Fields interpretable in superrosy groups with NIP (the non-solvable case)

Krzysztof Krupiński^{∗}

Abstract

Let G be a group definable in a monster model ℭ of a rosy theory satisfying NIP. Assume that G has hereditarily finitely satisfiable generics and 1 < U^{þ}(G) < ∞. We prove that if G acts definably on a definable set of U^{þ}-rank 1, then, under some general assumption about this action, there is an infinite field interpretable in ℭ. We conclude that if G is not solvable-by-finite and it acts faithfully and definably on a definable set of U^{þ}-rank 1, then there is an infinite field interpretable in ℭ. As an immediate consequence, we get that if G has a definable subgroup H such that U^{þ}(G) = U^{þ}(H) + 1 and G/ ⋂_{g∈G} H^{g} is not solvable-by-finite, then an infinite field interpretable in ℭ also exists.

0 Introduction

The paper is in some sense a continuation of [7]. But now we concentrate mainly on the existence of fields interpretable in groups. The general motivating question is

Question 1 For a given infinite, pure group ⟨G, ·⟩ (say of finite dimension, whatever the dimension means), does there exist an infinite field interpretable in ⟨G, ·⟩?

In this paper, the notion of dimension will be U^{þ}-rank, and so G will be rosy. Recall that in the stable (finite Morley rank) context U^{þ}-rank coincides with Lascar U-rank, and in the o-minimal context it coincides with the o-minimal dimension.

Notice that in order to have a positive answer to the above question, G cannot be abelian-by-finite. Indeed, otherwise ⟨G, ·⟩ would be 1-based of finite Lascar U-rank, so it would not interpret an infinite field (to see this, one should use the fact that abelian groups are 1-based together with some facts from [11, Chapters 2, 4], e.g. [11, Proposition 6.4]).

Assuming that G is not abelian-by-finite, one has three cases to consider:

∗Research supported by the Polish Government grant N201 032 32/2231 and by the NSF grant DMS-0300639

02000 Mathematics Subject Classification: 03C45, 03C60

0Key words and phrases: superrosy group, interpretable field, non independence property

(1) G is not solvable-by-finite,

(2) G is solvable-by-finite but not nilpotent-by-finite,

(3) G is nilpotent-by-finite but not abelian-by-finite.

[10, Corollary 5.1] tells us that the answer to Question 1 is positive whenever G is a non-abelian-by-finite group definable in an o-minimal structure.

When G is of finite Morley rank, the answer to Question 1 in Case (2) is positive. In Case (1), it is open, but there are some partial results (assuming that G is not a bad group). Finally, in Case 3 the answer is negative by [3], but there are also some positive partial answers, e.g. in [14, 6]. For example, [6, Theorem 2.3] tells us that the answer is positive if the commutator subgroup of G is torsion-free.

A general goal is to answer Question 1 in a more general context than finite Morley rank or o-minimal structures, namely for G being dependent and of finite U^{þ}-rank. For some technical reason, in this paper, we will additionally assume that G has hereditarily fsg. This still covers the situations when G is superstable of finite U-rank and when G is a definably compact group definable in an o-minimal expansion of a real closed field. The definitions of rosiness, NIP, fsg, and other relevant notions are given in Section 1.

The main result of [7] says that if G is a dependent group of finite U^{þ}-rank, then Question 1 has a positive answer in Case (2).

Of course, Baudisch’s theorem (see [3]) implies that the answer in Case (3) is negative for the class of dependent groups of finite U^{þ}-rank. However, a tiny modification of the proof of [6, Theorem 2.3] yields a positive answer under the additional assumption that the commutator subgroup of G is torsion-free. In fact, even more is true: every locally nilpotent, non-abelian group with icc on centralizers and with torsion-free commutator subgroup interprets an infinite field. Indeed, the proof of [6, Theorem 2.3] produces an integral domain F interpretable in ⟨G, ·⟩ whose additive group is a subgroup of G′. So, the field of fractions of F is also interpretable in ⟨G, ·⟩, and since G′ is torsion-free, this field is of characteristic 0 and so infinite.

In this paper, we concentrate on the existence of a field both in Case (1) and also in the presence of a definable action of a definable group on a definable set of U^{þ}-rank 1.

For the next three results, we work in ℭ^{eq} where ℭ is a monster model of a (rosy) theory T satisfying NIP.

One of the main conclusions of this paper is the following theorem concerning Case (1) and generalizing the part of Cherlin’s result (see [12, Corollary 3.28]) concerning the existence of a field (we do not generalize the part yielding the description of G).

Theorem 2 Suppose T is rosy with NIP, and G is a definable group having hereditarily fsg and with U^{þ}(G) = n + 1 < ∞. Assume that G has a definable subgroup H of U^{þ}-rank n, and G/ ⋂_{g∈G} H^{g} is not solvable-by-finite. Then there is an infinite field interpretable in ℭ [in ⟨G, ·⟩, if H is definable in ⟨G, ·⟩].

As in [12], in order to prove the above theorem, we investigate definable actions of definable groups on sets of U^{þ}-rank 1. However, in the finite Morley rank situation, it was sufficient to consider actions on strongly minimal sets, whereas in our case, we need to work with sets of U^{þ}-rank 1. This and the fact that we do not have definable connected components makes our situation more complicated and calls for different arguments.

The main effort is put into proving the following theorem, generalizing the part of the Hrushovski’s result (see [12, Theorem 3.27]) concerning the existence of a field (we do not generalize the part yielding the description of permutation groups). A similar result has also been proved for o-minimal structures in [8, Theorem 1.5]. In the whole paper, if G is a group acting on a set S and R ⊆ S, then G_{R} denotes the pointwise stabilizer of R; if R = {s}, then G_{s} := G_{R}.

Theorem 3 Assume T is rosy with NIP, and let G be a definable group having hereditarily fsg and with 1 < U^{þ}(G) < ∞. Assume that G acts definably on a definable set S of U^{þ}-rank 1 so that there is s ∈ S for which no finite index, definable subgroup of G_{s} has normalizer of finite index in G. Then there is an infinite field interpretable in ℭ; in fact, it is interpretable in the two-sorted structure with sorts G and S equipped with the group operation on G and the action of G on S.

In particular, assuming the first sentence of the theorem, if G acts definably on a definable set S of U^{þ}-rank 1 so that at least one orbit is infinite, and G does not have a definable subgroup H with U^{þ}(H) = U^{þ}(G) − 1 and whose normalizer is of finite index in G, then an infinite interpretable field exists.

From Theorem 3, we will get the following corollary, which implies Theorem 2.

