The power conjugacy problem in Higman-Thompson groups

An introduction to the universal algebra approach to Higman-Thompson groups (including Thompson's group $V$) is given, following a series of lectures by Graham Higman in 1973. In these talks, Higman outlined an algorithm for the conjugacy problem; which although essentially correct fails in certain cases, as we show here. A revised and complete version of the algorithm is written out explicitly. From this, we construct an algorithm for the power conjugacy problem in these groups. Python implementations of these algorithms can be found at [26].


Introduction
Thompson introduced the group now called "Thompson's group V ", and its subgroups F < T , in 1965, and so gave the first examples, namely V and T , of finitely presented, infinite, simple groups (see [CFP96,Tho]). McKenzie and Thompson [MT73] also used V to construct finitely presented groups with unsolvable word problem. Subsequently, Galvin and Thompson (unpublished) identified V with the automorphism group of an algebra, V 2,1 , studied by Jónsson and Tarski [JT61]. Higman [Hig74] generalised this construction, defining G n,r to be automorphism group of a generalisation V n,r of V 2,1 , for n ≥ 2 and r ≥ 1, and showed the commutator subgroup of G n,r to be a finitely generated, infinite, simple group, for all n ≥ 2. (G n,r is perfect when n is even, and its commutator subgroup has index 2 when n is odd.) The groups G n,r are the "Higman-Thompson" groups of the title. There are many isomorphic groups in this set: in fact the algebras V n,r and V n ′ ,r ′ are isomorphic if and only if n = n ′ and r ≡ r ′ mod n − 1; so G n,r ∼ = G n ′ ,r ′ if n = n ′ and r ≡ r ′ mod n − 1. Higman [Hig74] showed that there are infinitely many non-isomorphic groups G n,r and gave necessary conditions for such groups to be isomorphic. Recently Pardo [Pardo11] completed the isomorphism classification, showing that Higman's necessary conditions are also sufficient: that is G n,r ∼ = G n ′ ,r ′ if and only if n = n ′ and gcd(n − 1, r) = gcd(n ′ − 1, r ′ ). Higman-Thompson groups have been much studied and further generalised: we refer to [CFP96,Brin04,BGG11,MPN13,DMP14,BCR] for example.
In this paper we consider the conjugacy and power-conjugacy problems in Higman-Thompson groups. We use Higman's method, describing the groups G n,r in terms of universal algebra. This allows us to give a detailed description of the algorithm for the conjugacy problem; and to uncover a gap in the original algorithm proposed by Higman. To be precise, Lemma 9.6 of [Hig74] is false, and consequently the "orbit sharing" algorithm in [Hig74] does not always detect elements in the same orbit of an automorphism. The orbit sharing algorithm is crucial to the algorithm for conjugacy given in [Hig74], which may fail to recognise that a pair of elements of G n,r are conjugate.
Fortunately it is not difficult to complete the algorithm. We then extend these results to construct an algorithm for the power conjugacy problem. The third author has implemented the algorithms described in this paper in Python [D15]. In fact it was the process of testing this implementation which uncovered the existence of an orbit unrecognised in [Hig74]; and it became evident that the algorithms of [Hig74] are incomplete.
Note that other approaches to algorithmic problems in G n,r have been developed. For example [SD10] gives a different algorithm for the conjugacy problem in G 2,1 , using the revealing tree pairs of Brin [Brin04]. In [BGG11] the same methods are used to study the centralisers of elements of G n,1 for n ≥ 2. Again Belk and Matucci [BM07] gave a solution to the conjugacy problem in G 2,1 based on strand diagrams. On the other hand, Higman's methods were used by Brown [B87] to show that all the Higman-Thompson groups are of type F P ∞ , and have been extended to generalisations of Higman-Thompson groups, to prove finiteness properties, by Martinez-Perez and Nucinkis [MPN13].
In detail the contents of the paper are as follows. In order to make this account self-contained we begin the paper with an introduction to universal algebra. Section 2 outlines the universal algebra required, following Cohn's [Cohn91]. In Section 2.1 we introduce Ω-algebras; that is universal algebras with signature Ω. Sections 2.2 and 2.3 cover quotients of Ω-algebras, varieties of Ωalgebras and free Ω-algebras. We use this machinery in Section 3 to define the algebras V n,r and establish their basic properties, following the exposition of [Hig74].
The groups G n,r are defined in Section 4, as the automorphism groups of V n,r . Following [Hig74] we represent elements of G n,r as bijections between carefully chosen generating sets of the algebras V n,r . This is done in two stages beginning with the semi-normal forms of Section 4.1. There are many ways of representing a given automorphism in semi-normal form, but in Section 4.2 it is shown that this representation may be refined to a unique quasi-normal form. Furthermore, an algorithm is given which takes an automorphism and produces a quasi-normal form representation.
The solution to the conjugacy problem is based on an analysis of certain orbits of automorphisms in quasi-normal form, and we give a full account of this analysis in Sections 4.1 and 4.2. Here we follow [Hig74] except that, as pointed out above, there exist orbits of types not recognised there, which give automorphisms in quasi-normal a richer structure, as described here.
Section 5 contains the algorithm for the conjugacy problem. This involves breaking an automorphism down into well-behaved parts. It is shown that every element of G n,r decomposes into factors which are called periodic and regular infinite parts. The conjugacy problem for periodic and regular infinite components are solved separately and then the results recombined. The decomposition into these parts is the subject of Section 5.1 and here we give the main algorithm for the conjugacy problem, Algorithm 5.6. This algorithm depends on algorithms for periodic and regular infinite automorphisms: namely Algorithm 5.13 in Section 5.3 and Algorithm 5.27 in Section 5.4.
In Section 6 we turn to the power conjugacy problem. That is, given elements g, h ∈ G n,r the problem is to find all pairs of non-zero integers (a, b) such that g a is conjugate to h b . Again the problem splits into the periodic and regular infinite parts. The periodic part is straightforward, and reduces to the conjugacy problem; see Section 6.1. The algorithm for power conjugacy of regular infinite elements is Algorithm 6.13, in Section 6.3 and gives the main result of the paper Theorem 6.14: that the power conjugacy problem is solvable. On input g, h ∈ G n,r the algorithm returns a (possibly empty) set S consisting of all pairs of integers (a, b) such that g a and h b are conjugate; as well as a conjugator, for each pair. The examples given throughout the text are used as examples in [D15], from where these and other examples may be run through the third author's implementations of the algorithms. (To find Example x.y in [D15], follow the instructions in the documentation to install the program; then run from thompson.examples import example x y in a Python session.) Definition 2.2. Given an Ω-algebra (S, Ω), an Ω-subalgebra is an Ω-algebra (T, Ω) whose carrier T is a subset of S.
