Nielsen equivalence in GuptaSidki groups
Aglaia Myropolska^{*}^{*}*The author acknowledges the support of the Swiss National Science Foundation, grant 200021_144323.
Abstract
For a group generated by elements, the Nielsen equivalence classes are defined as orbits of the action of , the automorphism group of the free group of rank , on the set of generating tuples of .
Let be prime and the GuptaSidki group. We prove that there are infinitely many Nielsen equivalence classes on generating pairs of .
1. Introduction
Let be a finitely generated group. The rank of a group is the minimal number of generators of . Fix and let be the set of epimorphisms from the free group of rank to .
Consider the natural action of the group on : for and for we define
The orbits of this action are called systems (systems of transitivity). B.H. Neumann and H. Neumann, motivated by the study of presentations of finite groups, introduced systems in [NN51]. One of the main conjectures in this area, sometimes attributed to Wiegold, is that for every finite simple group there is only one system of transitivity when ^{†}^{†}†The classification of finite simple groups implies that every finite simple group can be generated by elements.. It is also not known whether there is only one orbit when the action of with is only considered.
It was proved by Nielsen [Nie18] that is generated by the following automorphisms, where is the basis of :
where , . The transformations above are called elementary Nielsen moves.
Observe that there is a onetoone correspondence between and the set of generating tuples of . The action of on the generating tuple is done by applying sequences of elementary Nielsen moves to by precomposition. For example, if then the set of generating tuples of coincides with and the elementary Nielsen moves induce elementary row operations on the matrices. It follows that the action of on is transitive.
The orbits of the action are called Nielsen (equivalence) classes on generating tuples of . In recent years the Nielsen equivalence classes became of particular interest as they appear as connected components of the Product Replacement Graph, whose set of vertices coincides with the set and the edges correspond to elementary Nielsen moves (see [Eva07, Lub11, Pak01] and Section 3 for more on this topic).
Before studying further the Nielsen equivalence, we point out its relation to the famous AndrewsCurtis conjecture [AC65]. Elementary Nielsen moves together with the transformations
where and , form the set of elementary AndrewsCurtis moves. Elementary AndrewsCurtis moves transform normally generating sets (sets which generate as a normal subgroup) into normally generating sets.
The AndrewsCurtis conjecture asserts that, for a free group of rank and a free basis of , any normally generating tuple of can be transformed into by a sequence of elementary AndrewsCurtis moves.
We say that two normally generating tuples of are AndrewsCurtis equivalent if one is obtained from the other by a finite chain of elementary AndrewsCurtis moves. The AndrewsCurtis equivalence corresponds to the actions of and of on normally generating tuples of . More generally, for a finitely generated group and , the above actions can be defined on the set of normally generating tuples of by precomposition. The orbits of this action are called the AndrewsCurtis equivalence classes in . The analysis of AndrewsCurtis equivalence for arbitrary finitely generated groups has its own importance to analyse potential counterexamples to the conjecture. A possible way to disprove the conjecture would be to find two normally generating tuples of such that their images in some finitely generated group are not AndrewsCurtis equivalent. The AndrewsCurtis equivalence was studied for various classes of groups, for instance, for finite groups in [BKM03, BLM05], for free solvable and free nilpotent groups in [Mya84], for the class of finitely generated groups of which every maximal subgroup is normal in [Myr13]. The class includes finitely generated nilpotent groups; moreover all Grigorchuk groups and GGS groups, e.g. GuptaSidki groups, belong to by [Per00, Per05]. In [AKT13] the result for GGS groups was generalized: the authors proved that all multiedge spinal torsion groups acting on the regular ary rooted tree, with odd prime, belong to .
Observe that, for a group in , a normally generating set of is, in fact, a generating set. Therefore, for groups in the partition (of the set of generating tuples) into Nielsen equivalence classes is a refinement of the partition into AndrewsCurtis classes. We further describe what is known about Nielsen equivalence for some groups in the class .
The most wellunderstood classification of Nielsen equivalence classes is known for finitely generated abelian groups (see [NN51, DG99, Oan11]). Namely, if is a finitely generated abelian group then the action of on is transitive when . Moreover, if then the number of Nielsen equivalence classes is finite and depends on the primary decomposition of (see Theorem 3.2 for details). It also follows from the latter papers that for any finitely generated abelian group there is only one system for any .
For a finitely generated nilpotent group the action of is transitive on when [Eva93]. However when the unicity of Nielsen equivalence class generally breaks down. For instance, Dunwoody [Dun63] showed that to every pair of integers and there exists a finite nilpotent group of rank and nilpotency class for which there are at least systems.
