Finite rank and pseudofinite groups

Question.

Are there infinite finitely generated groups which are pseudofinite?

Theorem 1.1.

There is no infinite finitely generated group which is elementarily equivalent to an ultraproduct of finite groups of a given Prüfer rank.

Fact 2.1.

Let G be a finitely generated pseudofinite group. Then the words wc,n and wn have finite width in G. Moreover, the subgroups G(n)=〈wc,n⁢(G)〉 and Gn=〈wn⁢(G)〉 are definable and have finite index.

Lemma 2.2.

Let G be a finitely generated pseudofinite group and assume that it contains an infinite definable definably simple non-abelian normal subgroup N. Then, the group N is a finitely generated pseudofinite group.

Proof.

Let G be a finitely generated pseudofinite group G and assume that it contains an infinite definably simple non-abelian normal subgroup N which is definable by some formula ϕ⁢(x,a¯). By Frayne’s theorem [2, Corollary 4.3.13], there exists some non-principal ultraproduct of an infinite family {Gi}i∈I of finite groups in which G is elementarily contained. Let a¯i be a representative of the equivalence class a¯ in Gi and let Ni be the set ϕ⁢(Gi,a¯i). Since by Łos’s theorem, almost every subset Ni is a normal subgroup of Gi, we may assume that every subset Ni is indeed a normal subgroup of Gi. Furthermore, as N is elementary equivalent to an ultraproduct of the family {Ni}i∈I, we may assume by [14, Corollary 2.10] that every subgroup Ni is a finite simple group.

The action of Gi by conjugation on each Ni gives rise to an embedding from Gi/CGi⁢(Ni) into the group of automorphisms Aut⁢(Ni) of Ni, yielding a subgroup containing the group of inner automorphisms Inn⁢(Ni) of Ni. It is clear that Inn⁢(Ni) is isomorphic to Ni since the latter is centerless. Furthermore, we know by [4, Chapter 4.B, p. 304] that the group Aut⁢(Ni)/Inn⁢(Ni) is solvable of derived length three, and then so is each Gi/(Ni⁢CGi⁢(Ni)).

Now, since the normal subgroup Ni⁢CGi⁢(Ni) is definable over a¯i, it can be expressed as a first-order property of the tuple a¯i that Gi modulo Ni⁢CGi⁢(Ni) is solvable of derived length three. Namely, it suffices to express that any element of Gi of the form wc,3⁢(x1,…,x8) belongs to Ni⁢CGi⁢(Ni). Hence, we obtain by Łos’s theorem that the pseudofinite group G/(N⁢CG⁢(N)) is also solvable of derived length three and so it is finite by Fact 2.1. Whence, the group N⁢CG⁢(N) is finitely generated and so is N⁢CG⁢(N)/CG⁢(N), which is clearly isomorphic to N, yielding the result. ∎

Fact 2.3.

There exists a formula ϕs⁢(x) in the pure language of groups that defines the solvable radical of any finite group.

Lemma 3.1.

If S is a finite simple non-abelian group of finite 2-rank r, then either |S| is r-bounded or S is a finite simple group of Lie type of r-bounded Lie rank over a finite field.

Proof.

We know by the Classification of Finite Simple Groups that a finite simple non-abelian group is either an alternating group, a simple group of Lie type or one of the finitely many sporadic groups.

It is clear that we can bound the size of all sporadic groups. Moreover, since the alternating group Alt⁢(n) contains an elementary abelian 2-subgroup of rank 2⋅⌊n/4⌋, the size of an alternating group of 2-rank r is also r-bounded.

Suppose that S is a finite simple group of Lie type of Lie rank n over the finite field 𝔽q and assume further that it has 2-rank r. To bound the Lie rank in terms of r, it suffices to assume that S is a classical simple group. In that case, we have that S contains PSLm⁢(𝔽q), where m=max⁡{2,⌊12⁢(n-1)⌋}. Thus, since Alt⁢(m) embeds into PSLm⁢(𝔽q) via permutation matrices, the paragraph above implies that m and hence n are r-bounded. ∎

Lemma 3.2.

A semisimple pseudofinite group of finite 2-rank has an infinite definable non-abelian simple normal subgroup which is a linear group.

Proof.