The intersection of any family of subalgebras is again a subalgebra. Hence, for any subset X of the set S we may define the subalgebra X generated by X to be the intersection of all subalgebras containing X. The subalgebra X may also be defined recursively: that is X is the subset of S such that (i) X ⊆ X , (ii) if y 1 , . . . , y n ∈ X then y 1 · · · y n f ∈ X , for all f ∈ Ω(n) and (iii) if s does not satisfy (i) or (ii) then s does not belong to X . Loosely speaking we might say that X is obtained from X by applying a finite sequence of operations of Ω. If the subalgebra generated by X is the whole of S, then X is called a generating set for (S, Ω).
If g is compatible with each f ∈ Ω, it is called a homomorphism from A = (S, Ω) to B = (S ′ , Ω). If a homomorphism g from A to B has an inverse g −1 which is again a homomorphism, g is called an isomorphism and then the Ω-algebras A = (S, Ω), B = (S ′ , Ω) are said to be isomorphic. An isomorphism of an algebra A = (S, Ω) with itself is called an automorphism and a homomorphism of an algebra into itself is called an endomorphism. A homomorphism is determined once the images of a generating set are fixed.

From a family
of Ω-algebras we can form the direct product P = m i=1 A i of Ω-algebras. Its set is the Cartesian product S of the S i , and the operations are carried out component wise. Thus, if π i : S → S i are the projections from the product to the factors then any f ∈ Ω of arity n is defined on S n by the equation (p 1 · · · p n f )π i = (p 1 π i ) · · · (p n π i )f, where p i ∈ S.
Let C be a class of Ω-algebras, whose elements we will call C-algebras. By a free C-algebra on a set X we mean a C-algebra F with the following universal property.
There is a mapping µ : X → F such that every mapping f : X → A into a C-algebra A can be factored uniquely by µ to give a homomorphism from F to A, i.e. there exists a unique homomorphism f ′ : F → A such that µf ′ = f .
In this case we say that X is a free generating set or a basis for F . If X is a subset of F then we shall always assume that µ is the inclusion map. Not every class has free algebras, but they do exist in the class under consideration here (see Proposition 2.16).
A free product is defined similarly, replacing the set X by a collection of C algebras. Given an indexing set I and for each i ∈ I an Ω algebra A i from C the free product A of {A i } i∈I , written A = * i∈I A i , is an Ω-algebra in C satisfying the following property.
There exist homomorphisms µ i : A i → A, for all i ∈ I, such that for any Ω-algebra B and homomorphisms f i : A i → B, for all i ∈ I, there exists a unique homomorphism f ′ : A → B such that µ i f ′ = f i , for all i.
Given collections {A i } i∈I and {B i } i∈I of Ω-algebras such that there exist free products A = * i∈I A i and B = * i∈I B i , then, by definition, there exist homomorphisms µ i : A i → A and µ ′ i : B i → B, for all i ∈ I. In this case, given homomorphisms f i : A i → B i , for all i ∈ I, the composition f i µ ′ i is a homomorphism from A i to B, so there exists a unique homomorphism f ′ : A → B, with µ i f ′ = f i µ ′ i , for all i ∈ I. We denote f ′ by * i∈I f i .

Congruence on an Ω-algebra
A relation between two sets S and R is defined to be a subset of the Cartesian product S × R. A mapping f : S → R is a relation Γ f ⊂ S × R with the properties that for each s ∈ S there exists r ∈ R such that (s, r) ∈ Γ f (everywhere defined) and if (s, r), (s, r ′ ) ∈ Γ f then r = r ′ (single valued). A relation Γ ⊂ S × R has an inverse Γ −1 , defined by and if ∆ ⊂ R × T is a relation then the composition Γ • ∆ of Γ and ∆ is defined by Γ • ∆ = {(s, t) ∈ S × T |(s, x) ∈ Γ and (x, t) ∈ ∆ for some x ∈ R}.
Given a set S the identity relation 1 S = {(s, s)|s ∈ S} and the universal relation S 2 = {(s, s ′ )|s, s ′ ∈ S} always exist. An equivalence on a set S is a subset Γ of S 2 with the properties Γ•Γ ⊂ Γ (transitivity): Γ −1 = Γ (symmetry) and 1 S ⊆ Γ (reflexivity). The equivalence class of s ∈ S is {s ′ ∈ S|(s, s ′ ) ∈ Γ} = {s}Γ. Given any subset U of S × S, the equivalence generated by U is E = {V ⊆ S × S|V is an equivalence and U ⊆ V }; that is, the smallest equivalence E on S containing U . It follows that E is {(a, b) ∈ S × S|there exists a 0 , . . . , a n such that a 0 = a, a n = b and (a i , a i+1 ) ∈ U }.
Of particular interest in the study of Ω-algebras are relations which are also subalgebras. Firstly, if A = (S, Ω) and B = (R, Ω) are Ω-algebras and Γ ⊂ S × R is a relation which is closed under the operations of Ω, as defined in A × B, then (Γ, Ω) is a subalgebra of A × B. In this case we abuse notation and say Γ is a subalgebra of A × B.
Lemma 2.4 ([Cohn91, Lemma 2.1, Chapter 1]). Let A, B, C be Ω-algebras and let Γ, ∆ be subalgebras of A × B, B × C respectively. Then Γ −1 is a subalgebra of B × A, Γ • ∆ is a subalgebra of A × C and if A ′ is a subalgebra of A, with carrier S ′ ⊆ S, then (S ′ Γ, Ω) is a subalgebra of B.
Let S and T be sets and f : S → T a mapping between them. The image of f is defined as SΓ f , and the kernel of f is defined as The latter is an equivalence on S; the equivalence classes are the inverse images of elements in the image (sometimes called the fibres of f ).
