Free subgroups of inverse limits of iterated wreath products of non-abelian finite simple groups in primitive actions

Lemma 2.1.

Let w=xi1ϵ1⁢xi2ϵ2⁢…⁢xinϵn be a reduced word of length n in Fk and let G be any group with a normal subgroup N. Consider (g1,…,gk)∈Gk and (b1,…,bk)∈Nk. Then

where aj=(bij)w¯j⁢(g1,…,gk) for j=1,…,n.

Lemma 2.2.

Let (G,Γ) be an element of Wn for some n≥2, with G=S⁢wrΔ⁡H, where (H,Δ)∈Wn-1. Then one of the following holds:

1. The action of G on Γ is the diagonal action.

2. Γ=ΣΔ for a suitable primitive S-set Σ, and the action of G on Γ is the product action.

3. G is isomorphic to a twisted wreath product S⁢wrΘ⁡X with X≃H, and the socle SΘ of G acts regularly on Γ.

Proof.

The socle of G is the base group B=SΔ, and it acts transitively on Γ. If the action of B is regular, then G is the semidirect product of B and Gγ for any γ∈Γ. Moreover, X=Gγ acts on the components of B in the same way as H. By [7], the group G is isomorphic to a twisted wreath product S⁢wrΘ⁡X in this case.

We may thus assume that B is not regular, so that Bγ≠1. Since |Δ|>1, we can use [4, Theorem 4.6.A] to see that one of the following cases must hold: either G is of diagonal type, or G is a primitive subgroup of U⁢wrΛ⁡Sym⁡(Λ) acting in product action and U is itself primitive with non-regular socle. In order to complete the proof, it suffices to show that U=S and Λ=Δ in the latter case.

But then, soc(U)Λ is the socle of G, so that the components of soc⁡(U) form a block under the action of G on the components of B. This action is primitive, so that soc⁡(U) must have just one component. Therefore U is isomorphic to a subgroup of Aut⁢(S) containing S, so that U/S is solvable. The normal closure UG of U contains B and it is easy to see that UG/B is solvable. However, H has no non-trivial solvable normal subgroups. This shows that UG=B and U=S. ∎

Lemma 2.3.

Using the above notation,

Proof.

The first claim holds true for n=1, because the symmetric group of degree 4 is solvable. If (S⁢wrΔ⁡H,Γ)∈𝒲n and ξ⁢(n-1)≤|Δ| by induction, then Lemma 2.2 implies that |Γ|≥5|Δ|≥5ξ⁢(n-1)=ξ⁢(n).

The second inequality holds true for n=3. For n≥4, induction gives

Example 2.4.

For each n≥1, there exists (Gn,Ωn)∈Wn such that |Ωn|=ξ⁢(n).

Proof.

For n=1 we choose G1=Alt⁢(5) and the natural Alt⁢(5)-set Ω1. Suppose that we have already found (Gn-1,Ωn-1)∈𝒲n-1 with |Ωn-1|=ξ⁢(n-1). Then Gn=Alt⁢(5)⁢wrΩn-1⁡Gn-1 acts primitively on Ωn=Ω1Ωn-1 in product action, and the degree of this representation is ξ⁢(n). ∎

Theorem 2.5.

Consider a reduced word w of length n>0 in the free group Fk of rank k. Let (G,Γ)∈Wd+n for some d∈N. Then, for any y=(y1,…,yk) in Gk∖Kdk, there exist g=(g1,…,gk)∈Gk and γ∈Γ such that

1. g≡y(modKd),

2. the elements γj=γ⁢wj⁢(g) are pairwise distinct for j=0,…,n.

In particular, w⁢(g)≠1.

Proof.

If k>n, then at least k-n variables do not appear in the reduced form of w. We may therefore assume that k≤n. It is also easy to check that the result is true for n≤2. Let n≥3 and assume by induction, that the statement holds for words of length at most n-1. By Lemma 2.2 the group G=S⁢wrΔ⁡H acts either in diagonal action or in product action, with respect to a suitable primitive representation of S, or its socle is regular. Notice that soc⁡(G)=Kd+n-1 is the base subgroup B of the wreath product. Another fact which will be used frequently in the sequel is that, if h≡y(modKd) for some h∈Hk, then b⁢h≡y(modKd) for every b∈Bk.

Our induction provides elements h=(h1,h2,…,hk)∈Hk and δ∈Δ such that h≡y(modKd) and such that the elements δi=δ⁢wi⁢(h) are pairwise distinct for i=0,…,n-1. Choose any faithful transitive S-set Ω and let G act on Ω×Δ in the natural way. The remark in the first paragraph of the proof of [2, Lemma 4] shows that we can find ε=(ω,δ)∈Ω×Δ and a subset ℰ⊆Bk containing at least

elements such that, whenever b∈ℰ, then ε⁢wi⁢(b⁢h)≠ε⁢wj⁢(b⁢h) for 0≤i<j≤n. In particular, wi⁢(b⁢h)≠wj⁢(b⁢h) for 0≤i<j≤n and b∈ℰ.

