Infinite 32 ‐generated groups

Every finite simple group can be generated by two elements, and Guralnick and Kantor proved that, moreover, every nontrivial element is contained in a generating pair. Groups with this property are said to be 32 ‐generated. Thompson's group V was the first finitely presented infinite simple group to be discovered. The Higman–Thompson groups Vn and the Brin–Thompson groups mV are two families of finitely presented groups that generalise V . In this paper, we prove that all of the groups Vn , Vn′ and mV are 32 ‐generated. As far as the authors are aware, the only previously known examples of infinite noncyclic 32 ‐generated groups are the pathological Tarski monsters. We conclude with several open questions motivated by our results.


Introduction
A group G is d-generated if it has a generating set of size d. We say that G is 3 2 -generated if G is 2-generated and, moreover, every nontrivial element of G is contained in a generating pair.
It is a remarkable fact that every finite simple group is 2-generated. This result relies on the classification of finite simple groups and the proof for the groups of Lie type was given by Steinberg in 1962 [30]. In that paper, Steinberg asked whether every finite simple group has the much stronger property of being 3 2 -generated. The alternating (and symmetric) groups of degree at least 5 had been known to be 3 2 -generated since the work of Piccard in 1939 [27]. In 2000, Guralnick and Kantor [19] resolved this long-standing question of Steinberg and proved the following.
Theorem. Every finite simple group is 3 2 -generated.
This theorem has motivated a great deal of recent work on the generation of finite groups (see Section 3 of Burness' recent survey article [12] for a good overview of this work).
It is easily seen that if G is a 3 2 -generated group, then every proper quotient of G is cyclic. The following conjecture, due to Breuer, Guralnick and Kantor [8], avers that this clearly necessary condition is indeed sufficient for finite groups. 3 2 -generated if and only if every proper quotient of G is cyclic.

Conjecture. A finite group G is
For recent work towards proving this conjecture, see [13,20]. The above necessary condition for 3 2 -generation is not sufficient for infinite groups. Indeed, the infinite alternating group A ∞ is simple but not finitely generated, let alone 3 2 -generated. Moreover, in [18] a simple group is exhibited that is finitely generated but not 2-generated, which solves Problem 6.44 in the Kourovka Notebook [23]. However, the authors are not aware of any 2-generated group with no noncyclic proper quotients that is not 3 2 -generated. Equally, the authors are not aware of any existing examples of noncyclic infinite 3 2 -generated groups, other than Tarski monsters. These latter groups are the infinite groups whose only proper nontrivial subgroups have order p for a fixed prime p; they are clearly simple and 3 2 -generated, and they were proved to exist for all p > 10 75 by Olshanskii in [25].
In this paper, we provide two natural infinite families of infinite 3 2 -generated groups. Thompson's group V , introduced in 1965, was the first known example of a finitely presented infinite simple group [31], and until 2000 [11] all known examples of such groups were close relatives of V . The group V naturally arises as a group of homeomorphisms of Cantor space C = {0, 1} N , and, together with its subgroups T and F and the generalisations we will soon introduce, it has been the subject of much research, finding connections with dynamics and the word problem, among much else (see [1,3,5,15,32], for example).
As observed by Mason [22], V is 2-generated, and we prove the following.
In fact, we will prove two theorems which generalise the above theorem in two different directions.
The Higman-Thompson group V n , for all n 2, is an infinite finitely presented group, which was introduced by Higman in [21]. The group V n has a natural action on n-ary Cantor space C n = {0, 1, . . . , n − 1} N , and the group V 2 is nothing other than Thompson's group V . The derived subgroup of V n equals V n when n is even and has index two when n is odd. In both cases, V n is simple. Mason proved that both V n and V n are 2-generated [22], and we prove the following.
Theorem 1. For all n 2, the Higman-Thompson groups V n and V n are 3 2 -generated.
In particular, the groups V n when n is odd give examples of infinite noncyclic 3 2 -generated groups that are not simple.
Theorem 2. For all m 1, the Brin-Thompson group mV is 3 2 -generated.
Our proofs owe a great deal to the Bleak-Quick perspective, introduced in [4], that mV (especially V ) is analogous to the finite symmetric group, a viewpoint that readily extends to V n . Indeed, in a sense, our proofs highlight that the 3 2 -generation of these groups is analogous to the 3 2 -generation of the finite symmetric groups. For instance, the groups V n and mV are generated by an infinite conjugacy class of involutions, known as transpositions, so called because of their similarity to the transpositions in the finite symmetric group. In mV and V 2k every element is a product of an even number of transpositions (even a transposition is such a product), so we think of these groups as analogous to both the alternating and symmetric groups. When n is odd, V n is the subgroup of V n containing the elements that are a product of an even number of transpositions.
Our proofs are concrete and constructive. Let G be one of V n , V n or mV (here and throughout, the reader is encouraged to think of V in the first instance). For every nontrivial element g ∈ G we will exhibit an element h ∈ G of finite order and prove that g, h = G. We prove that g, h = G by essentially showing that g, h contains all of the transpositions. We modify our approach slightly for V n , which is not generated by the transpositions when n is odd. To describe the element h ∈ G we will use the permutation-like notation of Bleak and Quick.
We will begin with some preliminary results in Sections 2 and 3, before turning to the proofs of Theorems 1 and 2 in Section 4. The authors expect that collecting together uniformly stated results on the generating sets for V n , V n and mV in Section 3 will be useful for others interested in studying these groups, as this information is currently spread across the literature in terms of various different descriptions of and notation for these groups. In Section 5, we conclude by posing several natural questions that are raised by our theorems.  The Higman-Thompson group V n is the group of bijections g ∈ Sym(C n ) for which there exists a basis pair, namely a bijection σ : A → B between two bases A and B of C n such that (uw)g = (uσ)w for all u ∈ A and all w ∈ C n .
The following records several important results about V n , which were proved by Higman when he introduced these groups [21].