Example 2.5 (Groups). Given a group homomorphism f : G → H, the (group-theoretic) kernel of f is a normal subgroup N ; and the different cosets of N in G are the fibres of f . So, the equivalence classes of kerf , in the definition above, are the cosets of N in G.
A congruence on an Ω-algebra A = (S, Ω) is an equivalence on S which is also a subalgebra of A 2 i.e. an equivalence Γ ⊂ S × S which is Ω-closed. From the above, 1 A and A 2 are congruences on A. Given any subset U ⊆ S × S the congruence generated by U is It follows that C is the smallest congruence on A containing U .
Let A be an Ω-algebra. By definition a congruence is an equivalence which admits the operations ω (ω ∈ Ω). Now each n-ary operator ω defines an n-ary operation on A: (a 1 , . . . , a n ) → a 1 · · · a n ω for a 1 , . . . , a n ∈ A. (1) By giving fixed values in A to some of the arguments, we obtain r-ary operations for r ≤ n. In particular, if we fix all the a j except one, say the ith, we obtain, for any n − 1 fixed elements a 1 , . . . , a n−1 ∈ A, a unary operation x → a 1 · · · a i−1 xa i · · · a n−1 ω; (2) and this applies for all i ∈ {1, . . . , n}. We say that the operation (2) is an elementary translation (derived from Ω by specialisation in A). Given a finite sequence τ 1 , . . . , τ n of elementary transformations the composition τ = τ 1 • · · · • τ n is also a unary operation on A, which we call a translation.
(In particular we allow n = 0 in this definition, so the identity map on A is a translation.) Proposition 2.6 ([Cohn81, Proposition 6.1, Chapter6]). An equivalence q on an Ω-algebra A is a congruence if and only if it is closed under all translations. More precisely, a congruence is closed under all translations, while any equivalence which is closed under all elementary translations is a congruence.
Remark 2.7. If U ⊆ S × S, then the congruence generated by U can be seen to consist of pairs (a, b) ∈ S × S such that there exist m ≥ 0, a 0 , . . . , a m ∈ S, and a translation τ with That is, there exist s 1 , . . . , s n−1 ∈ S, u 0 , . . . , u m ∈ S, and ω ∈ Ω(n) such that (u i , u i+1 ) ∈ U ∪ U −1 ∪ 1 S and setting a i = (s 1 , . . . , s j−1 , u i , s j , . . . , s n−1 )ω, The next two theorems explain the significance of congruences for Ω-algebras and will be used in the following section on free algebras and varieties.
Theorem 2.8 ([Cohn91, Theorem 2.2, Chapter 1]). Let g : A → B be a homomorphism of Ωalgebras. Then the image of g is a subalgebra of B and the kernel of g is a congruence on A.
Theorem 2.9 ([Cohn91, Theorem 2.3, Chapter 1]). Let A be an Ω-algebra and q a congruence on A. Then, there exists a unique Ω-algebra, denoted A/q, with carrier the set of all q-classes such that the natural mapping ν : A → A/q is a homomorphism.
The homomorphism ν in the previous theorem, which maps an element s of the carrier of A to its q-equivalence class, is called the natural homomorphism from A to A/q. The algebra A/q is called the quotient algebra of A by q.
Example 2.10. Given a group G and a normal subgroup N of G, the natural mapping G → G/N is a homomorphism.

Free algebras and varieties
Let X = {x 1 , x 2 , . . .} be a non-empty, finite or countably enumerable set, called an alphabet, and Ω an operator domain, with Ω ∩ X = ∅. We define an Ω-algebra as follows. An Ω-row in X is a finite sequence of elements of Ω ∪ X. The set of all Ω-rows in X is denoted W (Ω; X). The length of the Ω-row w = w 1 · · · w m where w i ∈ Ω ∪ X, is defined to be m and is written |w|. The carrier of our Ω-algebra is W (Ω; X), the set of Ω-rows. We define the action of elements Ω on W (Ω; X) by concatenation. First observe that if u and v are Ω-rows then the concatenation uv of u with v is also an Ω-row, and this may be extended to the concatenation of arbitrarily many Ω-rows in the obvious way. Now, if f ∈ Ω(n), and u 1 , . . . , u n ∈ W (Ω; X), then the the image of the n-tuple (u 1 , . . . , u n ) ∈ W (Ω, X) n under the operation f is the Ω-row u 1 · · · u n f . By abuse of notation we will refer to W (Ω; X) as an Ω-algebra.
It is clear that X ⊂ W (Ω; X) and we call the subalgebra generated by X the Ω-word algebra on X, denoted W Ω (X). Its elements are called Ω-words in the alphabet X. There is a clear distinction between Ω-rows which are Ω-words and those that are not. For example, if there is one binary operation f , then is a Ω-row which is also an Ω-word while x 1 f f x 2 f x 3 is an Ω-row which is not an Ω-word.
Moreover, each Ω-word can be obtained in precisely one way by applying a finite sequence of operations of Ω to elements of X.
Let A be an Ω-algebra. If in an element w of W Ω (X) we replace each element of X by an element of A we obtain a unique element of A. For |w| = 1, this is clear, so assume |w| > 1 and we will use induction on the length of w. We have w = u 1 · · · u n f (f ∈ Ω(n), u i ∈ W Ω (X)), where, by Proposition 2.12, the u i are uniquely determined once w is given. By induction each u i becomes a unique element a i ∈ A, when we replace the elements of X by elements of A. Hence w becomes a 1 · · · a n f ; a uniquely determined element of A.
This establishes the next theorem.
Theorem 2.13 ([Cohn91, Theorem 3.2, Chapter 1]). Let A be an Ω-algebra and let X be a set. Then any injective mapping θ : X → A extends, in just one way, to a homomorphismθ : W Ω (X) → A. That is, W Ω (X) is a free Ω-algebra, freely generated by X.
Corollary 2.14 ([Cohn91, Corollary 3.3, Chapter 1]). Any Ω-algebra A can be expressed as a homomorphic image of an Ω-word algebra W Ω (X) for a suitable set X. Here X can be taken to be any set mapping onto a generating set of A.
