set $X(g)$ contains infinitely many lines, so it has Morley rank at least 2 (Lemma 3.6) and by Proposition 4.13 it has Morley rank 2.

3. ${\mathrm{rk\,}}\Lambda _{X(g)}=2$.

By Lemma 4.3 the set $\Lambda _{X(g)}$ has Morley rank at most 2. Since $X(g)$ is infinite by part 2, for each positive integer $n$ we can find $n$ distinct elements $x_1,\ldots ,x_n$ in $X(g)$. By part 1 the set $\Lambda _i=\{\lambda \in \Lambda _{X(g)}\nobreakspace |\nobreakspace x_i\in \lambda \}$ is infinite for each $i$. We may assume that its Morley rank is 1 for each $i$. Then, since there are finitely many lines containing two distinct elements among $x_1,\ldots ,x_n$ (Lemma 3.3), the union $\bigcup _{i=1}^n\Lambda _i$ has Morley rank 1 and Morley degree at least $n$. This implies that $\Lambda _{X(g)}$ does not have Morley rank 1, so ${\mathrm{rk\,}}\Lambda _{X(g)}=2$.

4. Conclusion.

By part 2 the set $X(g)$ has Morley rank 2. Let $d$ be its Morley degree. Then $X(g)$ is the disjoint union of definable subsets $X_1,\ldots ,X_d$ of Morley rank 2 and Morley degree 1.

For each element $\lambda$ of $\Lambda _{X(g)}$, since $\lambda \cap X(g)$ is infinite and since $\lambda$ has Morley rank 1 and Morley degree 1, there is a unique $i\in \{1,\ldots ,d\}$ such that $\lambda \cap X_i$ is infinite, that is $\lambda \in \Lambda _{X_i}$. Thus, each $\lambda \in \Lambda _{X(g)}$ belongs to a unique definable set $\Lambda _{X_i}$ for $i\in \{1,\ldots ,d\}$. Hence $\Lambda _{X(g)}$ is the disjoint union of $\Lambda _{X_1},\ldots ,\Lambda _{X_d}$, and there exists $i\in \{1,\ldots ,d\}$ such that ${\mathrm{rk\,}}\Lambda _{X_i}=2$. Now the set ${X_i}$ satisfies the conditions (1) and (2) of the beginning of our proof, so $\widetilde{X_i}$ is a plane.

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5. A projective space?

In this section we analyze planes. We remember that by Theorem 4.14 the group $G$ has a plane, and that by Corollary 4.7 any plane has Morley degree 1. The initial goal of this section was to show that if $X$ and $Y$ are two distinct planes, then $\Lambda _X\cap \Lambda _Y$ has a unique element. However, along the way, we obtain our final contradiction.

Definition 5.1.

For each line $\lambda$ we consider the following subset of $\Lambda$:

$$\mathscr{L}(\lambda )=\{\, m\in \Lambda \mid \text{$\lambda \cap m$ is not empty}\,\}.$$

Lemma 5.2.

For any line $\lambda$ the set $\mathscr{L}(\lambda )$ is definable and it has Morley rank $3$ and Morley degree $1$.

Proof.

We consider the definable map $f:\lambda \times (G\setminus \lambda )\to \mathscr{L}(\lambda )\setminus \{\lambda \}$ defined by $f(x,g)=l(x,g)$. By Lemma 3.3 for each $m\in \mathscr{L}(\lambda )\setminus \{\lambda \}$, there is a unique element $x$ in $\lambda \cap m$. Moreover, for any $g\in G\setminus \lambda$, we have $f(x,g)=m$ if and only if $g\in m\setminus \{x\}$. Consequently, we have ${\mathrm{rk\,}}f^{-1}(m)={\mathrm{rk\,}}m=1$, and

$${\mathrm{rk\,}}\mathscr{L}(\lambda )={\mathrm{rk\,}}(\lambda \times (G\setminus \lambda ))-1=3.$$

Furthermore, since $\lambda$ and $G$ have Morley degree 1, the Morley degree of $\lambda \times G$ and $\lambda \times (G\setminus \lambda )$ is 1, and the Morley degree of $\mathscr{L}(\lambda )\setminus \{\lambda \}$ and $\mathscr{L}(\lambda )$ is 1 too (Fact 2.2).