Theorem 4 Assume T has NIP. Let G be a definable group having hereditarily fsg and with U^{þ}(G) < ∞. Assume that G acts faithfully and definably on a definable set S of U^{þ}-rank 1. If G is not solvable-by-finite, then there is an infinite field interpretable in ℭ; in fact, it is interpretable in the two-sorted structure with sorts G and S equipped with the group operation on G and the action of G on S. Moreover, if T is rosy, then U^{þ}(G) < ω.

It is worth mentioning that in the proofs of the above three theorems, the only places where the fsg assumption is used are the applications of [4, Theorem 2] saying that any dependent group of U^{þ}-rank 2 and satisfying hereditarily fsg is solvable-by-finite. So, if one was able to remove the fsg assumption from [4, Theorem 2], then it could be automatically removed from all the results of this paper.

On the other hand, keeping the fsg assumption in the above theorems and analyzing our proofs, one can check that the NIP assumption can be replaced by the condition that G has icc.

1 Definitions and basic observations

In this section, we work in ℭ^{eq} where ℭ is a monster model of a theory T in a language L.

First we recall some things about rosy theories. For details on rosy theories, the reader is referred to [1, 5, 9], and on rosy groups to [4].

T is rosy if there is a ternary relation | ⌣^{∗} on small subsets of ℭ^{eq} satisfying all the basic properties of forking independence in simple theories except for the Independence Theorem. Such a relation will be called an independence relation.

A formula δ(x, a) strongly divides over A if the formula is not almost over A and {δ(x, a′)}_{a′⊧tp(a/A)} is k-inconsistent for some k ∈ ℕ.

We say that δ(x, a) þ-divides over A if we can find some tuple c such that δ(x, a) strongly divides over Ac.

A formula þ-forks over A if it implies a (finite) disjunction of formulas which þ-divide over A.

We say that the type p(x) þ-divides over A if there is a formula in p(x) which þ-divides over A; þ-forking is similarly defined. We say that a is þ-independent from b over A, denoted a | ⌣_{A}^{þ} b, if tp (a/Ab) does not þ-fork over A.

In rosy theories, þ-independence is the weakest independence relation in the sense that a | ⌣_{C}^{∗} b implies a | ⌣_{C}^{þ} b for any independence relation | ⌣^{∗}.

In [5], the following has been noticed.

Fact 1.1 T is rosy iff | ⌣^{þ} has local character.

We define U^{þ}-rank by means of | ⌣^{þ} in the same way as Lascar U-rank is defined in terms of | ⌣. In rosy theories, this rank shares many nice properties of U-rank in stable theories. In particular, it satisfies Lascar Inequalities.

If D is A-definable, we put U^{þ}(D) = sup{U^{þ}(d/A) : d ∈ D}. If T is rosy, this definition, not depending on A, provides a good, general notion of dimension of a definable set. Without rosiness, it is not clear whether it depends on A (because it is not clear whether U^{þ}-rank is preserved under taking non-þ-forking extensions). So, whenever U^{þ}(D) is considered without rosiness, we assume that the set A over which D is defined is distinguished. Sometimes A is clear from the context, e.g. computing U^{þ}(ℭ), we have A = ∅. If ℭ is many-sorted, then U^{þ}(ℭ) is the supremum of the U^{þ}-ranks of all sorts. In particular, U^{þ}(ℭ) < ∞ means that the U^{þ}-rank of every sort is less than ∞.

Some nice properties of U^{þ}-rank in rosy theories are listed in Section 1 of [7]. But even without rosiness, we still have some good properties. For example, analyzing the proof of Lascar Inequalities [13, Theorem 5.1.6] and using [9, Lemma 2.1.6], we get that U^{þ}(ab/A) ≤ U^{þ}(a/bA) ⊕ U^{þ}(b/A) holds in any theory. Also, it follows easily that if a ∈ acl(Ab), then U^{þ}(b/A) ≥ U^{þ}(a/A).

Below we prove two basic facts, which are not written down explicitly anywhere. They are a piece of folklore that has already been used in [4] and [7], and will be used in this paper in order to skip the assumption of rosiness of the theory T in some results. The first remark is pretty obvious. The proof of the second one, however, turns out to be a bit more complicated and it leads to some basic questions formulated right below it.

Remark 1.2 Any theory with U^{þ}(ℭ) < ∞ is rosy.

Proof. By assumption, U^{þ}(a) < ∞ for every a ∈ ℭ. Thus, we get U^{þ}(a/A) < ∞ for any a ∈ ℭ and A ⊆ ℭ^{eq}. We know that U^{þ}(ab/A) ≤ U^{þ}(a/bA)⊕U^{þ}(b/A) holds in any theory (without rosiness). Therefore, for any finite tuple a in ℭ and any A ⊆ ℭ^{eq}, we have U^{þ}(a/A) < ∞.

Now if a = [a_{0}]_{E} is imaginary (a_{0} is a real tuple), then a ∈ acl^{eq}(a_{0}). So, U^{þ}(a/A) ≤ U^{þ}(a_{0}/A) for any A ⊆ ℭ^{eq}.

We have proved that every type in ℭ^{eq} has an ordinal U^{þ}-rank. This implies that there is no infinite sequence p_{0} ⊆ p_{1} ⊆ … of þ-forking extensions, and so | ⌣^{þ} has local character. We finish using Fact 1.1. ■

The theories of ordinal U^{þ}-rank (i.e. with U^{þ}(ℭ) < ∞) are called superrosy. We say that T is of U^{þ}-rank α if U^{þ}(ℭ) = α.

It has been noticed in [2] that rosiness is preserved under taking reducts. Our second remark is related to this observation.

Remark 1.3 Assume M with the universe M is interpretable in ℭ. Let L′ be the signature of M. Working in the language L′, we will use the subscript L′. Otherwise we work in the language L. U_{M}^{þ} stands for U^{þ}-rank computed in M.

(i) If T is (super)rosy, so is T h_{L′}(M). (ii) Assume T is rosy. Consider any m ∈ M^{eq} and A ⊆ M^{eq}. If U_{M}^{þ}(tp_{L′}(m/A)) = n < ω, then there is a completion p of tp_{L′}(m/A) over A in the sense of ℭ such that U^{þ}(p) ≥ n. In particular, if T′ is a reduct of T and T is of U^{þ}-rank n < ω, then T′ is of U^{þ}-rank at most n. (iii) The conclusion of (ii) holds with the assumption that T is rosy replaced by U^{þ}(M) < ∞. (iv) If U^{þ}(M) < ∞, then T h_{L′}(M) is superrosy.

Proof. Assume for simplicity that M is interpretable in ℭ over ∅. Of course, M^{eq} is also interpretable in ℭ over ∅.