By an identity or law over Ω in X we mean a pair (u, v) ∈ W Ω (X) × W Ω (X) or an equation u = v formed from such a pair. We say that the law (u, v) holds in the Ω-algebra A or that A satisfies the equation u = v if every homomorphism W Ω (X) → A maps u and v to the same element of A. This correspondence between sets of laws and classes of algebras establishes a pair of maps, with the following definitions.
• Given a set Σ of laws over Ω in X, form V Ω (Σ), the class of all Ω-algebras satisfying all the laws in Σ. This class V Ω (Σ) is called the variety generated by Σ.
• Given a class C of Ω-algebras we can form the set q = q(C) of all laws over Ω in X which hold in all algebras of C.
Thus we have a pair of maps V Ω and q; relating each variety of Ω-algebras to a relation q on W Ω (X) and vice-versa. We shall see below that q(C) is a congruence, but first we make a further definition. A subalgebra of an Ω-algebra A is called fully invariant if it is mapped into itself by all endomorphisms of A. A congruence Γ on A is said to be fully invariant if (u, v) ∈ Γ implies (uθ, vθ) ∈ Γ, for all endomorphisms θ of A. The fully invariant congruence generated by Γ is It follows that I is the smallest invariant congruence on A generated by Γ.
We claim that if C is a class of Ω-algebras then q(C) is a fully invariant congruence on W Ω (X). To see that q(C) is a congruence note first that in every class C of Ω-algebras: u = u for all u ∈ W Ω (X); if u = v holds then so does v = u; and if u = v and v = w then also u = w. Further, if u i = v i for i = 1, . . . , n, are laws holding in A and ω ∈ Ω(n), then u 1 · · · u n ω = v 1 · · · v n ω holds in A. Hence q(C) is indeed a congruence.
To see that q(C) is a fully invariant congruence, let (u, v) ∈ q(C) and let θ be any endomorphism of W Ω (X). If α : W Ω (X) → A, where A ∈ C, is any homomorphism, then so is θα, hence uθα = vθα. Thus the law uθ = vθ holds in A, so (uθ, vθ) ∈ q(C) and thus q(C) is a fully invariant congruence.
Cohn shows in addition that the map V Ω is a bijection with inverse q, and deduces the following theorem.
Given sets S and T and a relation Γ from S to T , we may use Γ to define a system of subsets of S, T , as follows. For any subset X of S we define a subset X * of T by and similarly, for any subset Y of T we define a subset Y * of S by We thus have mappings X → X * and Y → Y * of the power sets of S and T with the following properties: A pair of maps X → X * , from the power set 2 S of S to the power set 2 T of T , and Y → Y * , from 2 T to 2 S , satisfying (3-5) is called a Galois connection.
Theorem 2.15 ([Cohn91, Theorem 3.5, Chapter 1]). Let W = W Ω (X) be the Ω-word algebra on the alphabet X. The pair of maps Σ → V Ω (Σ) and C → q(C) forms a Galois connection giving a bijection between varieties of Ω-algebras and fully invariant congruences q on W Ω (X).
Proposition 2.16 ([Cohn91, Proposition 3.6, Chapter 1]). Let V be a variety of Ω-algebras and q the congruence on W Ω (X) (the Ω-word algebra generated by X) consisting of all the laws on V i.e. the fully invariant congruence q(V). Then W Ω (X)/q is the free V-algebra on X.
Suppose Σ is a set of laws over Ω in X and let V = V Ω (Σ) and q = q(V). Then Σ ⊆ q and, from Proposition 2.16, q is a fully invariant congruence and W Ω (X)/q is the free V-algebra. Now let p be the fully invariant congruence generated by Σ. Then, as Σ ⊆ q and q is a fully invariant congruence, we have p ⊆ q. Let A = W Ω (X)/p. Then A is an Ω-algebra, in which every law of Σ holds (as Σ ⊆ p). Thus A is a V-algebra. Then, from Proposition 2.16, the natural map X → A extends to a homomorphism W Ω (X)/q → A. It follows that q ⊆ p. Therefore p = q = q(V). We record this as a corollary which we shall use in Section 3 to construct Higman's algebras V n,r .
Corollary 2.17. Let Σ be a set of laws over Ω in X, let V = V Ω (Σ) and q = q(V). Then q is the fully invariant congruence generated by Σ.

The Higman Algebras V n,r
In this section we define the algebras which Higman called V n,r . Let n be an integer, n ≥ 2 and let A be an Ω-algebra, with carrier S and operator domain Ω = {λ, α 1 , . . . , α n }, such that a(α i ) = 1, for i = 1, . . . , n and a(λ) = n. We call the n-ary operation λ : S n → S a contraction and the unary operations α i : S → S descending operations. We define a map α : S → S n , which we shall call an expansion, by vα = (vα 1 , . . . , vα n ), for all v ∈ S. For any subset Y of S, a simple expansion of Y consists of substituting some element y of Y by the n elements of the tuple yα. A sequence of d simple expansions of Y is called a d-fold expansion of Y . A set obtained from Y by a d-fold expansion, d ≥ 0, is called an expansion of Y . For example, if x ∈ S then {xα 1 , . . . , xα n } is the unique simple expansion of {x} and the 2-fold expansions of {x} are the sets {xα 1 , . . . , xα i−1 , xα i α 1 , . . . , xα i α n , xα i+1 , . . . , xα n }, for 1 ≤ i ≤ n. Every d-fold expansion of Y has |Y | + (n − 1)d elements. Similarly, a simple contraction of Y consists of substituting n distinct elements {y 1 , . . . , y n } ∈ Y by the single element (y 1 , . . . , y n )λ. A set obtained from Y by applying a finite number of simple contractions is called a contraction of Y . From now on in this paper, Ω is fixed as above. Let X be a non-empty set and recall that the Ω-word algebra W Ω (X) is the free Ω-algebra on X.
From Proposition 2.16 and Corollary 2.17, it follows that q is the fully invariant congruence on W Ω (X) generated by Σ n and W Ω (X)/q is the free V n -algebra on X.
Definition 3.2. Let X be a non-empty, finite or countably enumerable, set of cardinality r and n ≥ 2 an integer. Then V n,r (X) is the the free V n -algebra W Ω (X)/q, where q = q(V n ) and V n = V Ω (Σ n ).