Brin-Thompson group mV
Write C = C 2 = {0, 1} N and X = X 2 = {0, 1} * . Fix m 1 and let u = (u 1 , . . . , u m ) ∈ X m . We write |u| = min{|u i | | 1 i m}. For an element w ∈ (w 1 , . . . , w m ) of either X m or C m , we write uw = (u 1 w 1 , . . . , u m w m ). In addition, we write The Brin-Thompson group mV is the group of bijections g ∈ Sym(C m ) for which there exists a basis pair, namely a bijection σ : A → B between two bases A and B of C m such that (uw)g = (uσ)w for all u ∈ A and all w ∈ C m . Part (i) of the following was proved by Brin [10], and (ii) is due to Bleak and Lanoue [2].  viewpoint, a basis of C is exactly the set of leaves of a finite binary rooted tree. Therefore, the basis pairs give the familiar tree pair notation for elements of V , namely (A, σ, B) where A and B are finite binary rooted trees and σ is a bijection between the leaves of A and B. When referring to tree pairs, we will number the leaves of the trees to manifest the bijection and we will adopt the convention that unlabelled leaves are fixed. Three elements α, β, γ ∈ V are represented by tree pairs in Figure 1. To demonstrate the action of these elements on C, note, for example, that
Remark 2.3. Although we do not use this fact, we note that the permutations are exactly the elements of G of finite order. For V n , this is [29,Lemma 17], and the same proof applies to mV .

Incomparability
Let (G, Γ, Ω) be (V n , C n , X n ) or (mV, C m , X m ). We say that u, v ∈ Ω are incomparable and write u ⊥ v if uΓ and vΓ are disjoint. The following two technical lemmas are straightforward.
Lemma 2.6. Let g ∈ G be nontrivial. Then there exists u ∈ Ω such that u ⊥ ug.
Proof. We assume that G = mV since the proof is similar and easier when G = V n . Let (A, σ, B) be a basis pair for g and let u = (u 1 , . then the proof is complete. Therefore, assume that u and v are not incomparable. Consequently, there exists 1 i m such that either u i is a proper prefix of v i or v i is a proper prefix of u i (otherwise u = v). We will assume that u i is a proper prefix of v i ; the proof works in the other case making the obvious changes.
is not a prefix of (u g) i = u i awb. Therefore, u and u g are incomparable, as required.