Consider any primitive faithful G-set Γ. We aim to show that, for each b∈ℰ, we can find some γ∈Γ such that the element xi⁢j=wi⁢(b⁢h)⁢wj⁢(b⁢h)-1 does not belong to Gγ for 0≤i<j≤n. Assume by way of contradiction that this does not hold for a certain b∈ℰ. It follows that Γ is the union of the sets

The number of these sets is (n+12). Hence there exists a pair (i,j) such that

Therefore xi⁢j fixes at least μ elements from Γ. The desired contradiction will now be derived separately in each of the three cases furnished by Lemma 2.2.

Case 1: Product action. In this case, G acts on Γ=ΣΔ, where Σ is a primitive faithful S-set. We shall try to estimate the size of the set of fixed points fixΓ⁢(g) for a non-trivial element g=f⁢x∈G (with f∈B, x∈H). Recall that g acts on Γ via (φ⁢g)⁢(δ)=φ⁢(δ⁢x-1)⁢f⁢(δ⁢x-1) for all φ:Δ→Σ.

Suppose first that x-1=(δ1⁢δ2⁢…⁢δℓ) is a non-trivial cycle. If φ:Δ→Σ is a fixed point for g, and if δ∈fixΔ⁢(x), then φ⁢(δ)∈fixΣ⁢(f⁢(δ)). Moreover, we have φ⁢(δi+1)⁢f⁢(δi+1)=φ⁢(δi) for i=1,…,ℓ-1 and φ⁢(δℓ)⁢f⁢(δℓ)=φ⁢(δ1).

Therefore, if we want to construct an element φ∈fixΓ⁢(g), then we should, first of all, choose φ⁢(δ)∈fixΣ⁢(f⁢(δ)) whenever δ∉suppΔ⁡(x). Then, in order to define φ on suppΔ⁡(x), we can choose freely the value φ⁢(δ1) (this allows |Σ| different choices), and then the value of φ on the other points δi is determined by the above relations. However, not every such choice is admissible, since the ℓ-th relation φ⁢(δℓ)⁢f⁢(δℓ)=φ⁢(δ1) might not hold. We conclude that

It is now an easy matter to work out the general case: If x has r non-trivial cycles in its action on Δ, then

Suppose that x≠1. Since n≥3, the group H is not an alternating group. Therefore [8, Corollary 3] yields suppΔ⁡(x)≥2⁢(|Δ|-1), whence

Now r≤|Δ|-|fixΔ⁢(x)|2. Hence

Suppose now that x=1 and g=f∈B. In this situation it is clear that

corresponding to the case when S is an alternating group in natural action, f has a single non-trivial entry and this entry is a 3-cycle.

As in the proof of [2, Lemma 3] at most one of the elements xi⁢j (say x0⁢n) has trivial top component, so that the set

has size |Γ0|≥|Σ||Δ|-(|Σ|-3)⁢|Σ||Δ|-1=3⁢|Σ||Δ|-1. An application of the pigeonhole principle yields that there exists a 2-set {i,j}≠{0,n} for which Γi⁢j has size at least

The inequality μ0≤|Σ||Δ|-|Δ|+1 now implies |Σ||Δ|-2≤(n+12)-1 . But this contradicts Lemma 2.3.

Case 2: Diagonal action. In this case, the set Γ can be identified with the set of right cosets in B=SΔ of the diagonal subgroup D of B. As before, we try to describe fixΓ⁢(g) for g=f⁢x∈G (with f∈B, x∈H). Recall that the action of G on Γ is induced from the product action of G on B with respect to the right regular representation of S on Σ=S.

If g fixes Dt (where t∈B), then there exists d∈S such that (t⁢f)⁢x=d¯⁢t, where d¯ indicates the element of D, whose entries are all equal to d. Thus,

for every δ∈Δ. In particular, f⁢(δ)=dt⁢(δ) whenever δ∈fixΔ⁢(x).

We shall now use this information in order to count the elements in fixΓ⁢(g). There are at most |S| possible choices for d. Once d has been selected, each t⁢(δ) (δ∈fixΔ⁢(x)) can be chosen in |CS⁢(d)| different ways. Let χ⁢(x) denote the set of x-orbits c in Δ with length ℓ⁢(c)>1. If c=(δ1⁢δ2⁢…⁢δℓ⁢(c))∈χ, then, for any choice of t⁢(δ1), the values t⁢(δi) (i=2,…,ℓ⁢(c)) are uniquely determined by the equation t⁢(δi)⁢f⁢(δi)=d⁢t⁢(δi+1). Moreover, also the equation

must hold, and it is not difficult to see, that this happens if and only if

Thus t⁢(δ1) must satisfy

Hence the number of possible choices for t⁢(δ1) is |CS⁢(dℓ⁢(c))|. Altogether we conclude that

If x≠1, then the same arguments as in Case 1 (product action) yield

On the other hand, if x=1 but f≠1, then we obtain that all values of f must lie in the conjugacy class of a single non-trivial element d∈S, and that we have |CS⁢(d)| choices for each of the values t⁢(δ) (δ∈Δ). Moreover, if t1 and t2 have been obtained in this way, then t1⁢t2-1∈CS⁢(d)Δ. In particular, D⁢t1=D⁢t2 if and only if t1⁢t2-1∈D∩CS⁢(d)Δ. Therefore,

Considering Γ0 and μ0 as in Case 1 (product action), we find that

and

This leads to the contradiction |S||Δ|-3⁢(|S|-1)≤(n+12)-1 (cf. Lemma 2.3).