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Lemma 5.3.

Let $X$ be a plane, and let $\lambda \in \Lambda _X$. Then $\mathscr{L}(\lambda )\cap \Lambda _X$ has Morley rank $2.$

Proof.

Since $\lambda$ belongs to $\Lambda _X$, the set $\lambda \cap X$ is infinite, and since $\lambda$ is a line, we have ${\mathrm{rk\,}}(\lambda \cap X)=1$. We consider the definable set

$$\mathscr{A}=\{(x,m)\in (\lambda \cap X)\times \Lambda _X\nobreakspace |\nobreakspace m\neq \lambda ,\nobreakspace x\in m\},$$

and the definable maps $p:\mathscr{A}\to \lambda \cap X$ and $q:\mathscr{A}\to \Lambda _X$ are defined by $p(x,m)=x$ and $q(x,m)=m,$ respectively. By Proposition 4.4, the set $p^{-1}(x)$ has Morley rank 1 for each $x\in \lambda \cap X$, so ${\mathrm{rk\,}}\mathscr{A}=1+{\mathrm{rk\,}}(\lambda \cap X)=2$.

Moreover, each $m\in \Lambda _X\setminus \{\lambda \}$ contains at most one element of $\lambda$ (Lemma 3.3), so $q$ is an injective map and its image has Morley rank ${\mathrm{rk\,}}\mathscr{A}=2$. But the image of $q$ is contained in $(\mathscr{L}(\lambda )\cap \Lambda _X)\setminus \{\lambda \}$, and we have ${\mathrm{rk\,}}\Lambda _X\leq 2$ (Lemma 4.3), hence $\mathscr{L}(\lambda )\cap \Lambda _X$ has Morley rank 2.

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Lemma 5.4.

Let $\lambda _1$ and $\lambda _2$ be two distinct lines. Then $\mathscr{L}(\lambda _1)\cap \mathscr{L}(\lambda _2)$ has Morley rank $2$ and Morley degree $1$.

Proof.

Let $A=\{(x,y)\in \lambda _1\times \lambda _2\nobreakspace |\nobreakspace x\not \in \lambda _1,\, y\not \in \lambda _2\}$, and let $f:A\to (\mathscr{L}(\lambda _1)\cap \mathscr{L}(\lambda _2))\setminus \{\lambda _1,\lambda _2\}$ be the map defined by $f(x,y)=l(x,y)$. This map is definable and bijective by Lemma 3.3. Since $\lambda _1$ and $\lambda _2$ are two lines, the sets $\lambda _1\times \lambda _2$ and $A$ have Morley rank 2 and Morley degree 1, and since $f$ is a definable bijection, $\mathscr{L}(\lambda _1)\cap \mathscr{L}(\lambda _2)$ has Morley rank 2 and Morley degree 1.

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Proposition 5.5.

If $X$ and $Y$ are two distinct planes, then $\Lambda _X\cap \Lambda _Y$ has at most one element.

Proof.

Suppose toward a contradiction that $\lambda _1$ and $\lambda _2$ are two distinct elements of $\Lambda _X\cap \Lambda _Y$. By Lemma 5.3, the sets $\mathscr{L}(\lambda _1)\cap \Lambda _X$ and $\mathscr{L}(\lambda _2)\cap \Lambda _X$ have Morley rank 2. But $\Lambda _X$ has Morley rank 2 and Morley degree 1 by Proposition 4.10, hence $\mathscr{L}(\lambda _1)\cap \mathscr{L}(\lambda _2)\cap \Lambda _X$ has Morley rank 2. In the same way, $\mathscr{L}(\lambda _1)\cap \mathscr{L}(\lambda _2)\cap \Lambda _Y$ has Morley rank 2. Thus, since $\mathscr{L}(\lambda _1)\cap \mathscr{L}(\lambda _2)$ has Morley rank 2 and Morley degree 1 (Lemma 5.4), the set $\Lambda _X\cap \Lambda _Y$ has Morley rank 2.