Convenient shorthand
Let (G, Γ, Ω) be (V n , C n , X n ) or (mV, C m , X m ). For g ∈ G and u ∈ Ω, define an element g [u] ∈ G as follows: for each w ∈ Γ, Write noting that G [u] acts on uΓ and

Bertrand's postulate
In 1852, Chebyshev proved Bertrand's postulate: for all integers k > 3, there exists a prime number p such that k < p < 2k. Indeed, by the prime number theorem, for all ε > 0, there exists k 0 > 0 such that for all k > k 0 , there exists a prime number p such that k < p < (1 + ε)k. We will use the following useful nonasymptotic bound due to Nagura [24].
Theorem 2.7. Let k 25. Then there exists a prime p such that k < p < 6 5 k.

Generating symmetric and alternating groups
The following is [22,Lemma 2]; our proof is based on, but shorter than, the original. Here, and throughout, for group elements x and y we write x y = y −1 xy and [x, y] = x −1 y −1 xy.
Proof. To simplify the details of the proof, let us assume that n 9; the result can be easily verified (for example, in Magma [6]) when n ∈ {7, 8}. Write x = (1 2 · · · b) and y = (a a + 1 · · · n). Since b a the subgroup x, y S n is transitive. For all 0 < j < n − b, Therefore, x, y is a primitive group of degree n 9 containing an element supported on at most four points, so x, y A n (see [16,Example 3.3.1]).

Generating sets
This section is dedicated to recording important generating sets of the Higman-Thompson and Brin-Thompson groups. To aid the reader, let us first comment on the prototype V .