When no ambiguity arises we refer to V n,r (X) as V n,r .
Remark 3.3. In [Hig74, Section 2] Higman defines a standard form over X to be one of the finite sequences of elements of X ∪ {α 1 , . . . , α n , λ} specified by the following rules.
(ii) If w 1 , . . . , w n are standard forms then so is w 1 · · · w n λ, unless there is a standard form u such that w i = uα i for i = 1, . . . , n.
(iii) No sequence is a standard form unless this follows from (i) and (ii).
The set of standard forms is made into an Ω-algebra by defining the operations α 1 , . . . , α n , λ as follows.
Higman then goes on to prove that the algebra of standard forms is a free V n -algebra, freely generated by X ([Hig74, Lemma 2.1]). This follows in our case from the definition above, and the remarks following it, together with Lemma 3.4 below. To prove Lemma 3.4 one of several standard arguments, to prove statements of this form in algebras of various types, may be followed: and we omit the details. Let y be the minimal length representative of its equivalence class in V n,r i.e. y is a standard form. Then the length of of the equivalence class of y is the length of y, denoted |y|, and the λ-length of the equivalence class of y is the number of times the symbol λ occurs in y.
Now that we have a concrete description of the free algebra V n,r in the variety V n , we recall those results of [Hig74, Section 2], required in the sequel. 1. Every expansion of B is a basis of V n,r (X).
2. Every contraction of B is a basis of V n,r (X).

Proof.
1. Let Y be a d-fold expansion of B, where d ≥ 0. Arguing by induction, we assume that every d-fold expansion of B is a basis of V n,r and show that any simple expansion of Y is also a basis. Let y ∈ Y and let Y ′ be the simple expansion Since y = yα 1 · · · yα n λ, the set Y ′ generates V n,r . It remains to show that Y ′ is a basis for V n,r .
Henceθ * also extends θ. Furthermore, any other homomorphism which extends θ must equal θ * , since any such map must be defined on Y in the same way as θ * .
2. This is proved in the same way as 1 The final statement of Corollary 3.13 forms a partial converse to this lemma, for finite bases. Mostly we work with bases for V n,r (X) which are expansions of X, so we make the following definition.
Lemma 3.7. Let B be an A-basis and Y a finite basis for V n,r (X). If B ⊆ Y A then B is an expansion of Y .
Proof. Since Y is finite, there exists an expansion of Y contained in B A . Let d be minimal such that a d-fold expansion of Y is contained in B A and let W be a d-fold expansion of Y which is contained in B A . Let w ∈ W ; so w = bΓ, for some b ∈ B and Γ ∈ A * . As B ⊆ Y A , we have b = y∆, for some y ∈ Y and ∆ ∈ A * ; so w = y∆Γ. Also, as w ∈ W , there exists y ′ ∈ Y such that w = y ′ Γ ′ , as part of an expansion of Y . As Y is a basis it follows that y = y ′ and ∆Γ = Γ ′ . If Γ = 1 then Γ = Γ 0 α j , for some α j ∈ A and Γ 0 ∈ A * , and as W is an expansion of Y it follows that y∆Γ 0 α i ∈ W , for all i ∈ {1, . . . , n}.
This contradicts the choice of d and W , so Γ = 1 and w ∈ B. Hence W ⊆ B.
If A = (S, Ω) is an Ω-algebra with carrier S then we may form the A-algebra (S, A) and the {λ}-algebra (S, {λ}), where the elements of A and {λ} have actions inherited from A. We call these, respectively, the A-algebra and {λ}-algebra of A. A subset U of V n,r is said to be A-closed if uα i ∈ U , for all α i ∈ A, and an A-closed subset is called an A-subalgebra of (the A-algebra of) V n,r . Similarly W ⊆ V n,r is called a {λ}-subalgebra (of the {λ}-algebra of V n,r ) if it is {λ}-closed : that is if wλ ∈ W , for all w ∈ W .
Definition 3.8. Let Y be a subset of V n,r . The A-subalgebra generated by Y is denoted Y A . The {λ}-subalgebra generated by Y is denoted Y λ .
The free monoid on a set L is denoted L * . If Y is a subset of V n,r (X) then Y A * = {yΓ | y ∈ Y, Γ ∈ A * } is A-closed, and it follows that Y A = Y A * . If, in addition, Y ⊆ X A then yΓ is a standard form, for all y ∈ Y and Γ ∈ A * . In the sequel we write Y A λ for (Y A ) λ .
A word Γ ∈ A * is called primitive if it is not a proper power of another word; that is, if Γ is non-trivial and Γ ∈ {∆} * , for some ∆ ∈ A * , then Γ = ∆. In particular, for each word Γ ∈ A * , there exists a unique primitive word Λ such that Γ ∈ {Λ} * .
Proposition 3.10 ([Lot83], Proposition 1.3.2, Chapter 1). Two words Γ, ∆ ∈ A * commute if and only if they are powers of the same word. More precisely, the set of words commuting with a word Γ ∈ A * is a monoid generated by a single primitive word.
Lemma 3.11 ([Hig74, Section 2, Lemma 2.2]). Let Y be a subset of V n,r and let W be the Ωsubalgebra of V n,r generated by Y . Then 1. W = Y A λ and 2. for all w ∈ W , the set w A \ Y A is finite. Proof.
1. Let w ∈ W . Then there exists a finite subset Y 0 of Y such that w belongs to the subalgebra W 0 of V n,r generated by Y 0 . Let Z be an expansion of X such that |Z| ≥ |Y 0 |. Choose a surjection β of Z onto Y 0 . As V n,r is freely generated by Z we may extend β to a homomorphism from V n,r to W 0 . Let w 0 be the preimage of w under this homomorphism and let l be the λ-length of the standard form of w 0 over Z. By a straightforward induction on l it is apparent that w 0 ∈ Z A λ . Hence the image w of w 0 in W 0 belongs to Y 0 A λ ⊆ Y A λ , as required.
2. As in the previous part of the proof, we may assume that W is freely generated by Y . Let w ∈ W and let l be the λ-length of the standard form of w over Y . Then wα i1 · · · α ir ∈ Y A , whenever r ≥ l. Hence, the only elements of the set difference w A \ Y A are those of the form wα i1 · · · α ir with r < l, and there are only finitely many of these since we only have n choices for each α ij .
