Compact groups with many elements of bounded order

Conjecture 1.1 ([5, Conjecture 3], [7, Problem 14.53]).

Let G be a profinite group such that the set of solutions of the equation xn=1 has positive Haar measure. Then G has an open subgroup H and an element t such that all elements of the coset tH have order dividing n.

Lemma 2.1.

Let ξ be in L2⁢(G). Then the map

is continuous.

Proof.

It is a special case of [1, (2.41) Proposition]. ∎

Theorem 2.2.

Let G be a compact group, and let ξ1⁢…,ξn be elements in the unit ball of L∞⁢(G). Then the map

is continuous.

Proof.

The Cauchy–Schwarz inequality in the Hilbert space L2⁢(G) can be stated in the following integral form:

The left translate of a function f on G by an x∈G, Lx⁢f, is defined by

For xk,yk, k=1,…,n, in G, and 1≤s≤n, put

If the index of the latter product is empty, we put 1 instead of the product. By the latter convention,

It follows from the Cauchy–Schwarz inequality, (2.1), and the fact that ∥Fs∥2≤1,

whence by Lemma 2.1 the continuity of Ψ follows. ∎

Theorem 2.3.

Let G be a compact group. Suppose that A1,…,An are measurable subsets of G with positive Haar measure. Then the map

is non-zero and continuous. In particular, if A is a measurable subset with positive Haar measure, then for any positive integer k there exists an open subset U of G containing 1 such that mG⁢(A∩u1⁢A∩⋯∩uk⁢A)>0 for all u1,…,uk∈U.

Proof.

Applying Theorem 2.2, we conclude that Λ is continuous. Take ξk as χAk for k=1,…,n in Theorem 2.2, and note that, by Fubini’s theorem, we have

Since ϵ:=𝔪G(A)>0, it follows from the continuity of the map Λ corresponding to Ai:=A (i=1,…,k+1) that there exists an open subset

such that 𝔪G⁢(A∩u1⁢A∩⋯∩uk⁢A)≥ϵ>0 for all (1,u1,…,uk)∈𝔘. Now U:=U0∩U1∩⋯∩Uk, where

are open subsets of G containing 1, has the required property. ∎

Remark 2.4.

The statement of Theorem 2.3 is conjectured by the authors and proposed in [10] by the second author. We are guided by the comments of other people on [10] not only to write a detailed proof for Theorem 2.3 but also to give Theorem 2.2.

Remark 2.5.

Following [2], a subset X of a group G is called large if ⋂a∈Fa⁢X is not empty for any finite non-empty subset F⊆G. We call a subset X of a group Grelatively k-large with respect to a subset M of G for some k∈ℕ if ⋂a∈Fa⁢X is not empty for any subset F⊆M with |F|=k. Thus, by Theorem 2.3, for every k∈ℕ, each measurable subset of a compact group with positive Haar measure is relatively k-large with respect to an open subset Uk containing 1.

Theorem 3.1.

Let G be a compact group having an automorphism (not necessarily continuous) α such that the set X={x∈G∣xα=x-1} is measurable and of positive Haar measure. Then G contains an open normal abelian subgroup. In particular, there exists an open abelian subgroup A of G such that t⁢A⊆X for some t∈X.

Proof.

It follows from Theorem 2.3 that there exists an open subset U of G containing 1 such that X∩u1⁢X∩u2⁢X∩u3⁢X is of positive Haar measure for all u1,u2,u3∈U. Since U is open and 1∈U, it follows from [4, Theorem 4.5] that there exists an open subset V⊆U such that 1∈V, V=V-1={v-1∣v∈V} and V2:={v1v2∣v1,v2∈V}⊆U (see [4, Theorem 4.5]). Now suppose that a,b are arbitrary elements of V. Thus

has positive Haar measure, and in particular, it is non-empty. It follows that there exists x∈X such that b⁢x,a⁢x,a⁢b⁢x are all in X. Let β be the automorphism y↦x⁢yα⁢x-1. The fact that α is an automorphism that inverts x and bx gives us

so bβ=b-1. Similarly, aβ=a-1 and (a⁢b)β=(a⁢b)-1. Hence

showing that a and b commute, and so the subgroup H generated by V is abelian. Since H is a subgroup with non-empty interior, H is open. Now take the core K of H in G, which is open normal and abelian. This completes the proof of the first part.