Higman-Thompson groups
Fix n 2. The first result extends Brin's observation that V is generated by transpositions. This is proved in the proof of [17,Theorem 8.7]. Indeed, more is shown in that proof. If n is even, then (u v) = n−1 a=0 (ua va), so every element of V n is a product of an even number of transpositions. In contrast, if n is odd, then a transposition cannot be written as the product of an even number of transpositions, from which we deduce the following.  For k 0, write X (k) n = {x ∈ X n | |x| = k}. First we claim that every product of two transpositions is a product of some number of double transpositions. Let u 1 , v 1 , u 2 , v 2 ∈ X n such that u 1 ⊥ v 1 and u 2 ⊥ v 2 . Observe that there exist k 0 and x 1 , n \ {x 2 }| = n k − 1 is even, the above equation expresses the element (u 1 v 1 )(u 2 v 2 ) as a product of n k double transpositions. Therefore, every product of two transpositions is a product of double transpositions, so by Lemma 3.2, V n is generated by the double transpositions.
Since V n is generated by the double transpositions, it is certainly generated by the even permutations in V n . This implies that V n is also generated by the three-cycles, that is elements It is via the following result that we use Lemmas 3.1 and 3.2.
Proof. We begin with part (i). Write Let v, w ∈ X n such that v ⊥ w and |v|, |w| k.
If v 1 = w 1 , then fix x ∈ X n \ {v 1 , w 1 } with |x| = k and note that Therefore, in all cases, (v w) ∈ H. Since every transposition is a product of transpositions of the form (v w) where |v|, |w| k, we conclude, by Lemma 3.1, that H = V n .
We will now assume that n is odd and consider part (ii). Write Let v, w, z ∈ X n be pairwise incomparable with |v|, |w|, |z| k.
Next assume that exactly one of u 0 and u 1 is contained in {v 1 , z 1 , w 1 }. Fix x ∈ X n \ {u 0 , u 1 , v 1 , w 1 , z 1 } with |x| = k. Without loss of generality, we can assume that u 0 = v 1 and We may now assume, without loss of generality, that v 1 = w 1 = z 1 . In this case, Without loss of generality, we can assume that u 0 = v 1 and u 1 = w 1 . If z 1 ∈ {v 1 , w 1 }, then It remains to assume, without loss of generality, that z 1 = w 1 . In this final case, Therefore, in all cases (v w z) ∈ K. By Remark 3.3, this implies that K = V n , as required.
Recall that X where we only define δ when n is odd. In the following proposition, we use notation such as δ [u] , which we introduced in Section 2.6.
Proposition 3.5. Let u ∈ X n with |u| 1 and let k = |u| + 1. Then Proof. For this proof, it will be convenient to write X We will assume that u a = ua if a ∈ [n]. In addition, letδ be either δ [u] or δ [u] and letS be either S Claim. There exists g ∈ S ,δ such that u N −1 g = u N −1 , u N g = u N and u 1 g = v.
Proof of Claim. We induct on |v|. If |v| = k, then v ∈ X (k) n , so there is clearly an element g ∈S such that u N −1 g = u N −1 , u N g = u N and u 1 g = v. Now let l > k and assume that the claim is true for all v ∈ X n such that |v| < l. Assume |v| = l and write v = wa where a ∈ [n]. By induction, there exists h ∈ S ,δ such that u N −1 h = u N −1 , u N h = u N and u 1 h = w. First assume that nk > 4 and note that u 1 (u 2 u 1 u 0 )δ a+1 = u 1 a. Therefore, v = wa = u 1 ah = u 1 (u 2 u 1 u 0 )δ a+1 h = u 1 g, where g = (u 2 u 1 u 0 )δ a+1 h is an element of S ,δ for which u N −1 g = u N −1 and u N g = u N . When nk = 4, note thatS = S (2) 2 and, arguing as above, we see that g = (u 1 u 0 )δ a+1 h ∈ S ,δ fixes u N −1 and u N and satisfies u 1 g = v. This proves the claim.
Let us now record some consequences of the claim. [u] . Now δ [u] , being an (n + 1)-cycle, is an odd permutation, so δ [u] ∈ V n and again we conclude that A (k+1) n , δ [u] = V n . This proves (i). We now turn to (ii), where we assume that n is odd. Similar to above ( n , δ [u] = V n , noting that δ [u] , being an (n + 2)-cycle, is an even permutation. Moreover, A This completes the proof.
We will now use Proposition 3.5 to obtain two generating sets for V n and V n that will play important roles in the proof of Theorem 1. We begin with an infinite generating set reflecting the self-similar nature of V n .
Proposition 3.6. Let A = {x 1 , . . . , x } be a basis. Then Proof. Let We begin with (i). Write Let v ∈ X n such that x 1 0 ⊥ v and |v| k. We will prove that ( If i > 1, then Therefore, by Lemma 3.4(i), H = V n . We now assume that n is odd and turn to (ii). Write Let v ∈ X n such that x 1 0 ⊥ v, x 1 1 ⊥ v and |v| k. We will prove that ( Now assume that i > 1. Write w = w 0 w 1 where w 0 ∈ [n] and fix a ∈ [n] \ {w 0 }. Then Therefore, by Lemma 3.4(ii), K = V n .
We now turn to generating pairs. Let us introduce some notation for particular types of elements that we will often use.
Notation 3.7. Let d 1 and k log n d , where · is the usual ceiling function. Define where N = n k − 1 and a i is the n-ary expansion of i with exactly k digits. We will write σ n (d) = σ n (d, log n d ) and τ n (d) = τ n (d, log n d ).
Fix primes p and q satisfying 1 4 n 3 p < 1 2 n 3 and 3 4 n 3 < q < n 3 . (3.5) If n 5, then Theorem 2.7 guarantees that this is possible and it is straightforward to find suitable primes when 1 r < n 4. When n = 2 we insist that p = 2. (The primes p and q depend on n, but for clarity of notation we do not give them a subscript of n.) Let ζ = σ n (p, 3) and β = τ n (q, 3).  We can now state our preferred generating pairs for the Higman-Thompson groups.

Brin-Thompson groups
This section very closely mirrors the previous. Fix m 1. Let us comment on our notation for mV . We adopt the convention of Quick [28], that for a ∈ {0, 1}, a denotes (a, a, . . . , a) ∈  X m and a i denotes (ε, . . . , a, . . . , ε) ∈ X m , where the a is in the ith position (and ε is the empty word). If u, v ∈ X m , then u.v denotes componentwise concatenation. (The dot in this notation is necessary to distinguish otherwise ambiguous expressions; for example, in 2V we write 0.0 1 = (00, 0) and 00 1 = (00, ε). ) Brin proved the following in [10].
Lemma 3.11. The group mV is generated by the set of transpositions.
By arguing as in the proof of Lemma 3.4(i) and Proposition 3.6(i), we obtain the following; we omit the details.
Remark 3.18. Observe that the α and β have coprime order and each have a fixed point in mX (3) . When m = 1, these are exactly the elements from Figure 1.