Since $\Lambda _X\cap \Lambda _Y$ is infinite, the set $U=\bigcup (\Lambda _X\cap \Lambda _Y)$ has Morley rank at least 2 by Lemma 3.7, and since $U$ is contained in $\bigcup \Lambda _X$, its Morley rank is exactly 2 (Lemma 4.3). Now the set $Z=\{g\in G\nobreakspace |\nobreakspace {\mathrm{rk\,}}(\mathscr{L}(g)\cap \Lambda _X\cap \Lambda _Y)=1\}$ has Morley rank 2 by Lemma 4.9. But Proposition 4.4 says that $Z$ is contained in $X\cap Y$, hence $X\cap Y$ has Morley rank 2 and Lemma 4.8 gives $X=Y$, a contradiction.

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From now on, we try to show that the set $\Lambda _X\cap \Lambda _Y$ has exactly one element. However, the final contradiction will appear earlier.

Corollary 5.6.

Let $X$ be a plane, and let $(a,b)\in G\times G$. Then the following assertions are equivalent:

•

$aXb=X.$

•

$a\Lambda _X b=\Lambda _X.$

•

$aXb\cap X$ has Morley rank $2$.

Proof.

We note that $aXb$ is a plane and that $a\Lambda _X b=\Lambda _{aXb}$. If $aXb\cap X$ has Morley rank 2, then $aXb=X$ by Lemma 4.8, and if $aXb=X$, then we have $a\Lambda _X b=\Lambda _{aXb}=\Lambda _X$. Moreover, if $a\Lambda _X b=\Lambda _X$, then we have $\Lambda _{aXb}=\Lambda _X$ and $aXb=X$ by Proposition 5.5, so $aXb\cap X=X$ has Morley rank 2.

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By Fact 2.4, if $A$ is a Borel subgroup distinct from $B$, then ${\mathrm{rk\,}}(ABA)=3$. The following result is slightly more general, and its proof is different.

We recall that if a group $H$ of finite Morley rank acts definably on a set $E$, then the stabilizer of any definable subset $F$ of $E$ is defined to be

$${\mathrm{Stab}}F=\{h\in H\nobreakspace |\nobreakspace {\mathrm{rk\,}}((h\cdot F)\Delta F)<{\mathrm{rk\,}}(F)\},$$

where $\Delta$ stands for the symmetric difference. It is a definable subgroup of $H$ by 5, Lemma 5.11.

Lemma 5.7.

Let $A$ and $C$ be two Borel subgroups distinct from $B$. Then ${\mathrm{rk\,}}(ABC)=3$.

Proof.

We consider the action of $G$ on itself by left multiplication. Then we have $b\cdot BC=BC$ for each $b\in B$, so $B$ is contained in ${\mathrm{Stab}}(BC)$.

We assume toward a contradiction that $C$ is contained in ${\mathrm{Stab}}(BC)$. Since $BC$ has Morley rank 2 and Morley degree 1 (Fact 2.4), we have ${\mathrm{rk\,}}(cBC\setminus BC)\leq 1$ for each $c\in C$, and since ${\mathrm{rk\,}}C=1$, we obtain ${\mathrm{rk\,}}(CBC\setminus BC)\leq 2$ and ${\mathrm{rk\,}}(CBC)=2$, contradicting Fact 2.4. Consequently, $C$ is not contained in ${\mathrm{Stab}}(BC)$, and since ${\mathrm{Stab}}(BC)$ contains $B$, Fact 2.3 implies that ${\mathrm{Stab}}(BC)=B$.