For the second part, since |G:K| is finite, t⁢K∩X has positive Haar measure for some t∈X. Thus A:={a∈K∣ta∈X} has positive Haar measure. The set A is the subset of elements of K inverted by the automorphism γ of G, where γ:y↦t-1⁢yα⁢t; it follows that A is a subgroup; for, given a,b∈A, then

and the last expression is the inverse of a⁢b-1 since a and b commute, so a⁢b-1∈A. Since X is closed, A is closed, and it is open since A has positive Haar measure. This completes the proof. ∎

Lemma 4.1.

Let a, b and x be elements of a group such that

Then [a,b,b]=1.

Proof.

See the proof of [3, Proposition 4.3] and the proof of [6, Proposition 1]. ∎

Lemma 4.2.

There exists a positive integer k such that every group generated by a symmetric subset X=X-1 containing 1 satisfying the property [x,y,y]=1 for all x,y∈Xk:={x1⋯xk∣x1,…,xk∈X} is nilpotent of class at most 3.

Proof.

It is enough to show that every 4-element subset of X generates a nilpotent group of class at most 3. Since 2-Engel groups are nilpotent of class at most 3 [9, Corollary 3, page 45], F4〈[a,b,b]∣a,b∈F4〉 is nilpotent of class at most 3, where F4 is the free group of rank 4 on the free generators x1,x2,x3,x4. Now it follows from [8, Lemma 1.43, page 32 and its Corollary, page 33] that 〈[a,b,b]∣a,b∈F4〉 is the normal closure of a finite number of its elements. Then there exist elements a1,…,as,b1,…,bs∈F4 such that

Let k be a positive integer such that

Therefore, any group satisfying the laws [ai,bi,bi]=1 for all i=1,…,s is nilpotent of class at most 3, and since ais and bis are the product of at most k free generators xjs and their inverses, the proof is now complete. ∎

Lemma 4.3.

There exists a positive integer ℓ such that every group generated by a symmetric subset X=X-1 containing 1 satisfying the property [x,y,y]=1 for all x,y∈Xℓ is a 2-Engel group.

Proof.

Let ℓ:=max{k,2}, where k is a positive integer mentioned in the statement of Lemma 4.2. By Lemma 4.2, G:=〈X〉 is nilpotent of class at most 3. Suppose that g and h are arbitrary elements of G. Then g=x1⁢⋯⁢xt and h=y1⁢⋯⁢yt for some xi,yj∈X. Now we may write [g,h,h] as follows:

This follows from the facts that G is nilpotent of class at most 3 and [x,y,y]=1 for all x,y∈X. Now, since ℓ≥2, [x,y⁢z,y⁢z]=1 for all x,y,z∈X. It follows that [x,y,z]⁢[x,z,y]=1. Therefore, [g,h,h]=1, and so G is 2-Engel. ∎

Theorem 4.4.

Let G be a compact group, and let α be a not necessarily continuous automorphism of G such that α3=1 and the set X={x∈G∣xα2⁢xα⁢x=1} is measurable. If X has positive Haar measure, then G contains an open normal 2-Engel subgroup.

Proof.

By Lemma 2.3, there exists an open subset U of G such that

is of positive Haar measure for all u1,…,u7∈U. Since U is open and 1∈U, it follows from [4, Theorem 4.5] that there exists an open subset V⊆U such that 1∈V, V=V-1 and V2⁢ℓ⊆U. Now suppose that a1,…,aℓ and b1,…,bℓ are arbitrary elements of V. Let a=a1⁢⋯⁢aℓ and b=b1⁢⋯⁢bℓ. Thus

has positive Haar measure, and in particular, it is non-empty. It follows that there exists x∈X such that b⁢x,a⁢x,a-1⁢x,a⁢b-1⁢x,b⁢a-1⁢x,a⁢b⁢x,b-1⁢a-1⁢x are all in X. Working in the semidirect product G⋊〈α〉, since α3=1, it follows that a∈X if and only if (a⁢α)3=1. Therefore, (x⁢α)3=(b⁢x⁢α)3=1 and

Now Lemma 4.1 implies that [a,b,b]=1, and it follows from Lemma 4.3 that H:=〈V〉 is 2-Engel. Since V is open, H is an open subgroup. Now take the core of H in G, which is normal open and 2-Engel. This completes the proof. ∎

Remark 4.5.

However, the statements of Theorem 4.4 and [6, Theorem 1] are not directly comparable as “largely splitting of order 3” has been replaced with “splitting of order 3 on a set of positive Haar measure”, but thanks to Theorem 2.3, the proof has quite a similar structure.

Corollary 4.6.

Let G be a compact group such that the set X={x∈G∣x3=1} has positive Haar measure. Then G contains an open normal 2-Engel subgroup.

Proof.

Take α as the identity automorphism in Theorem 4.4. ∎