Claim Let a, A, B be from M^{eq} and suppose that tp_{L′}(a/AB) þ-divides over A in the sense of M. Then there are a′ and B′ in M^{eq} such that tp_{L′}(a′B′/A) = tp_{L′}(aB/A), and for any realization a′′ of tp_{L′}(a′/AB′) one has that tp_{L}(a′′/AB′) þ-divides over A in the sense of ℭ.

Proof of Claim. By the assumption, there is a formula φ(x, y) ∈ L′ and b ∈ M^{eq} such that φ(x, b) ∈ tp_{L′}(a/AB) and φ(x, b) þ-divides over A in the sense of M. Thus, there is c ∈ M^{eq} such that tp_{L′}(b/Ac) is non-algebraic and {φ(x, b′) : b′ ⊧ tp_{L′}(b/Ac)} is k-inconsistent for some k. Hence, there is b′ ⊧ tp_{L′}(b/Ac) such that tp_{L}(b′/Ac) is non-algebraic. We conclude that φ(x, b′) þ-divides over A in the sense of ℭ.

To finish, choose any automorphism f of M fixing A pointwisely and mapping b to b′, and put a′ = f(a), B′ = f[B]. □

We argue that under the assumption of (ii) or (iii), T h_{L′}(M) is rosy. The case when T is rosy follows easily using local þ-ranks (see [2, Corollary 5.3]). Now consider the case when T is not necessarily rosy, but U^{þ}(M) < ∞. Then the U^{þ}-rank of M equipped with the induced structure (i.e. with predicates for the traces of all ∅- definable subsets of the appropriate sorts of ℭ) is also less than ∞, and hence by Remark 1.2, M with the induced structure is rosy. Thus, by [2, Corollary 5.3], T h_{L′}(M) is rosy.

Now we give an inductive, with respect to n, proof of (ii) and (iii). The case n = 0 is obvious. For the induction step, assume U_{M}^{þ}(tp_{L′}(a/A)) ≥ n + 1 for some a, A in M^{eq} and n ∈ ω. Then there is B ⊇ A such that a is þ-dependent on B over A in the sense of M and U_{M}^{þ}(tp_{L′}(a/B)) ≥ n. Extending B and using rosiness of T h_{L′}(M) (in order to have that taking non-þ-forking extensions preserves U^{þ}-rank), we can assume that tp_{L′}(a/B) þ-divides over A in the sense of M. By the Claim, we get that there are a′B′ ⊧ tp_{L′}(aB/A) such that for any a′′ ⊧ tp_{L′}(a′/B′), tp_{L}(a′′/B′) þ-divides, and hence þ-forks over A in the sense of ℭ. So, by the inductive hypothesis, U^{þ}(tp_{L}(a′′/A)) ≥ n + 1 for some a′′ ⊧ tp_{L′}(a′/B′).

The proof of (i) is similar. As in [2], using local þ-ranks, we get that T h_{L′}(M) is rosy. Now, suppose for a contradiction that T is superrosy, but T h_{L′}(M) is not superrosy. Using rosiness of T h_{L′}(M), we get an infinite sequence {∅} = B_{0} ⊆ B_{1} ⊆ B_{2} ⊆ ⋯ ⊆ M^{eq} and a ∈ M such that tp_{L′}(a/B_{i+1}) þ-divides over B_{i} in the sense of M. Using recursively the Claim, we find a_{i}, i ∈ ω, and an increasing sequence of sets B′_{i}, i ∈ ω, in M^{eq} so that:

(a) a_{0} = a,

(b) a_{i}B′_{i} ⊧ tp_{L′}(aB_{i}),

(c) tp_{L′}(a_{i}/B′_{i}) ⊆ tp_{L′}(a_{i+1}/B′_{i+1}),

(d) for every a′_{i+1} ⊧ tp_{L′}(a_{i+1}/B′_{i+1}), tp_{L}(a′_{i+1}/B′_{i+1}) þ-divides, and hence þ-forks over B′_{i} in the sense of ℭ.

By compactness and (c), there is a′ ∈ M realizing all types tp_{L′}(a_{i}/B′_{i}), i ∈ ω. Thus, by (d), U^{þ}(a′) = ∞, a contradiction with superrosiness of ℭ. The proof of (iv) is almost the same as the proof of (i) ■

It is not clear, however, whether the above remark holds when n is an infinite ordinal (in particular, whether taking reducts always decreases U^{þ}-rank, even for infinite ordinal ranks). A similar remark can also be proved for SU-rank in simple theories, and one can also ask whether taking reducts necessarily decreases SU-rank (even for infinite ordinal ranks).

In Section 2, T will be assumed to have the non independence property (NIP).

Definition 1.4 We say that T has the NIP if there is no formula φ(x, y) and sequence ⟨a_{i}⟩_{i<ω} such that for every w ⊆ ω, there is b_{w} such that ⊧ φ(a_{i}, b_{w}) iff i ∈ w.

In the paper, whenever H and G are groups, then H < G means that H is a (not necessarily proper) subgroup of G.

Recall that we say that a definable group G has icc (the uniform chain condition on intersections of uniformly definable subgroups) if for every formula φ, there is n_{φ} ∈ ω such that any chain of intersections of φ-definable subgroups of G has length at most n_{φ}.

Recall that the centralizer connected component of a definable group G is the intersection of all centralizers C(g), g ∈ G, of finite index in G. The group G is said to be centralizer connected if it is equal to its centralizer connected component. If we have icc (or only icc on centralizers), then the centralizer connected component is a definable, finite index subgroup of G. In the proof of Theorem 2.1, we will need an easy fact that if G is centralizer connected and Z(G) is finite, then G/Z(G) is centerless.

For this paper, an important consequence of NIP and rosiness is [4, Proposition 1.7]:

Fact 1.5 Suppose T is rosy and has NIP. Then any definable group G has icc.

We will also use [7, Theorem 3]:

Fact 1.6 Assume T has NIP and G is a definable group of finite U^{þ}-rank. Assume that G is solvable-by-finite but not nilpotent-by-finite. Then there is an infinite field interpretable in ⟨G, ·⟩.

In [7, Theorem 3], rosiness of T was additionally assumed, but we see, using Remark 1.3, that it is not necessary.

One more property that we are going to assume is fsg (finitely satisfiable generics). As was mentioned at the end of the introduction, we do not use this property anywhere in our proofs except for the applications of [4, Theorem 2]. That is why, in this paper, we restrict ourselves only to giving the definition of fsg and recalling [4, Theorem 2].