Assume that G is a semisimple pseudofinite group of 2-rank at most r. Thus, it is elementarily equivalent to an ultraproduct of an infinite family {Fi}i∈I of finite groups with respect to a non-principal ultrafilter 𝒰 on the set I. After shrinking I and 𝒰 if necessary, we assume that all Fi are semisimple and of 2-rank at most r. Notice that the latter property is clearly expressible by a first-order sentence asserting the non-existence of r+1 elements of order 2 which commute pairwise.

Now, set Soc⁢(Fi) to denote the socle of Fi, the subgroup of Fi generated by all minimal normal subgroups. Thus

where each Si,j is a non-abelian simple normal subgroup of Fi. Note that for every i we have ki≤r by the Feit–Thompson theorem, since each Si,j contains an element of order 2 and any two Si,j and Si,l commute. Furthermore, by Lemma 3.1 there is a positive integer δ⁢(r), depending only on r, such that either |Si,j|≤δ⁢(r) or Si,j is a simple group of Lie type of Lie rank at most δ⁢(r). On the other hand, as we are assuming that G is infinite, the size of the groups Fi is unbounded with respect to 𝒰. Hence, since CFi⁢(Soc⁢(Fi))=1, the group Fi acts faithfully by conjugation on Soc⁢(Fi) and so it embeds into the group of automorphisms of Soc⁢(Fi). Thus, the size of the socle Soc⁢(Fi) must be unbounded with respect to 𝒰 as well. Consequently, as there is only a finite number of possible Lie ranks, there is some τ≤δ⁢(r) and a subset I0 in 𝒰 such that for each i∈I0 some Si,ji is a simple group of a fixed Lie type of Lie rank τ and moreover, the size of all these Si,ji is unbounded with respect to 𝒰.

Next, we see that every Si,ji with i∈I0 can be defined in a uniform way. To do so, we use [5, Theorem 4.2], which asserts the existence of a positive integer m=m⁢(τ) such that any finite simple group of a fixed Lie type of Lie rank τ is equal to the set (C∪C-1)*m for any non-trivial conjugacy class C with |C|>m. Hence, using this result, we find a formula ψ⁢(x,y) in the language of groups such that for some non-trivial element ai in Si,ji we have that ψ⁢(Fi,ai)=Si,ji and such that the Si,ji-conjugacy class Ci of any non-trivial element of Fi satisfying ψ⁢(x,ai) generates Si,ji in m many steps, i.e. Si,ji=(Ci∪Ci-1)*m, provided that |Ci|>m.

By Łos’s theorem, we can find an element a in G such that the subset ψ⁢(G,a) is an infinite definable normal subgroup of G and that ψ⁢(G,a)=(C∪C-1)*m for any ψ⁢(G,a)-conjugacy class C of every non-trivial element of ψ⁢(G,a) with |C|>m. In fact, since ψ⁢(G,a) is a definably simple group, any ψ⁢(G,a)-conjugacy class C of a non-trivial element of ψ⁢(G,a) must be infinite, as otherwise the centralizer of C in ψ⁢(G,a) would be a definable (finite index) normal subgroup, yielding a contradiction. As a consequence, it then follows that the group ψ⁢(G,a) is simple.

Finally, for the second part of the statement, note that ψ⁢(G,a) is elementarily equivalent to the ultraproduct ∏𝒰Si,ji and so it is elementarily embeddable into an ultrapower of ∏𝒰Si,ji by Frayne’s theorem [2, Corollary 4.3.13]. By a result of Point [11], such an ultrapower is isomorphic to a simple group of Lie type over a pseudofinite field. Hence, the simple group ψ⁢(G,a) is a linear subgroup. ∎

Remark 3.3.

In fact, the proof of the lemma yields the existence of a simple group of Lie type by Ryten’s results. Namely, we have shown that the simple group ψ⁢(G,a) is elementarily equivalent to the ultraproduct ∏𝒰Si,ji of simple groups Si,ji of a fixed Lie type. Therefore, applying [3, Proposition 2.14 (i)], we get that ψ⁢(G,a) is a simple group of Lie type.

Corollary 3.4.

Any pseudofinite group of finite Prüfer rank is virtually elementarily equivalent to an ultraproduct of finite solvable groups.

Proof.