Proofs
In this section, we will prove Theorems 1 and 2. In the proof of Theorems 1 and 2 we will fix the group G as one of V n , V n or mV and the element g as a nontrivial element of G, then will construct an element h, which we prove satisfies g, h = G. To help the reader visualise the element h that we will construct in this proof, we begin this section with an example where, in a special case, we construct this element h, depict it via a tree pair and outline the general idea of why g, h = G.
Example 4.1. Let g = (00 01) ∈ V . Let x = α [00] and y = β [01] . We have given the tree pairs for g and the product xy in Figures 2 and 3. Since the subgroup g, xy stabilises the partition {00C, 01C, 1C} we deduce that g, xy is not the entire group V . The key idea is to modify xy to obtain an element h such that g, h = V .
For long binary sequences, we adopt notation such as 0 3 1 2 0 4 for 000110000. Let  Let us make some observations about the element h. First note that x, y, z 0 and z 1 have orders 6, 7, 5 and 11, respectively. In particular, these elements have coprime orders. Moreover, these elements have disjoint support and therefore commute. Consequently, each of these elements is a suitable power of h and, accordingly, is contained in g, h .
We now outline the argument for why g, h = V . Proposition 3.9 implies that We now prove Theorems 1 and 2.
Proof of Theorems 1 and 2. Let (G, Γ, Ω) be one of V n , V n or mV . Let g ∈ G be nontrivial. If G is V n or V n , then let (Γ, Ω) = (C n , X n ), and if G is mV , then (Γ, Ω) = (C m , X m ). By Lemma 2.6, we may fix u ∈ Ω such that u ⊥ ug. By Lemma 2.5, we may fix w 1 , . . . , w ∈ Ω such that W = {u, ug, w 1 , . . . , w } is a basis for Γ. For clarity of notation, write v = ug.
We will now define an element h ∈ G, which we will prove satisfies g, h = G. Bear in mind that the definition of h in Example 4.1 is a special case of the construction we are about to describe.
Let y = β [v] . Let a, b ∈ Ω (3) be fixed points of α and β, respectively (see Remarks 3.10 and 3.18). Fix 0 i , recalling that + 2 is the size of the basis W. We now define an element z i . Write u i = uau i where u i is the n-ary (where n = 2 if G = mV ) expansion of i with exactly log n ( + 1) digits. Let p 1 < p 2 < p 3 · · · be the prime numbers strictly greater than q (recall that we defined the prime q in (3.5) in the case of V n and V n and in (3.9) in the case of mV ). Write w 0 = vb and p 0 = 5. Define where σ is σ n or σ n as appropriate. Observe that z i is a p i -cycle.
We split into three cases depending on whether G is V n , V n or mV . The arguments in each of these cases are quite similar.

Questions
In this section, we present some open problems, which connect our main results with the existing work on finite groups. In the introduction we remarked that a finite group is conjectured to be 3 2 -generated if and only if every proper quotient is cyclic, and that while this conjecture is false for arbitrary groups, there are no known 2-generated counterexamples. This leads to the following natural question. (The authors were recently made aware that this question has attracted attention on MathOverflow [26].) Question 1. Is there a 2-generated group with no noncyclic proper quotients that is not 3 2 -generated? Following [7], the spread of a group G, written s(G), is the greatest nonnegative integer k such that for all nontrivial elements x 1 , . . . , x k there exists y ∈ G such that x 1 , y = · · · = x k , y = G. (5.1) Write s(G) = ∞ if there is no such maximum. Therefore, s(G) 1 is equivalent to G being 3 Question 4. Is there an infinite noncyclic group G such that s(G) > 0 but u(G) = 0?
Following [14], a subset S ⊆ G is a total dominating set for G if for all x ∈ G there exists y ∈ S such that x, y = G. Therefore, u(G) > 0 if and only if G has a total dominating set consisting of conjugate elements, a so-called uniform dominating set.
Question 5. Is there an infinite noncyclic group with a finite total dominating set?
Of course, the authors are particularly interested in whether the Higman-Thompson or Brin-Thompson groups provide affirmative answers to these questions.