Lemma 3.12 ([Hig74, Section 2, Lemma 2.4]). Let x be a set of size r ≥ 1 and let X ⊆ V n,r (x) be an expansion of x. If U is a subset of V n,r (x) contained in X A , then the following are equivalent: Moreover, if Y in 1 is a basis for V n,r then Z in 3 is an expansion of Y .
Proof. Firstly, let U = X A ∩ Y A . Since U is the intersection of A-closed sets, it is also A-closed. By lemma 3.11, X A \ Y A is finite and therefore X A \ U is finite. So 1 implies 2. Secondly, assume that U is A-closed and X A \ U is finite. We will prove 3, and the final statement of the Lemma, by induction on the size of |X A \ U |. If |X A \ U | = 0, then 3 holds with Z = X. Moreover, in this case it follows from Lemma 3.7 that Z is an expansion of Y . Otherwise, |X A \ U | > 0 and we choose an element w ∈ X A \ U of greatest length (|w| is maximal). Then the set U * = U ∪ {w} is A-closed and |X A \ U * | = |X A \ U | − 1. By induction, there is an expansion Z * of X such that U * = Z * A . If Y is a basis then, in addition, Z * is an expansion of Y . The element w belongs to Z * , otherwise w would have the form w = zα i1 · · · α ir , where z ∈ Z * and r > 0, and hence z ∈ U * \ {w} = U . However, U is A-closed and so this would imply that w ∈ U , a contradiction. If we take then this is again an expansion of X (and of Y in the case that Y is a basis) and by the choice of w we have wα i ∈ U , for all i. Therefore U = Z A , 2 implies 3 and the final statement holds if Y is a basis.
Finally, if U = Z A for some expansion Z of X, then U = X A ∩ Y A , with Y = Z, and so 3 implies 1.
Corollary 3.13 (cf. [Hig74, Corollary 1, page 12]). Let B and C be finite free generating sets for V n,r (X). Then B and C have a common expansion Z, which may be chosen such that Z A = B A ∩ C A . In particular, every finite basis of V n,r (X) may be obtained from X by an expansion followed by a contraction.
Proof. Let f be an isomorphism from V n,r (X) to V n,|B| (B) mapping b ∈ B ⊆ V n,r (X) to b ∈ V n,|B| (B), for all b ∈ B. Let C ′ = Cf , so C ′ is a basis for V n,|B| (B). From Lemma 3.12, B and C ′ have a common expansion Z ′ such that B A ∩ C ′ A = Z ′ A . Then B and C have common expansion Z = Z ′ f −1 , and the remainder of the first statement of the lemma follows. The final statement follows on taking B to be an arbitrary finite free generating set and C = X. Proof. If r ≡ s mod n − 1 then it follows from Lemma 3.5 that V n,r ∼ = V n,s . Conversely, let θ be an isomorphism from V n,r (X) to V n,s (Y ), where X and Y are sets of size r and s, respectively. Then Xθ is a basis of V n,s (Y ) of size r. From Corollary 3.13, there is a common expansion Z of Xθ and Y . If Z is a d-fold expansion of Xθ and an e-fold expansion of Y then r + (n − 1)d = |Z| = s + (n − 1)e, so r ≡ s mod (n − 1), as claimed.
We could henceforth restrict to V n,r , where 1 ≤ r ≤ n − 1. However, we do not need to do this for what follows here, and it is convenient to allow arbitrary positive values of r, and multiple instances of the same algebra.
Definition 3.15. Let u, v be elements of V n,r . Then, u is said to be a proper initial segment of v if v = uΓ for some non-trivial Γ ∈ A * . If u = v or u is a proper initial segment of v then u is called an initial segment of v . Then ( †) implies that V ⊆ U . Also, U is A-closed and B A \ U consists of initial segments of the elements of the finite set V , so it is finite. Thus, by Lemma 3.12, there is an expansion Z of B such that U = Z A . Therefore, U ⊆ Z A , and this implies that V ⊆ Z (for an element of Z A \ Z has a proper initial segment in Z ⊆ U so it can not be in V by the definition of U ). Hence, V is contained in an expansion of B.
2. If V is an expansion of B then ( †) is satisfied and for each u ∈ B A there exists v ∈ V such that one of u, v is an initial segment of the other.
Suppose V satisfies ( †) and for each u ∈ B A there exists v ∈ V such that one of u, v is an initial segment of the other. By Part 1, V is contained in an expansion Z of B. If V = Z then there is an element z ∈ Z \ V and hence by the hypothesis there exists v ∈ V such that one of v or z is an initial segment of the other. But no element of Z can be an initial segment of another, so this is a contradiction and hence V = Z.
3. If V is a set of free generators for the subalgebra it generates then ( †) is satisfied.
Suppose ( †) is satisfied. If V is not a free generating set then the same is true of some finite subset V 0 and clearly ( †) is also satisfied with V replaced by V 0 . Then V 0 ⊆ B 0 A for some finite subset B 0 of B. As ( †) holds, it follows from Part 1 that V 0 is a subset of an expansion Z 0 of B 0 . However, this means that V 0 is a subset of a basis of V n,r , a contradiction.
Corollary 3.17. Let Y i be a finite basis for V n,r , for i = 1, . . . , m. Then there is a unique minimal Proof. For m = 2, from Corollary 3.13 we have a common expansion Z of Y 1 and Y 2 such that where we assume inductively that Z is the unique minimal expansion of Y 1 , . . . , Y m−1 . From the previous paragraph there exists a unique minimal expansion W of Z and Y m such that W A = V . It follows that the result holds for Y 1 , . . . , Y m and hence by induction for all m.
Corollary 3.18. Let Y be a finite basis and let B be an A-basis of V n,r (X). If Y ⊆ B A then Y is an expansion of B: i.e. Y is an A-basis.