We assume toward a contradiction that ${\mathrm{rk\,}}(ABC)\neq 3$. Since ${\mathrm{rk\,}}(BC)=2$, we have ${\mathrm{rk\,}}(ABC)=2$ and $ABC$ is a disjoint union of finitely many definable subsets $E_1,\ldots ,E_k$ of Morley rank 2 and Morley degree 1. For each $a\in A$, the set $aBC$ has Morley rank ${\mathrm{rk\,}}(BC)=2$ and Morley degree 1, so there exists a unique $i\in \{1,\ldots ,k\}$ such that ${\mathrm{rk\,}}(aBC\cap E_i)=2$. Since $A$ is infinite, there are $i\in \{1,\ldots ,k\}$ and two distinct elements $a$ and $a'$ of $A$ such that ${\mathrm{rk\,}}(aBC\cap E_i)={\mathrm{rk\,}}(a'BC\cap E_i)=2$. Since $E_i$ has Morley degree 1, the Morley rank of $aBC\cap a'BC$ is 2, and we obtain ${\mathrm{rk\,}}(a'^{-1}aBC\cap BC)=2$. But $BC$ has Morley degree 1, hence $a'^{-1}a$ belongs to ${\mathrm{Stab}}(BC)=B$. Thus $a'^{-1}a$ belongs to $A\cap B=\{1\}$ (Fact 2.3(4)), contradicting that $a$ and $a'$ are distinct. So we have ${\mathrm{rk\,}}(ABC)=3$, as desired.

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Corollary 5.8.

Let $A$ and $C$ be two distinct Borel subgroups. Then ${\mathrm{rk\,}}(BA\cap BC)=1$.

Proof.

We may assume $A\neq B$ and $C\neq B$. By Fact 2.4, we have

$$1={\mathrm{rk\,}}B\leq {\mathrm{rk\,}}(BA\cap BC)\leq {\mathrm{rk\,}}(BA)=2.{}$$

We assume toward a contradiction that ${\mathrm{rk\,}}(BA\cap BC)=2$. Since $BC$ has Morley rank 2 and Morley degree 1 (Fact 2.4), the set $E=BC\setminus BA$ has Morley rank at most 1. Consequently, $EA$ has Morley rank at most ${\mathrm{rk\,}}E+{\mathrm{rk\,}}A=2$, and since $(BA\cap BC)A\subseteq BA$ has Morley rank 2, we obtain ${\mathrm{rk\,}}(BCA)= {\mathrm{rk\,}}(EA\cup (BA\cap BC)A)=2$, contradicting that $BCA$ has Morley rank 3 (Lemma 5.7).

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Lemma 5.9.

For any plane $X$, we have $BX\neq X$ and $XB\neq X$.

Proof.

We assume toward a contradiction that $BX=X$ for a plane $X$. Let $x\in X$. Since $X$ is a plane, Proposition 4.4 gives ${\mathrm{rk\,}}(\mathscr{L}(x,X)\cap \Lambda _X)=1$, so $\mathscr{L}(x,X)\cap \Lambda _X$ is infinite. But each line containing $x$ has the form $B^ux$ for $u\in G$, hence there exist $u\not \in B$ and $v\not \in B$ such that $B^u\neq B^v$, and such that $B^ux$ and $B^vx$ belong to $\mathscr{L}(x,X)\cap \Lambda _X$. In particular, there is a cofinite subset $S$ of $B$ such that $S^ux$ and $S^vx$ are contained in $X$.

Now, since $BX=X$, the sets $BS^ux$ and $BS^vx$ are contained in $X$. By Fact 2.4 the set $BB^{u}$, and so $Bu^{-1}B$, has Morley rank 2, and since $Bu^{-1}(B\setminus S)$ is a finite union of lines, the set $Bu^{-1}(B\setminus S)$ has Morley rank 1 (Lemma 3.6), and $Bu^{-1}S$ has Morley rank 2. Thus, the sets $BS^ux=Bu^{-1}Sux$ and $BS^vx=Bv^{-1}Svx$ are subsets of $X$ of Morley rank 2, and since the Morley degree of $X$ is 1, the set $BS^ux\cap BS^vx$ has Morley rank 2. This implies that ${\mathrm{rk\,}}(BB^u\cap BB^v)=2$, contradicting Corollary 5.8. Now we have $BX\neq X$ and in the same way, we show that $XB\neq X$.