Definition 1.7 A definable group G defined by a formula G(x) has finitely satisfiable generics (or fsg) if there is a global type p containing G(x) and a model M ≺ ℭ of cardinality less than the degree of saturation of ℭ, such that for all g, gp is finitely satisfiable in M (i.e. each formula in gp defines a set which intersects M). We say that G has hereditarily fsg if every definable subgroup of G also has fsg.

It is easy to check that if G has fsg and N is a definable, normal subgroup of G, then G/N also has fsg.

Fact 1.8 Assume T has NIP and G is a definable group satisfying hereditarily fsg. If U^{þ}(G) = 2, then G is solvable-by-finite.

2 Getting fields

In this section, T T T is a theory with NIP, and we work in ℭ^{eq} where ℭ is a monster model of T. Whenever we have a definable group G acting definably on a definable set S, M will denote the two-sorted structure with sorts G and S equipped with the group operation on G and the action of G on S.

First we will show a variant of Theorem 3, which will allow us to prove all three theorems from the introduction.

Theorem 2.1 Assume T is rosy, and let G be a definable group having hereditarily fsg and with 1 < U^{þ}(G) < ∞. Assume that G acts definably on a definable set S of U^{þ}-rank 1 so that there is s ∈ S for which no infinite, definable subgroup H of G_{s} of finite index in the intersection of stabilizers of some points in S and with U^{þ}(G/H) ≤ 2 has normalizer of finite index in G. Then there is an infinite field interpretable in ℭ.

The following strengthening of Theorem 2.1 can be proved in almost the same way as Theorem 2.1 is proved below.

Theorem 2.2 Assume T is rosy, and let G be a definable group having hereditarily fsg and with 1 < U^{þ}(G) < ∞. Assume that G acts definably on a definable set S of U^{þ}-rank 1 so that there is s ∈ S for which no infinite, definable in M^{eq} subgroup H of G_{s} of finite index in the intersection of stabilizers of some points in S and with U^{þ}(G/H) ≤ 2 has normalizer of finite index in G. Then there is an infinite field interpretable in M.

Proof of Theorem 2.1. The proof is by induction on U^{þ}(G). First consider the case U^{þ}(G) = 2. By Fact 1.8, G is solvable-by-finite (this is the only place in the proof where the fsg assumption is used). So, using Fact 1.6, either we get a field interpretable in the pure group G and we are done, or else G is nilpotent-by-finite. Since by rosiness and NIP we have icc on centralizers, using [7, Corollary 3.3(ii)] and replacing G by a definable subgroup of finite index, we can assume that G is nilpotent and centralizer connected. Then Z(G) is infinite (otherwise G/Z(G) would be centerless, a contradiction). Since U^{þ}(G) = 2 > U^{þ}(S), G_{s} is infinite. So, by assumption, [G : G_{s}] must be also infinite, and so Gs is infinite. Replacing S by Gs, the action becomes transitive. Since G_{S} ⊲ G, by assumption, G_{S} is finite. Thus, replacing G by G/G_{S}, we can assume that the action is additionally faithful.

Notice that there is no non-trivial z ∈ Z(G) and x ∈ S with zx = x. Otherwise z(gx) = (zg)x = (gz)x = g(zx) = gx for every g ∈ G, and so z ∈ G_{S} by transitivity of the action. This is a contradiction with faithfulness.

We have proved that G_{s} ∩ Z(G) = {e}. But N(G_{s}) ⊇ ⟨G_{s}, Z(G)⟩ and both G_{s} and Z(G) are infinite, which implies that U^{þ}(N(G_{s})) = 2. Thus, [G : N(G_{s})] is finite, a contradiction.

Now assume that U^{þ}(G) ≥ 3 and the theorem holds for groups of smaller U^{þ}-rank.

As in the base step, we see that Gs is infinite. So, replacing S by Gs, we can assume that the action is transitive. Since G_{S} ⊲ G, by assumption, we get U^{þ}(G/G_{S}) ≥ 3. So, replacing G by G/G_{S}, we can assume that the action of G on S is also faithful. This gives us easily that for any infinite definable subgroup H of G, there is t ∈ S with Ht infinite. Indeed, otherwise by icc, {e} = H_{S} = ⋂_{t∈S} H_{t} is an intersection of finitely many subgroups H_{t}, all of finite index in H. So, H is finite, a contradiction.

Claim 1 U^{þ}(G) < ω.

Proof of Claim 1. Suppose for a contradiction that U^{þ}(G) ≥ ω. We have that [G : G_{s}] is infinite, so U^{þ}(Gs) = 1, and hence U^{þ}(G/G_{s}) = U^{þ}(Gs) = 1. Thus, U^{þ}(G_{s}) + 1 = U^{þ}(G_{s}) + U^{þ}(G/G_{s}) ≤ U^{þ}(G) ≤ U^{þ}(G_{s}) ⊕ U^{þ}(G/G_{s}) = U^{þ}(G_{s}) + 1. So, we get U^{þ}(G) = U^{þ}(G_{s}) + 1. Therefore,

U^{þ}(G) > U^{þ}(G_{s}) ≥ ω.

Since G_{s} is infinite, there is s_{1} ∈ S with G_{s}s_{1} infinite and so of U^{þ}-rank 1. As above, we get

U^{þ}(G_{s}) > U^{þ}(G_{ss_{1}}) ≥ ω.

We continue this procedure and obtain s_{i} ∈ S, i ≥ 1, so that

U^{þ}(G) > U^{þ}(G_{s}) > U^{þ}(G_{ss_{1}}) > U^{þ}(G_{ss_{1}s_{2}}) > …,

a contradiction. □

Let H be the collection of all definable, finite index subgroups H of G_{s} with the property that for any g ∈ G, if [H : H ∩ H^{g}] < ω, then H = H^{g}.

Claim 2 H is nonempty.

Proof of Claim 2. By icc, the intersection of all G_{s}^{g} such that [G_{s} : G_{s} ∩ G_{s}^{g}] < ω is a definable, finite index subgroup of G_{s}. Denote this subgroup by H. We see that if [H : H ∩ H^{g}] < ω, then H = H ∩ H^{g}, and so H < H^{g}. This implies H^{g^{−1}} < H. On the other hand, U^{þ}(H^{g^{−1}}) = U^{þ}(H), so by Lascar Inequalities for groups, [H : H^{g^{−1}}] < ω. Thus, H^{g^{−1}} = H, so H = H^{g}. We have seen that H ∈ H. □

Consider any H ∈ H. It is infinite, so there is t ∈ S with Ht infinite. Since U^{þ}(S) = 1, there are only finitely many infinite orbits on S under H; denote them by o_{0}, …, o_{n}. For i = 0, …, n, put

H_{i} = H_{o_{i}} ⊲ H

and

δ(H) = min_{0≤i≤n} U^{þ}(H_{i}).