Proof. As Y ⊆ B A and Y is a basis, Y satisfied †, from Lemma 3.16.3. If u ∈ B A then u ∈ Y A λ , so for some Γ, ∆ ∈ A * , y ∈ Y , we have uΓ = y∆. As u ∈ B A and y ∈ Y ⊆ B A there exist b, b ′ ∈ B and Λ, Λ ′ ∈ A * such that bΛ = u and b ′ Λ ′ = y, so bΛΓ = b ′ Λ ′ ∆, and therefore b = b ′ and ΛΓ = L ′ ∆. Now, bΛΓ = bΛ ′ Γ so either u = bΛ is an initial segment of y = βΛ ′ , or vice-versa. Hence, from Lemma 3.16.2, Y is an expansion of B.
Proof. If no element of Y is a proper initial segment of an element of Z then, from Corollary 3.13, Y ⊆ Z A . Then Lemma 3.16 implies that Y is an expansion of Z. However, Y and Z are both d-fold expansions of B and thus Y = Z. This competes the proof.
Lemma 3.20. Let u ∈ V n,r and let d be a non-negative integer.
1. If v ∈ V n,r then u = v if and only if uΓ = vΓ, for all Γ ∈ A * of length d.
2. If S is a subalgebra of V n,r then u ∈ S if and only if uΓ ∈ S, for all Γ ∈ A * of length d.

Proof.
1. If u = v then uΓ = vΓ for all Γ ∈ A * of length d.
We shall show that given If d = 0 this holds trivially. We will use induction on d. Assume that d > 0 and that for all Suppose then that u, v ∈ V n,r and uΓ = vΓ for all Γ of length d. In this case we will show that for any ∆ ∈ A * of length d − 1 we have u∆ = v∆. In fact, if ∆ has length d−1 then ∆α i has length d, for i = 1, . . . , n. Therefore, u(∆α i ) = v(∆α i ) and we obtain u∆ = (u∆)α 1 · · · (u∆)α n λ = (v∆)α 1 · · · (v∆)α n λ = v∆. This applies to all ∆ of length d − 1, as required. From the inductive hypothesis u = v.
2. The proof is similar to that of part 1.

The Higman-Thompson groups G n,r
In this section we define the groups which form the object of study in this paper. Throughout the remainder of the paper, we assume that n ≥ 2, and that V n, When we discuss automorphisms of V n,r we assume that they are given by listing the images of a free generating set of V n,r . Assume ψ is an automorphism of V n,r defined by the map ψ : Y → Z, where Y and Z freely generate V n,r . If we expand y ∈ Y and form the free generating set Y ′ = Y \ {y} ∪ {yα 1 , . . . , yα n }, then yα i ψ = yψα i = zα i for i = 1, . . . , n. Thus, when we expand Y the automorphism ψ induces an expansion Z ′ of Z such that Y ′ ψ = Z ′ . Hence, if Y and Z are not expansions of x then by taking suitable expansions we may replace them by bases Y ′ and Z ′ contained in x A , and define the automorphism by a map from Y ′ to Z ′ . From Lemma 3.7 it follows that we may always describe an automorphism by a bijection between A-bases.
As bijections between bases are not particularly easy to read we represent automorphisms using pairs of rooted forests; as follows. An n-ary rooted tree is a tree with a single distinguished root vertex of degree n, such that all other vertices have degree n + 1 or 1. If a vertex v is at distance d ≥ 1 from the root then the n vertices incident to v and not on the path to the root are its children. Vertices of degree 1 are called leaves. An n-ary rooted tree is said to be A-labelled if the edges joining a vertex v to its n children are labelled with the elements α i ∈ A, so that two edges joining v to different children are labelled differently. An A-labelled, r-rooted, n-ary forest is a disjoint union of r rooted, A-labelled, n-ary trees.
Let T be a finite, A-labelled, r-rooted, n-ary forest, and let x = {x 1 , . . . , x r } be the set of roots of the r trees which make it up. We may now identify elements of the subtree T i with root x i , recursively, with elements of {x i } A ⊆ V n,r . We identify the root vertex x i with the corresponding element of V n,r (x). If a vertex v of T i , of degree n + 1, corresponds to an element x i Γ, for some Γ ∈ A * , the child joined to v by the edge labelled α j corresponds to x i Γα j , j = 1, . . . , n. Carrying out this correspondence for each subtree T i , we identify each node of T with a uniquely determined element of x A . Furthermore, by construction, the leaves of T correspond to an expansion of x. We use such trees to represent automorphisms as in the following example.
Example 4.1. Let n = 2, r = 1, x = {x} and let ψ be the element of G 2,1 corresponding to the bijective map between A-bases Y = {xα 2 1 , xα 1 α 2 , xα 2 } and Z = Y ψ = {xα 1 , xα 2 α 1 , xα 2 2 } given by The A-labelled binary trees corresponding to these bases are shown below. The labelling of edges is not shown, but edges from a vertex to its children are always ordered from left to right in the order α 1 , . . . , α n . Thus the leaves of the left hand tree correspond to Y and the leaves of the right hand tree to Z. The numbering below the leaves determines the mapping ψ; by taking leaf labelled j on the left to leaf labelled j on the right.
For an arbitrary automorphism, described as a bijection between A-bases, we generalise this example in the obvious way. Hig74]). The Higman-Thompson group G n,r is the group of Ω-algebra automorphisms of V n,r .
That is, any other expansion of X with this property is an expansion of Y .
Proof. For each i, Xψ −1 i is a generating set for V n,r , but may not be a subset of X A . Let Definition 4.4. Let {ψ 1 , . . . , ψ k } be a finite subset of G n,r and let X be an A-basis of V n,r . The expansion Y of X given by Lemma 4.3 is called the minimal expansion of X associated to {ψ 1 , . . . , ψ k }.

Semi-normal forms
Let ψ ∈ G n,r , let X be an A-basis of V n,r , and y ∈ V n,r . The ψ-orbit of y is the set {yψ n |n ∈ Z}. We consider how ψ-orbits intersect the A-subalgebra X A . To this end an X-component of the ψ-orbit of y is a maximal subsequence C of the sequence (yψ i ) i=∞ i=−∞ such that all elements of C are in X A . More precisely, C must satisfy 1. if yψ p and yψ q belong to C, where p < q then yψ k belongs to X A , for all k such that p ≤ k ≤ q; and 2. C is a maximal subset of the ψ-orbit of y for which 1 holds.
Note: X-components are what Higman, in [Hig74], refers to as "orbits in X A ". First we distinguish the five possible types of X-component of ψ by giving them names.