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Corollary 5.10.

For any plane $X$, the stabilizer of $X$ for the action of $G$ on itself by left multiplication is finite.

Proof.

By Corollary 5.6, we have ${\mathrm{Stab}}X=\{a\in G\nobreakspace |\nobreakspace aX=X\}$. If ${\mathrm{Stab}}X$ is infinite, then it contains a Borel subgroup, contradicting Lemma 5.9.

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Proposition 5.11.

Let $X$ be a plane. Then for each plane $Y$, there exist a unique $a\in G$ and a unique $b\in G$ such that $Y=aX=Xb$.

Proof.

We fix $\alpha \in G$, and we consider the following definable subset of $\Lambda$:

$$A=\{(\lambda _1,\lambda _2)\in \Lambda \times \Lambda\nobreakspace |\nobreakspace \alpha \in \lambda _1\cap \lambda _2,\,\lambda _1\neq \lambda _2\}.$$

We show that $A$ has Morley rank 4 and Morley degree 1. Let $U=\{(x,y)\in G\times G\nobreakspace |\nobreakspace y\not \in l(x,\alpha )\}$. Then $U$ is a generic definable subset of $G\times G$, and it has Morley rank 6 and Morley degree 1. Let $f:U\to A$ be the definable surjective map defined by $f(x,y)=(l(x,\alpha ),l(y,\alpha ))$. Since each line has Morley rank 1, the preimage of each $(\lambda _1,\lambda _2)\in A$ has Morley rank ${\mathrm{rk\,}}\lambda _1+{\mathrm{rk\,}}\lambda _2=2$, and the set $A$ has Morley rank ${\mathrm{rk\,}}U-2=4$ and Morley degree 1 (Fact 2.2).

For each plane $P$, we consider the following definable set:

$$A_P=\{(\lambda _1,\lambda _2)\in \Lambda \times \Lambda\nobreakspace |\nobreakspace \alpha \in \lambda _1\cap \lambda _2,\,\lambda _1\neq \lambda _2,\, \exists a\in G,\,a^{-1}\lambda _1\in \Lambda _P,\,a^{-1}\lambda _2\in \Lambda _P \}.$$

We show that the set $A_X$ is a generic definable subset of $A$. Indeed, for each $a\in \alpha X^{-1}$, we have $\alpha \in aX$ and ${\mathrm{rk\,}}(\mathscr{L}(\alpha )\cap \Lambda _{aX})=1$ by Proposition 4.4, so the definable set

$$L_{aX}=\{(\lambda _1,\lambda _2)\in \Lambda _{aX}\times \Lambda _{aX}\nobreakspace |\nobreakspace \alpha \in \lambda _1\cap \lambda _2,\,\lambda _1\neq \lambda _2\}{}$$

has Morley rank $2\ {\mathrm{rk\,}}(\mathscr{L}(\alpha )\cap \Lambda _{aX})=2$. But $\alpha X^{-1}$ has Morley rank ${\mathrm{rk\,}}X=2,$ and it follows from Proposition 5.5 that $L_{aX}\cap L_{bX}=\emptyset$ for any two elements $a$ and $b$ of $\alpha X^{-1}$ such that $aX\neq bX$. Moreover, for each $a\in \alpha X^{-1}$, there are finitely many elements $b\in \alpha X^{-1}$ such that $aX=bX$ (Corollary 5.10). Hence the set $A_X=\bigcup _{a\in \alpha X^{-1}}L_{aX}$ has Morley rank ${\mathrm{rk\,}}\alpha X^{-1}+2=4$, and it is a generic definable subset of $A$.