From now on, we choose H ∈ H with maximal possible δ(H) (we can do it by Claims 1 and 2).

Claim 3 Assume δ(H) ≤ U^{þ}(H) − 2. Then there is i such that H/H_{i} acting on o_{i} satisfies the assumptions of the theorem. Thus, since U^{þ}(H/H_{i}) ≤ U^{þ}(H) < U^{þ}(G), by the inductive hypothesis, there is an infinite, interpretable field.

Proof of Claim 3. Let H_{i_{1}}, …, H_{i_{k}} be all H_{i}’s for which U^{þ}(H_{i}) = δ(H). Suppose that for every j = 1, …, k, H/H_{i_{j}} acting on o_{i_{j}} does not satisfy the assumption of the theorem (it is enough to show that this leads to a contradiction). Then for every j = 1, …, k, there is a nonempty S_{j} ⊆ o_{i_{j}} and an infinite, definable, finite index subgroup K_{j}/H_{i_{j}} of (H/H_{i_{j}})_{S_{j}} = H_{S_{j}}/H_{i_{j}} such that [H/H_{i_{j}} : N_{H/H_{i_{j}}} (K_{j}/H_{i_{j}})] is finite. This implies [H : N_{H}(K_{j})] < ω. So, by an argument similar to the proof of Claim 2, we can find L ∈ H such that L < ⋂_{j=1}^{k} N_{H}(K_{j}) (namely, define L as the intersection of all (⋂_{j=1}^{k} N_{H}(K_{j}))^{g} such that [⋂_{j=1}^{k} N_{H}(K_{j}) : ⋂_{j=1}^{k} N_{H}(K_{j})∩(⋂_{j=1}^{k} N_{H}(K_{j}))^{g}] < ω.). By the definition of H, L ⊲ H.

We will be done if we prove that δ(L) > δ(H) because this will be a contradiction with maximality of δ(H).

Since [H : L] < ω, each o_{i} is a union of finitely many infinite orbits o_{0}^{i} ∪ ⋯ ∪ o_{n_{i}}^{i} under L. For i = 0, …, n and j = 0, …, n_{i}, define

L_{j}^{i} = L_{o_{j}^{i}}.

We need to prove that for any i ∈ {0, …, n} and j ∈ {0, …, n_{i}}, U^{þ}(L_{j}^{i}) > δ(H). So, consider any i, j as above.

Case 1 i ∉ {i_{1}, …, i_{k}}. Put H_{j}^{i} = H_{o_{j}^{i}}. Then H > H_{j}^{i} > H_{i}. Since [H : L] < ω, we also have that L∩H_{j}^{i} = L_{j}^{i} has finite index in H_{j}^{i}. Thus, U^{þ}(L_{j}^{i}) ≥ U^{þ}(H_{i}) > δ(H) (the last inequality is true because of the assumption of Case 1).

Case 2 i = i_{l} for some l ∈ {1, …, k}. Then δ(H) = U^{þ}(H_{i}), so it is enough to show that U^{þ}(L_{j}^{i}) > U^{þ}(H_{i}).

Take any x ∈ S_{l}. Then x ∈ o_{m}^{i} for some m ∈ {0, …, n_{i}}.

Subclaim U^{þ}(L_{m}^{i}) > U^{þ}(H_{i}).

Proof of Subclaim. By the choice of K_{j}’s, [K_{l} : H_{i}] ≥ ω. On the other hand, since [H : L] < ω and K_{l} < H, we have [K_{l} : L ∩ K_{l}] < ω. From these two observations, we obtain

(∗) U^{þ}(L ∩ K_{l}) > U^{þ}(H_{i}).

As L normalizes K_{l}, L∩K_{l} ⊲L. We also know that L∩K_{l} stabilizes x. Therefore, for any h_{1} ∈ L and h_{2} ∈ L ∩ K_{l}, we have h_{2}h_{1}x = h_{1}h′_{2}x = h_{1}x for some h′_{2} ∈ L ∩ K_{l}. Moreover, Lx = o_{m}^{i}. Hence, L ∩ K_{l} < L_{m}^{i}. In virtue of (∗), the proof of the Subclaim is completed. □

There is h ∈ H such that ho_{m}^{i} ∩ o_{j}^{i} ≠ ∅. Since L ⊲ H, we easily get ho_{m}^{i} = o_{j}^{i}. So, for any g ∈ L, we have the following equivalences:

g ∈ L_{m}^{i} ⟺ (∀y ∈ o_{m}^{i})(hgy = hy) ⟺ (∀z ∈ o_{j}^{i})(g^{h}z = z) ⟺ g^{h} ∈ L_{j}^{i}.

This means that (L_{m}^{i})^{h} = L_{j}^{i}. Hence, by the Subclaim, U^{þ}(L_{j}^{i}) = U^{þ}(L_{m}^{i}) > U^{þ}(H_{i}), which completes the proof of Claim 3. □

By Claim 3, in order to finish the proof of the theorem, it remains to show that the condition δ(H) = U^{þ}(H)−1 leads to a contradiction. So, assume δ(H) = U^{þ}(H)−1. Then U^{þ}(H_{i}) = U^{þ}(H) − 1 = U^{þ}(G) − 2 for every i = 0, …, n.

Claim 4 There are infinite, definable normal subgroups L_{i} of H for i = 0, …, n such that:

(i) L_{i} is a finite index subgroup of H_{i}, (ii) for any g ∈ G, L_{i}^{g} acts trivially on go_{i}, (iii) for any i, j ∈ {0, …, n} and g ∈ G, if [L_{i} : L_{i} ∩ L_{j}^{g}] < ω, then L_{i} = L_{j}^{g}.

Proof of Claim 4. We define L_{i} as the intersection of all groups H_{j}^{g} for which [H_{i} : H_{i} ∩H_{j}^{g}] is finite. By icc, each L_{i} is definable and of finite index in H_{i}. Since H_{i} ⊲H, we see that L_{i} ⊲ H. Thus, (i) and so (ii) holds.