Let G be a pseudofinite group of finite Prüfer rank, and consider the definable characteristic subgroup ϕs⁢(G) of G. Set G¯ to denote G/ϕs⁢(G), a semisimple group of finite Prüfer rank, and suppose that it is infinite. By Lemma 3.2, we know that G¯ contains an infinite simple normal subgroup N which is a linear group. Hence, G¯ is solvable-by-finite by Platonov’s theorem, see for instance [15, Theorem 10.9], and consequently it is finite, a contradiction. Therefore, we obtain that G¯ must be finite, which yields the result by Fact 2.3. ∎

Proposition 3.5.

There is no finitely generated pseudofinite group G such that G/ϕs⁢(G) has finite 2-rank and ϕs⁢(G) has finite upper rank.

Proof.

Suppose, to get a contradiction, that G is a finitely generated pseudofinite group that is a counterexample to the proposition, and consider the definable characteristic subgroup ϕs⁢(G) of G. Set G¯ to denote G/ϕs⁢(G), a finitely generated semisimple group of 2-rank r. The proof is divided into two parts:

First part (the semisimple case). We aim to prove that G¯ is finite. Assume otherwise that G¯ is infinite. Thus, it contains an infinite definable simple normal subgroup N which is also a linear group by Lemma 3.2. Additionally, it is finitely generated by Lemma 2.2. Hence, by Malcev’s theorem it is residually finite, see for instance [15, Corollary 4.4], and consequently it is finite, a contradiction. Therefore, we obtain that G¯ must be finite, as desired.

Second part (the solvable-like case). After the first part, the subgroup ϕs⁢(G) has finite index and so it is a finitely generated pseudofinite group of finite upper rank. Hence, we may assume that G and ϕs⁢(G) coincide, i.e. the group G is elementarily equivalent to a non-principal ultraproduct of an infinite family {Hi}i∈I of finite solvable groups.

The subgroups G(n) forming the derived series of G, as well as the subgroups Gn generated by the nth powers, are all definable and have finite index in G, by Fact 2.1. As each Hi is solvable and the words wc,1 have finite width in G(n) also by Fact 2.1, an easy application of Łos’s theorem yields that G(n+1) is properly contained in G(n). In particular, the group G has arbitrarily large finite quotients.

We claim that each finite quotient of G is solvable. To do so, consider a finite index normal subgroup N of G and notice that it is definable, since it contains the finite index subgroup Gk, where k=[G:N], which is also definable. Let χ⁢(x,y¯) be a formula in the pure language of groups such that χ⁢(G,c¯)=N for some finite tuple c¯ from G. By Łos’s theorem, for almost all indices i∈I (with respect to 𝒰), we can find some tuple c¯i in Hi such that χ⁢(Hi,c¯i) is a normal subgroup of Hi of index k. Hence, since each Hi is solvable, the kth term Hi(k) of the derived series of Hi must be contained in χ⁢(Hi,c¯i) and so G(k) is contained in N by Łos’s theorem.

To conclude, consider the characteristic subgroup G0 of G defined as the intersection of all finite index normal subgroups of G. Note that each finite quotient of G/G0 is solvable and has Prüfer rank at most r by assumption. Thus, using the main result from [13], we obtain that G/G0 is nilpotent-by-abelian-by-finite. Therefore, the quotient G/G0 is solvable and hence it must be finite, a contradiction. This finishes the proof. ∎

Corollary 3.6.

There is no finitely generated pseudofinite group of finite Prüfer rank.

Proof of Theorem 1.1.

Let G be an infinite finitely generated group. To get a contradiction suppose that there is an infinite family {Gi}i∈I of finite groups of Prüfer rank r such that G is elementarily equivalent to ∏𝒰Gi for some non-principal ultrafilter 𝒰 on I. Consider Wilson’s formula ϕs⁢(x). Note that Gi/ϕs⁢(Gi) has 2-rank at most r and then, so does the pseudofinite group G/ϕs⁢(G) by Łos’s theorem. Thus, the previous proposition yields that ϕs⁢(G) has finite index in G. Hence, after replacing G by ϕs⁢(G) if necessary, we may assume that each Gi is a solvable group of Prüfer rank at most r. Moreover, as remarked in the solvable-like case of the proof of the previous proposition, we have that every finite index subgroup N of G is definable, and so G has upper rank at most r. Therefore, applying Proposition 3.5 once more, we obtain the desired contradiction. ∎