In the same way, $A_Y$ is a generic definable subset of $A$, so there exists $(\lambda _1,\lambda _2)\in A_X\cap A_Y$. Thus there exist two elements $u$ and $v$ of $G$ such that two distinct lines $\lambda _1$ and $\lambda _2$ belong to $\Lambda _{uX}\cap \Lambda _{vY}$, and we obtain $uX=vY$ by Proposition 5.5, so $Y=aX$ for $a=v^{-1}u$. In the same way, there exists $b\in G$ such that $Y=Xb$.

We show the uniqueness of $a$ and $b$. Let $S=\{g\in G\nobreakspace |\nobreakspace gX=X\}$. It is a finite subgroup of $G$ by Corollary 5.10. For each $\alpha \in G$, the previous paragraph gives $\beta \in G$ such that $\alpha X=X\beta$. Then, for each $s\in S$, we have $s(\alpha X)=s(X\beta )=X\beta =\alpha X$, and we obtain $s^\alpha X=X$ and $s^\alpha \in S$. Thus any element $\alpha \in G$ normalizes the finite subgroup $S$, and since $G$ is a simple group, $S$ is trivial. This proves the uniqueness of $a$, and in the same way we obtain the uniqueness of $b$.

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By the previous result, the set of planes is $\mathscr{P}=\{aX\nobreakspace |\nobreakspace a\in G\}$, and it identifies with $G$. Thus, the set of planes is uniformly definable and has Morley rank 3.

Lemma 5.12.

There exists $a\in G$ such that $X^a\neq X$.

Proof.

We assume toward a contradiction that $X^a=X$ for each $a\in G$. Then for each $uBv\in \Lambda _X$ and each $a\in G$, we have

$$(uBv)^a\in \Lambda _X^a=\Lambda _{X^a}=\Lambda _X.$$

Since ${\mathrm{rk\,}}\Lambda _X=2$ (Proposition 4.10), the line $uBv$ is a Borel subgroup (Lemma 3.4), and by conjugacy of Borel subgroups we obtain $\Lambda _X=\mathscr{B}$. Now we have $\bigcup \Lambda _X=G$, so ${\mathrm{rk\,}}\bigcup \Lambda _X=3$, contradicting Lemma 4.3.

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From now on, we are ready for the final contradiction. Initially, it was more complicated, but Poizat proposed a simplification by introducing the inverted plane.

Proof.

First we note that for each plane $Y$, the set $y^{-1}Y$ is a plane containing 1, and the set $Y^{-1}$ is a plane too. We fix a plane $X$ containing 1. By Proposition 5.11, there is a bijective map $\mu :G\to G$ defined by $xX=X\mu (x)$, and $\mu$ is definable since the set $\mathscr{P}$ of planes is uniformly definable. Moreover, for each $(a,b)\in G\times G$, we have $X\mu (ab)=abX=aX\mu (b)=X\mu (a)\mu (b)$, so $\mu$ is an automorphism of $G$.

Since $X=\widetilde{X}$ contains $1$, there are infinitely many Borel subgroups in $\Lambda _X$. Let $B_1$ and $B_2$ be two distinct Borel subgroups belonging to $\Lambda _X$. Then $B_1$ and $B_2$ belong to $\Lambda _{X^{-1}}$ too, and we have $X=X^{-1}$ by Proposition 5.5. In the same way, since the plane $x^{-1}X$ contains 1 for each $x\in X$, we have $x^{-1}X=(x^{-1}X)^{-1}=X^{-1}x=Xx$ for each $x\in X$. Thus $\mu (x^{-1})=x$ for each $x\in X$, and since $X=X^{-1}$, we obtain $\mu ^2(x)=x$ for each $x\in X$.

But $X$ is a definable subset of $G$ of Morley rank 2, hence $G$ is generated by $X$, and $\mu$ is an involutive automorphism of $G$. Thus $\mu$ is the identity map by Fact 2.5, contradicting Lemma 5.12.

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Remark 5.13.

After Lemma 5.12 we were ready for a new step to provide a structure of projective space over $G$, which was the initial goal of our section. Indeed, in the first version of this paper, we have shown that, if $X$ and $Y$ are two distinct planes, then $\Lambda _X\cap \Lambda _Y$ has a unique element.