To see (iii), consider any i, j ∈ {0, …, n} and g ∈ G such that [L_{i} : L_{i} ∩ L_{j}^{g}] < ω. Then the property [H_{i} : L_{i}] < ω implies [H_{i} : L_{i}∩L_{j}^{g}] < ω. We also know that L_{i}∩L_{j}^{g} is an intersection of groups of the form H_{k}^{a}. Therefore, L_{i} = L_{i} ∩ L_{j}^{g}, i.e. L_{i} < L_{j}^{g}. Thus, L_{i}^{g^{−1}} < L_{j}. Since U^{þ}(L_{i}^{g^{−1}}) = U^{þ}(L_{i}) = U^{þ}(H_{i}) = U^{þ}(H) − 1 = U^{þ}(L_{j}), we get [L_{j} : L_{i}^{g^{−1}}] < ω. On the other hand, L_{i}^{g^{−1}} is an intersection of groups of the form H_{k}^{a}. Hence, L_{j} = L_{i}^{g^{−1}}, i.e. L_{i} = L_{j}^{g}. □

By transitivity of the action of G on S, we can choose g ∈ G so that gs ∈ o_{0}. Since o_{0} is an infinite orbit under H and Hs = {s}, we see that g ∉ N(H). But H ∈ H, so we conclude that [H : H ∩ H^{g}] is infinite.

Now we will show that for every i ∈ {0, …, n},

(!) [L_{0} : L_{0} ∩ L_{i}^{g}] ≥ ω and [L_{i}^{g} : L_{0} ∩ L_{i}^{g}] ≥ ω.

If any of the above conditions is false, then [L_{0} : L_{0} ∩L_{i}^{g}] < ω or [L_{i} : L_{i} ∩L_{0}^{g^{−1}}] < ω. Thus, by Claim 4, L_{0} = L_{i}^{g}. It follows that ⟨H, H^{g}⟩ < N(L_{0}). Since [H : H ∩ H^{g}] ≥ ω, we get [N(L_{0}) : H^{g}] ≥ ω. Hence, U^{þ}(N(L_{0})) ≥ U^{þ}(H^{g}) + 1 = U^{þ}(G). So,

[G : N(L_{0})] < ω.

On the other hand, the fact that H_{0} = G_{{s}∪o_{0}} ∩ H has finite index in G_{{s}∪o_{0}} and [H_{0} : L_{0}] < ω implies that

[G_{{s}∪o_{0}} : L_{0}] < ω.

Moreover, U^{þ}(L_{0}) = U^{þ}(G) − 2 ≥ 1, so

U^{þ}(G/L_{0}) = 2 and L_{0} is infinite.

All these three observations together give us a contradiction with the assumption of the theorem. So, (!) has been proved.

Notice that there are only finitely many orbits on S under H. We know that there are finitely many infinite orbits, so we need to check that there are finitely many finite orbits. For any a ∈ G, we have G_{as} = G_{s}^{a}. Thus, we see that Has being finite is equivalent to any of the following:

[H : H ∩ G_{s}^{a}] < ω ⟺ [H : H ∩ H^{a}] < ω ⟺ H = H^{a} ⟺ a ∈ N(H),

and the last condition implies as ∈ N(H)s. By the assumption of the theorem, [G : N(H)] ≥ ω. But also U^{þ}(H) = U^{þ}(G) − 1. Thus, [N(H) : H] < ω. So, from the previous computation, there are only finitely elements in S with finite orbits under H.

By the above observation, we can choose i ∈ {0, …, n} so that o_{0} ∩ go_{i} ≠ ∅. Choose an element g_{1}s ∈ o_{0} ∩ go_{i}.

Recall that two subgroups G_{1} and G_{2} of a given group are said to be commensurable (symbolically G_{1} ∼ G_{2}) if [G_{1} : G_{1} ∩ G_{2}] and [G_{2} : G_{1} ∩ G_{2}] are finite. ∼ is always an equivalence relation.

Since gs ∈ o_{0}, L_{0} < G_{s}^{g}. Also, H^{g} is a finite index subgroup of G_{s}^{g}. Hence,

(!!) L_{0} ∼ L_{0} ∩ H^{g}.

Since g_{1}s ∈ o_{0}, L_{0} < G_{s}^{g_{1}}. So, as above

(!!!) L_{0} ∼ L_{0} ∩ H^{g_{1}}.

Since g_{1}s ∈ go_{i} and L_{i}^{g} acts trivially on go_{i}, we get L_{i}^{g} < G_{s}^{g_{1}}. Therefore,

(!!!!) L_{i}^{g} ∼ L_{i}^{g} ∩ H^{g_{1}}.

We know that L_{0}^{g_{1}} ⊲ H^{g_{1}}. So, ⟨L_{0} ∩ H^{g_{1}}, L_{i}^{g} ∩ H^{g_{1}}⟩ < N(L_{0}^{g_{1}}). In particular, by (!!!!), L_{i}^{g} ∼ L_{i}^{g} ∩ H^{g_{1}} < N(L_{0}^{g_{1}}) ∩ H^{g}. Similarly, by (!!) and (!!!), L_{0} ∼ L_{0} ∩ H^{g_{1}} ∩ H^{g} < N(L_{0}^{g_{1}}) ∩ H^{g}. Using the last two observations together with (!), we get [N(L_{0}^{g_{1}}) ∩ H^{g} : L_{0} ∩ H^{g_{1}} ∩ H^{g}] ≥ ω. On the other hand, U^{þ}(L_{0} ∩ H^{g_{1}} ∩ H^{g}) = U^{þ}(L_{0}) = U^{þ}(H) − 1 = U^{þ}(H^{g}) − 1. So, we conclude that

(⋄) [H^{g} : N(L_{0}^{g_{1}}) ∩ H^{g}] < ω.

We claim that

(⋄⋄) [H^{g} : H^{g} ∩ H^{g_{1}}] ≥ ω.

Suppose it is not true. Then [H : H ∩ H^{g^{−1}g_{1}}] < ω. So, H = H^{g^{−1}g_{1}}, which implies H^{g} = H^{g_{1}}. Hence, H^{g}(g_{1}s) = g_{1}Hs = {g_{1}s}. But g_{1}s ∈ go_{i}, so there is t ∈ o_{i} such that g_{1}s = gt, and so H^{g}(g_{1}s) = gHt = go_{i} is infinite. This is a contradiction.

Of course, we have H^{g_{1}} ∪ (N(L_{0}^{g_{1}}) ∩ H^{g})) ⊆ N(L_{0}^{g_{1}}). So, by (⋄) and (⋄⋄), we get [N(L_{0}^{g_{1}}) : H^{g_{1}}] ≥ ω, and so [N(L_{0}) : H] ≥ ω. Since U^{þ}(H) = U^{þ}(G) − 1, we see that [G : N(L_{0})] < ω. It was shown in the proof of (!) that this gives us a contradiction with the assumption of the theorem. So, the proof of the theorem is completed. ■

Proof of Theorem 2.2. The proof is almost the same as the proof of Theorem 2.1. The only difference is that we have to work with objects definable in M^{eq} (whereas U^{þ}-ranks should still be computed in ℭ). ■

Before we turn to further results, let us make a few comments concerning Theorem 2.1.