As a generalization of finite nilpotent groups, we consider the family of GuptaSidki groups where is odd prime. The group is a group of rank acting on the rooted ary tree, every quotient of which is finite and, therefore, nilpotent (GuptaSidki groups were defined in [GS83b]; the reader can find the definition in Section ). It was shown by Pervova [Per05] that the groups belong to the class . This property was the main ingredient in [Myr13] for proving that there is only one Nielsen equivalence class for for . Moreover, for a group belonging to the class , it is relevant to analyse Nielsen equivalence classes in the quotient , where is the Frattini subgroup of . Namely, if there are two generating tuples of which are not Nielsen equivalent, then their preimages in are generating tuples of which also are not Nielsen equivalent (see the section on the class in [Myr13] for details). Using this argument and also the fact that for there are, by Theorem 3.2, Nielsen classes on generating pairs of the quotient , we conclude that there are at least Nielsen classes in for .
For the question on the transitivity of the action of on is more subtle. In this paper we show in particular that, although there is only one Nielsen class on generating pairs of , the action of is not transitive on . A natural question then is how many orbits this action has.
There are numerous examples of groups with infinitely many Nielsen classes when . These groups can be found among fundamental groups of certain knots ([Zie77, HW11]), onerelator groups ([Bru76]), relatively free polynilpotent groups (see [MN13] and references therein) and many others. We show that for the GuptaSidki group, with prime, there are infinitely many Nielsen classes when . To the author’s knowledge this is the first known examples of torsion groups that have this property.
Theorem 1.1.
Let be prime and the GuptaSidki group. Then there are infinitely many Nielsen equivalence classes on generating pairs of .
The GuptaSidki group being a subgroup of , the group of automorphisms of the regular ary rooted tree , has natural quotients by , the level stabilizer subgroups. These quotients are finite nilpotent generated groups with growing nilpotency class. The latter is true since the limit of these quotients in the space of marked generated groups is the GuptaSidki group itself, which is not finitely presentable [Sid87]. In the last part of the paper we show that for each the quotient group of the GuptaSidki group has the property that the action is not transitive. Note that there is only one Nielsen equivalence class in , the abelianization of each . It would be interesting to realize whether the number of Nielsen classes grows with but this for the moment remains an open question. An affirmative answer on this question, in particular, would imply that there were infinitely many Nielsen equivalence classes in . Notice, however, that the proof of Theorem 1.1 does not rely on Proposition 1.2.
Proposition 1.2.
Let be the GuptaSidki group and the level stabilizer subgroups of . Set . Then the action is not transitive for any .
Acknowledgement. The author would like to thank Pierre de la Harpe, Tatiana Nagnibeda and Said Sidki for stimulating discussions on this work, and Laurent Bartholdi for valuable suggestions during the conference “Growth in Groups” in Le Louverain.
2. Preliminaries on groups acting on rooted trees
Let with be a finite set. The vertex set of the rooted tree is the set of finite sequences over ; two sequences are connected by an edge when one can be obtained from the other by rightadjunction of a letter in . The top node (the root) is the empty sequence , and the children of are all the for . A map is an automorphism of the tree if it is bijective and it preserves the root and adjacency of the vertices. An example of an automorphism of is the rooted automorphism , defined as follows: for the permutation , set . Geometrically it can be viewed as the permutation of subtrees just below the root . Denote by the group of automorphisms of the tree .
Let . Denote by the subgroup of consisting of the automorphisms that fix the sequence , i.e.
And denote by the subgroup of consisting of the automorphisms that fix all sequences of length , i.e.
Notice an obvious inclusion . Moreover, observe that for any the subgroups are normal and of finite index in . We therefore have a natural epimorphism between finite groups
(1) 
for any .
The examples of groups acting on rooted trees include groups of intermediate growth, such as the Grigorchuk group [Gri80] and the GuptaSidki groups [GS83a]. We define the latter family of groups below.
Fix prime and . Let be the cyclic permutation on . Let belong to and belong to . Denote by the rooted automorphism of defined by
Denote by the automorphism of defined by
The GuptaSidki group is the group of automorphisms of the tree generated by and and we will write
To shorten the notation for the element we will simply write
More generally, for any element we can write for some and .
3. Nielsen equivalence in GuptaSidki groups
For a finitely generated group and , we define the Nielsen graph^{3}^{3}3Also called the Extended Product Replacement Graph. as follows:

the set of vertices consists of generating tuples, i.e.

two vertices are connected by an edge if one of them is obtained from the other by an elementary Nielsen move.
Observe that the graph is connected if and only if the action of on is transitive.
Recall, that for a finitely generated group the Frattini subgroup is defined as the intersection of all maximal subgroups of . Equivalently, the Frattini subgroup of contains all the nongenerators, i.e. the elements which can be removed from any generating set. The latter implies the following lemma.
Lemma 3.1 ([Eva93]).
Let be a group generated by and let . Then .
As it was explained in the Introduction, for groups in class (the class of finitely generated groups all maximal subgroups of which are normal), the number of connected components of is bounded below by the number of connected components of . Since all maximal subgroups of , the GuptaSidki group, are normal, it follows that the quotient is abelian. Moreover, any generating set of the quotient can be lifted up to the generating set of [Myr13]. Therefore is a quotient of of rank ; we deduce that . Using the following theorem, we find the number of connected components of the Nielsen graph .
Theorem 3.2 ([Nn51, Dg99, Oan11]).
Let be a finitely generated abelian group with the primary decomposition with and . Then and

is connected if .

if , i.e. , then is connected;

otherwise if or then is connected and if then has connected components,
where is the Euler function (the number of positive integers less than which are coprime with ).
It follows from Theorem 3.2 and the arguments before that for the Nielsen graph has at least connected components.
To prove Theorem 1.1 we use an observation by Nielsen (sometimes also attributed to Higman, see lemma 3.3) as well as an analysis on conjugacy classes in the GuptaSidki group.
Lemma 3.3 (Nielsen).
Let and be two Nielsen equivalent generating pairs of a group . Then the commutator is conjugate either to or to .
The proof of this lemma is a straightforward calculation of commutators of the pairs obtained from by the elementary Nielsen moves.
In order to show that two elements are not conjugate in , the GuptaSidki group, sometimes we use the finite quotients by the th level stabilizers. Consider a natural epimorphism
The finite quotient can be seen as a subgroup of with
and
Recall that two elements are conjugate in the symmetric group if and only if their cycle types are the same. Therefore if for two elements their images and have different cycle types in then, in particular, they are not conjugate in . Below all computations in were done using GAP.
Example 3.4.
The elements and are not conjugate in . Indeed,
and its cycle type differs from the one of .
Let be the GuptaSidki group. Set and for all set . The fact that follows from [GS84].
Proposition 3.5.
The elements , and are not pairwise conjugate in for any such that .
Proof.
We prove the following two claims in order to conclude the proposition:
Claim 1.
is not conjugate to for any .
Claim 2.
and are not conjugate for and .
The claims will be proved by contradiction. We compute that and .
Proof of Claim . Assume that and are conjugate, then there exists for some integer , such that . Observe that because is not conjugate neither to nor to . Moreover is not conjugate to in therefore can be conjugate only to . We will prove that it is not the case. For this it is enough to show that and are not conjugate in . We will show it by induction assuming that
and then will show that (*) is indeed the case.
Suppose that and are conjugate in then there exists for some integer such that .