Case 3: The socle acts regularly. Now we have Γ=B and G=B⁢X, where B=SΔ acts regularly on Γ and X is a point stabilizer. In order to avoid confusion, the action of s∈G on γ∈Γ shall be denoted by γ.s. As before, we consider g=f⁢x∈G with f∈B, x∈X and try to count the number of its fixed points. Notice that x might not belong to H. Therefore x=b⁢h with b∈B, h∈H.

Now γ∈fixΓ⁢(g) if and only if γ=γ.g=γ.(f⁢b⁢h) if and only if γ.h-1=γ⁢f⁢b. The latter means that

If h fixes δ, then f⁢(δ)⁢b⁢(δ)=1 and γ⁢(δ) can take any value in S. If (δ1⁢δ2⁢…⁢δm) is an h-orbit, then γ⁢(δi+1) is determined by γ⁢(δi) via

but we can choose γ⁢(δ1). With r denoting the number of non-trivial h-orbits in Δ, we obtain

as in Case 1 (product action) for elements g satisfying h≠1. However, any non-trivial g∈G either has no fixed-point in Γ or satisfies h≠1, because B acts regularly on Γ. In particular,

a contradiction to Lemma 2.3. ∎

Corollary 2.6.

In the notation of Theorem 2.5, let

where (H,Δ1)∈Wd and S1,…,Sn are non-abelian finite simple groups. If |Si|=si for all i and |G|=r⁢|H|, then the number of k-tuples g∈Gk satisfying the statement of Theorem 2.5 is at least

Proof.

Clearly, G=Kd⁢H. Suppose first that w⁢(y)≠1. Then, for every b∈Kdk, the k-tuple g=b⁢y satisfies w⁢(g)≠1 too. The number of such tuples is rk. Suppose next, that w⁢(y)=1. Following the recursive procedure in the proof of Theorem 2.5, the required elements are constructed in n steps. In the i-the step, every k-tuple constructed so far, can be extended in at least si|Δi|⁢(1-1si) different ways. Since ∏i=1nsi|Δi|=|Kd|, the claim follows in this case too. ∎

Theorem 3.1.

If S is a set of non-abelian finite simple groups, and if W∈W⁢(S), then the set W⁢(k) is dense in Wk and its complement is a meagre subset of Wk.

Proof.

Consider a non-trivial w∈Fk. The evaluation map μw:Wk→W, which sends every h∈Wk to w⁢(h)∈W, is continuous. Moreover, W is a Hausdorff space. Therefore μw-1⁢(1) is closed and C⁢(w)=Wk∖μw-1⁢(1) is open.

Let n be the length of w and choose a non-trivial element y∈Wk. There exists d such that y∉Kdk. An application of Theorem 2.5 gives h∈Wk such that h≡y(modKd) and w⁢(h)∉Kd+n. In particular, w⁢(h)≠1, so that h∈C⁢(w). On the other hand h⁢Kd=y⁢Kd, so that the intersection of the open neighborhood y⁢Kd of y with C⁢(w) is non-empty. This shows that C⁢(w) is dense in Wk. Since Wk is a compact Hausdorff space, Baire’s theorem yields that W⁢(k) is dense, being a countable intersection of open dense subsets. Finally,

Since each C⁢(w) is open and dense, Wk∖C⁢(w) is nowhere dense. This proves the claim. ∎

Corollary 3.2.

Suppose that S is a set of non-abelian finite simple groups. Then every W∈W⁢(S) contains 2ℵ0 free subgroups of rank 2ℵ0.

Proof.

The group W is metrizable and complete. Therefore it is a Polish space. Since W is not discrete, [6, Theorem 1] yields that W contains a dense free subgroup. The claim now follows from [6, Corollary 3]. ∎

Theorem 3.3.

Let S be a set of non-abelian finite simple groups, and let W∈S. Then the set Wk∖W⁢(k) has measure zero with respect to the normalized Haar measure of W.

Questions 3.4.

Suppose that the profinite group W is formed from iterated wreath products as above, but without further restrictions to the nature of the faithful Wn-sets (i.e., we do not necessarily require that these sets are primitive).

1. Is it still true then that Wk∖W⁢(k) is meagre?

2. Is it still true then that Wk∖W⁢(k) has measure zero with respect to the normalized Haar measure?