Notice that, using only icc instead of the full NIP assumption, the above inductive proof always reduces the situation to a smaller U^{þ}-rank, and finally leads to the case U^{þ}(G) = 2. Then we get a field via Fact 1.6 or even [4, Theorem 4.5]. Since in [4, Theorem 4.5], it is enough to assume icc instead of NIP, we see that in Theorems 2.1 and 2.2, NIP can be also replaced by icc. Looking at the proofs that follows, this will easily imply that in Theorems 2, 3, 4, one can also replace NIP by icc (keeping the fsg assumption).

Notice that if we strengthen the assumption of Theorem 2.1 by dropping the condition U^{þ}(G/H) ≤ 2, then the existence of an infinite field interpretable in M follows easily from this theorem and Remark 1.3. To see this, it is enough to show that G acting on S considered in M satisfies the assumptions of the theorem. By Remark 1.3, it is clear that M is rosy with NIP, and working in M, we have that U_{M}^{þ}(G) < ∞, U_{M}^{þ}(S) = 1, G has hereditarily fsg, the action of G on S is definable, and no infinite, definable subgroup H of G_{s} of finite index in the intersection of stabilizers of some points in S has normalizer of finite index in G. The only remaining assumption is that U_{M}^{þ}(G) > 1. However, since by the assumption of the theorem G_{s} is infinite (otherwise 1 = U^{þ}(S) ≥ U^{þ}(Gs) = U^{þ}(G/G_{s}) ≥ 2, a contradiction) and [G : G_{s}] ≥ ω, we get U_{M}^{þ}(G) ≥ U_{M}^{þ}(G_{s}) + U_{M}^{þ}(G/G_{s}) ≥ 2.

Now we will apply Theorem 2.2 to get Theorem 3. In fact, we will use the weaker version of Theorem 2.2 in which the condition U^{þ}(G/H) ≤ 2 is dropped, and we will prove the following strengthening of Theorem 3.

Theorem 2.3 Assume T is rosy, and let G be a definable group having hereditarily fsg and with 1 < U^{þ}(G) < ∞. Assume that G acts definably on a definable set S of U^{þ}-rank 1 so that there is s ∈ S for which no finite index, definable in M^{eq} subgroup of G_{s} has normalizer of finite index in G. Then there is an infinite field interpretable in M.

Proof. By the assumption of the theorem, [G : G_{s}] ≥ ω. So, Gs is infinite. Hence, replacing S by Gs, we can assume that the action of G on S is transitive. Also by the assumption, U^{þ}(G/G_{S}) > 1. Indeed, if U^{þ}(G/G_{S}) = 1, then [G_{s} : G_{S}] < ω, which together with G_{S} ⊲ G contradicts the assumption. So, replacing G by G/G_{S}, we can assume that G acts faithfully on S. As in the proof of Theorem 2.1, this implies that U^{þ}(G) < ∞.

Let G be the collection of all finite index subgroups of G which are definable in M^{eq}.

Consider any H ∈ G. Since [G : H] < ω, Hs is infinite. By the assumption that U^{þ}(S) = 1, there are only finitely many infinite orbits under H; denote them by o_{0}, …, o_{n}. For i = 0, …, n, put

H_{i} = H_{o_{i}} ⊲ H

and

Δ(H) = max_{0≤i≤n} U^{þ}(H_{i}).

From now on, we choose H ∈ G with maximal possible Δ(H). Take i with Δ(H) = U^{þ}(H_{i}).

By the assumption of the theorem, Δ(H) < U^{þ}(H) − 1. Indeed, since G acts transitively on S, there is g ∈ G such that s ∈ go_{i}. As G_{s} > H_{i}^{g} ⊲ H^{g} and [G : H^{g}] < ω, by the assumption of the theorem, we get [G_{s} : H_{i}^{g}] ≥ ω, and so Δ(H) = U^{þ}(H_{i}) = U^{þ}(H_{i}^{g}) < U^{þ}(G) − 1 = U^{þ}(H) − 1.

Since H/H_{i} and the action of H/H_{i} on S are definable in M^{eq}, the following claim will finish the proof.

Claim H/H_{i} acting on o_{i} satisfies the assumption of Theorem 2.2.

Proof of Claim. Of course, U^{þ}(H/H_{i}) ≤ U^{þ}(H) = U^{þ}(G) < ω and U^{þ}(H/H_{i}) = U^{þ}(H) − U^{þ}(H_{i}) = U^{þ}(H) − Δ(H) > 1. Moreover, U^{þ}(o_{i}) = 1. Thus, if the assumption of Theorem 2.2 fails, there is a nonempty S_{i} ⊆ o_{i} and an infinite, definable in M^{eq}, finite index subgroup K_{i}/H_{i} of (H/H_{i})_{S_{i}} = H_{S_{i}}/H_{i} such that [H/H_{i} : N_{H/H_{i}}(K_{i}/H_{i})] is finite. This implies [H : N_{H}(K_{i})] < ω. Put N = N_{H}(K_{i}). We see that N ∈ G.

We will be done if we prove that Δ(N) > Δ(H) because this will be a contradiction with maximality of Δ(H).

Since [H : N] < ω, we have that o_{i} is the union of finitely many infinite orbits o_{0}^{i} ∪ ⋯ ∪ o_{n_{i}}^{i} under N. Take any s_{i} ∈ S_{i}. Then s_{i} ∈ o_{j}^{i} for some j. Put N_{j}^{i} = N_{o_{j}^{i}}.

We have that K_{i}⊲N and K_{i} stabilizes s_{i}. Hence, K_{i} stabilizes o_{j}^{i}, and so K_{i} < N_{j}^{i}. On the other hand, [K_{i} : H_{i}] ≥ ω. So, U^{þ}(N_{j}^{i}) > U^{þ}(H_{i}). Thus, we conclude that Δ(N) > Δ(H). ■

Observe that in the above proof, one can use Theorem 2.1 (even with the stronger assumption obtained by dropping the condition U^{þ}(G/H) ≤ 2) instead of Theorem 2.2. Indeed, it is clear because applying Remark 1.3 at the beginning of the above proof, we can assume that ℭ = M.

Now, we will show that if one considers the assumption U^{þ}(G) > 1 in a stronger sense, namely that for every (not only for some) set A of parameters over which G is defined, there is g ∈ G with U^{þ}(g/A) > 1, then the assumption that T is rosy can be eliminated from Theorem 2.3. Indeed, by Remark 1.3, M is rosy, and working in M, all assumptions of Theorem 2.3 except for U_{M}^{þ}(G) > 1 are clearly satisfied. So, it remains to show that U_{M}^{þ}(G) > 1. By the assumption of the theorem, [G : G_{s}] is infinite. Moreover, there is a definable bijection between G/G_{s} and o(s). So, computing U^{þ}-rank over a set of parameters over which everything is defined, U^{þ}(G/G_{s}) = 1. Thus, G_{s} is infinite because otherwise, computing U^{þ}-rank over some set of parameters, U^{þ}(G) = 1, a contradiction. So, working in M, G has an infinite, definable subgroup of infinite index (namely G_{s}). Therefore, U_{M}^{þ}(G) ≥ 2.