If then and it follows that .

If then and it follows that .

If then and it follows that .
By assumption (*) elements and are not conjugate in and we deduce that is not conjugate to in modulo assumption (*).
Proof of the assumption (*): and are not conjugate in for any .

The assumption holds for . To see this, look at the action of and on the 4th level of the tree, see Example 3.4.

Suppose (*) is true for .

Consider and suppose it is conjugate to . Then there exists with such that . Since is not conjugate neither to nor to in then . Therefore . We obtain the contradiction with the step of induction.
Proof of Claim . We will prove Claim modulo Assumption (*) and (**) below and then in the end prove that both assumptions indeed hold.
Assumption (*): for any such that the elements and are not conjugate in .
Assumption (**): for any the element is not conjugate to or in .
We prove Claim by contradiction. Suppose that there exists such that
or equivalently
(2) 
Observe that is not conjugate to , and . To see this, look at the quotient and notice that the images of and are not conjugate in . Therefore can not be conjugate to . Moreover, it follows from Assumption (**) that in equation (2).
To obtain the contradiction it is sufficient to show that is not conjugate to . Suppose they are conjugate, then there exists with such that

If then and it follows that .

If then and it follows that .

If then and it follows that .
By Assumption (*), elements and are not conjugate in and we deduce that and are not conjugate in modulo assumptions (*) and (**).
Proof of the assumption (*) Without loss of generality suppose that . Suppose and are conjugate. Then there exists with such that
Since is not conjugate to or to we conclude that and hence and are conjugate. Continuing in the same way, we deduce that the elements and are conjugate. We obtain a contradiction since is not conjugate to or to (to see this it is enough to look at the action of these elements on the 4th level of the tree) or to .
Proof of the assumption (**) To see that is not conjugate to or , it is enough to look at the action of these elements on the third level of the tree and to see that they have different cycle types, hence they are not conjugate in the quotient . And for , the action of on the third level is trivial therefore it is enough to look at the action of , and on the third level to see that they have different cycle types and therefore not conjugate in . ∎
Let be the GuptaSidki group for prime. Set and for set . The fact that follows from [GS84].
Proposition 3.6.
For any and the elements and are not conjugate in .
Proof.
By contradiction, suppose that there exists an element
with such that
or, in other words,
(3) 
Suppose . Observe that is not conjugate to , , , , , and to . To see this, look at the quotient , and notice that the image of is not conjugate to the images of the elements above. Therefore must be conjugate to , in other words there exists with such that
where , , and otherwise.
It follows that the following system of equations holds:
where is the th power of the permutation and, for each , denotes the image of under .
After solving the system one obtains that
which gives us a contradiction to .
In view of equation and that , in order to obtain a contradiction to the initial assumption that is conjugate to , it is enough to show that is not conjugate to . Without loss of generality suppose that .
Suppose by contradiction that is conjugate to , i.e. there exists with such that
Observe that is not conjugate to , and . Hence and therefore is conjugate to . We repeat the same arguments times to conclude that and are conjugate. Observe that is not conjugate to , , , and . The contradiction then follows and we deduce that is not conjugate to which concludes the proof.
∎
We are now able to deduce that there are infinitely many Nielsen equivalence classes on generating pairs of the GuptaSidki group for any prime.
Proof of Theorem 1.1.
Proof of Proposition 1.2.
First, we show that the graph is not connected. Consider two pairs and in . Since and , it follows that is also a generating pair of by Lemma 3.1.
Denote the images of and in the finite quotient by and . Clearly the pairs and are generating. If they are Nielsen equivalent then by Nielsen criterion (Lemma 3.3) their commutators and must be conjugate in and, in particular, their cycle types must be the same. We will obtain the contradiction with the latter.
We calculate the commutators respectively :