Now, using Theorem 3, we will prove Theorem 4.

Proof of Theorem 4. By Remark 1.3, we see that M is superrosy and U_{M}^{þ}(S) = 1. So, it is enough to prove Theorem 4 under the assumption that T is rosy.

To see that U^{þ}(G) < ω, we use icc and faithfullness of the action in the same way as in the proof of Theorem 2.1.

In order to get the existence of a field, we argue by induction on U^{þ}(G). Suppose the theorem is true for groups of U^{þ}-rank smaller than U^{þ}(G). By Remark 1.3, we can assume that ℭ = M (but even without this reduction, the argument below works using Theorem 2.3 insead of Theorem 3). Consider two cases.

Case 1 There is a definable (in M^{eq}) subgroup H of G such that U^{þ}(G) = U^{þ}(H)+1 and [G : N(H)] < ω. If H is solvable-by-finite, then using icc and [7, Remark 3.3(i)], H has a definable in ⟨H, ·⟩, normal subgroup H_{0} of finite index which is solvable. By icc, H_{1} := ⋂_{g∈N(H)} H_{0}^{g} is a finite index, definable (in M^{eq}) subgroup of H_{0} whose normalizer contains N(H). So, replacing H by H_{1}, we can assume that H is solvable. But U^{þ}(N(H)/H) = 1, so N(H)/H is solvable-by-finite in virtue of Fact 1.8 (in fact, it is even abelian-by-finite by [4, Theorem 1]). Thus, G is solvable-by-finite, a contradiction.

We have proved that H is not solvable-by-finite, and so H acting on S satisfies the assumptions of the theorem. On the other hand, U^{þ}(H) < U^{þ}(G). So, by the inductive hypothesis, we get an infinite field interpretable in M.

Case 2 Case 1 does not hold. Since G is not solvable-by-finite, U^{þ}(G) > 1. By faithfullness and icc, there is s ∈ S for which Gs is infinite, i.e. [G : G_{s}] ≥ ω. Then s witnesses that G acting on S satisfies the assumption of Theorem 3. So, we get an infinite field interpretable in M. ■

In fact, even Theorem 2.1 is strong enough to get Theorem 4. To see this, one should repeat the above proof of Theorem 4 modifying Case 1 in the following way: There is an infinite, definable subgroup H of G such that 1 ≤ U^{þ}(G) − U^{þ}(H) ≤ 2 and [G : N(H)] < ω.

Proof of Theorem 2. Put Z = ⋂_{g∈G} H^{g}. It is clear that G/Z acting on G/H by (aZ) · (gH) = agH satisfies the assumption of Theorem 4. ■

It was mentioned in the introduction that some of our theorems generalize the appropriate results about the finite Morley rank case. Now, we will explain why.

We claim that Theorem 3 generalizes the part of [12, Theorem 3.27] concerning the existence of a field. More precisely, we will show that the assumption of [12, Theorem 3.27] implies the assumption of Theorem 3.

Recall that in [12, Theorem 3.27], T is stable and G is a definable, transitive group of permutations of a strongly minimal set S. It is checked right before [12, Theorem 3.27] that the Morley rank of G is finite. Thus, U^{þ}(G) < ω. In order to get a field, it is also assumed in [12] that the Morley rank of G is at least 2. Choose any s ∈ S. Then G_{s} must be infinite. On the other hand, [G : G_{s}] is also infinite. Hence, U^{þ}(G) ≥ 2. So, in order to show that in this situation, the assumption of Theorem 3 is satisfied, it is enough to prove that there is no definable subgroup H of G_{s} such that [G : N(H)] < ω. Suppose for a contradiction that such an H exists. By transitivity of the action, strong minimality of S and the fact that [G : G^{0}] < ω, we get that G^{0} acts transitively on S. Thus, N(H) also acts transitively on S. Take any non-trivial h ∈ H. Then for any g ∈ N(H), we have h(gs) = g(h^{g^{−1}}s) = gs as h^{g^{−1}} ∈ G_{s}. So, h ∈ G_{S} = {e}, a contradiction.

Another remark from the introduction says that Theorem 2 generalizes the part of [12, Corollary 3.28] concerning the existence of a field. Indeed, in [12, Corollary 3.28], one has a simple group G of finite Morley rank n > 0 with a definable subgroup H of rank n − 1. Then G/H is strongly minimal. So, U^{þ}(G/H) = 1, and hence U^{þ}(G) = U^{þ}(H) + 1 < ω. Moreover, Z := ⋂_{g∈G} H^{g} is normal in G, so it is trivial as G is simple. Thus, G/Z = G. Simplicity of G also implies that G is not solvable-by-finite, and we see that the assumption of Theorem 2 is satisfied.

We finish the paper with some open questions concerning the structure of groups and their possible U^{þ}-ranks.

[12, Theorem 3.27] in the finite Morley rank case and [8, Theorem 1.5] in the o-minimal case describe the structure of the permutation groups in terms of an interpretable field. Notice that by the argument from the base induction step of the proof of Theorem 2.1, if U^{þ}(G) = 2 in Theorem 3, then G is not nilpotentby-finite. So, by [4, Theorem 4.5], after passing to a definable subgroup of finite index and quotienting by its finite center, G is definably the semidirect product of the additive and multiplicative groups of an algebraically closed field interpretable in ℭ, and moreover G = G^{00}. However, a description of G in Theorem 3 in the case U^{þ}(G) = 3 remains an open problem.

The assumptions of Theorem 2 or 4 imply that U^{þ}(G) ≥ 3, and the structure of G in both these theorems is also unknown.

Moreover, as in the finite Morley rank or o-minimal case, one could try to prove that the assumption of Theorem 3 together with transitivity and faithfulness of the action imply that U^{þ}(G) = 2 or U^{þ}(G) = 3. Similarly, in Theorem 4, and in Theorem 2 with the stronger assumption that G is simple, one could try to prove that U^{þ}(G) = 3.

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Instytut Matematyczny Uniwersytetu Wrocławskiego pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland. e-mail: kkrup@math.uni.wroc.pl and Mathematics Department University of Illinois at Urbana-Champaign 1409 W. Green Street, Urbana, IL 61801, USA.