1. Complete infinite X-components. For any y in such an X-component, yψ i belongs to X A for all i ∈ Z, and the elements yψ i are all different.
2. Complete finite X-components. For any y in such an X-component, yψ i = y for some positive integer i, and y, yψ, . . . , yψ i−1 all belong to X A .
3. Right semi-infinite X-components. For some y in the X-component, yψ i belongs to X A for all i ≥ 0, but yψ −1 does not. The elements yψ i , i ≥ 0, are then, of course, necessarily all different.
4. Left semi-infinite X-components. For some y in the X-component, yψ −i belongs to X A for all i ≥ 0, but yψ does not. The elements yψ −i , i ≥ 0, are then, of course, necessarily all different.

5.
Incomplete finite X-components. For some y in the X-component and some non-negative integer i we have y, yψ, . . . , yψ i belonging to X A but yψ −1 and yψ i+1 do not.
Let ψ ∈ G n,r , let X be an A-basis of V n,r , let Y be the minimal expansion of X A associated to ψ and let Z = Y ψ. Then, as discussed above, Y and Z are both expansions of X. From Lemma 3.12, both X A \ Z A and X A \ Y A are finite. Furthermore, as |Y | = |Z|, both X and Y are d-fold expansions, for some d, Thus, if u ∈ X A \ Z A then u ∈ X A ψ, so uψ −1 ∈ X A and hence u is an initial element either of an incomplete finite X-component or of a right semi-infinite X-component i.e. in an X- Let u be an initial element of an incomplete finite X-component O. By the above, u ∈ X A \ Z A and by definition of an incomplete finite X-component, there is some non-negative integer k such that u, uψ, . . . , uψ k all belong to X A but uψ k+1 does not. Since uψ k is the terminal element of the incomplete finite X-component O, we have uψ k ∈ X A \ Y A . Therefore, the initial elements of incomplete finite X-components in X A \ Z A and terminal elements of incomplete finite X-components in X A \ Y A pair up.
Given that the initial and terminal elements of the incomplete finite X-components must be in one-to-one correspondence, all other elements of Hence there are as many right semi-infinite X-components as left semi-infinite X-components.
The above is summarised in a lemma.
Lemma 4.6 ([Hig74, Lemma 9.1]). Let ψ be an element of G n,r and let X be an A-basis of V n,r .
Then there are only finitely many X-components of ψ of type (3), (4) and (5) and there are as many of type (3) as of type (4). If Y is the minimal expansion of X A associated to ψ and Z = Y ψ then • u is an initial element of an orbit of type (3) or (5) if and and only if u ∈ X A \Z A and • u is a terminal element of an orbit of type (4) or (5) if and and only if u ∈ X A \Y A .
Definition 4.8 ([Hig74, Section 9]). An element ψ of G n,r is in semi-normal form with respect to the A-basis X if no element of X A is in an incomplete finite X-component of ψ.
Lemma 4.9 ([Hig74, Lemma 9.2]). Let ψ ∈ G n,r and let X be an A-basis of V n,r . There exists an expansion of X with respect to which ψ is in semi-normal form.
Proof. Let ψ ∈ G n,r . We prove the lemma by induction on the number of elements in X A which belong to an incomplete finite X-component. Note first that from Lemma 4.6 it follows that there are only finitely many elements of X A , which belong to incomplete finite X-components.
If there are no such elements then we are done. Suppose then that there exists an element u in X A which belongs to an incomplete finite X-component. Thus, there exist y ∈ X and Γ ∈ A * such that u = yΓ and some minimal m, k ∈ N 0 such that uψ −(m+1) , uψ k+1 ∈ X A . It follows that yψ −(m+1) , yψ k+1 ∈ X A , so that y is also in an incomplete finite X-component. Let X ′ = X\{y} and let X ′′ = X ′ ∪ {yα 1 , . . . , yα n }. Then X ′′ is an expansion of X and X A \ X ′′ A = {y} so the number of elements of X ′′ A in an incomplete finite X ′′ -component is one less than the number of elements of X A in an incomplete finite X-component. Hence, by induction, there exists an expansion of X with respect to which ψ is in semi-normal form.
Remark 4.10. Continuing the discussion above Lemma 4.6, observe that if u ∈ X A and u / ∈ Y A ∪ Z A then u is both the initial an terminal element of an X-component of ψ; so {u} constitutes an incomplete finite X-component. Therefore, in implementing the argument of Lemma 4.9, to find a semi-normal form for ψ, we may pass immediately to a minimal expansion containing all elements of Example 4.11. Let n = 2, r = 1, x = {x} and let ψ be the automorphism of Example 4.1.
The following, which follows directly from the definitions, summarises the possibilities for the intersection with X A of the orbit of an element under an automorphism in semi-normal form.
Corollary 4.14. Let ψ be an element of G n,r in semi-normal form with respect to the A-basis X, let v ∈ V n,r and let O v be the ψ-orbit of v. Then O v has one of the following six types.
Remark 4.15. As can be seen from Example 4.17 below, there are automorphisms for which orbits of the final type in this list exist. In fact we shall show in Section 4.2 that there exist automorphisms which have such orbits with respect to every semi-normal form. This means that [Hig74, Lemma 9.6] is false and that the algorithms described in [Hig74] for determining if two elements belong to a single orbit, and for conjugacy of automorphisms, [Hig74, Lemma 9.7 and Theorem 9.3], are incomplete.
Definition 4.16. Let ψ be an element of G n,r in semi-normal form with respect to the A-basis X, and let O be a ψ-orbit of the type given in Corollary 4.14 type 6. Then O is called a pond orbit with respect to X. The sub-sequence of O consisting of elements not in X A is called a pond.
Lemma 4.18 ([Hig74, Lemma 9.3]). Let ψ be an element of G n,r in semi-normal form with respect to the A-basis X. Suppose that x is an element of X. Then one of the following holds.
(A) There exists Γ ∈ A * such that xΓ is in a complete finite X-component. In this case x itself belongs to a complete finite X-component, which consists of elements of X, and we say x is of type (A).
(C) x is not of type (A) or (B) above and there exists some z ∈ X of type (B) and non-trivial ∆ ∈ A such that xψ i = z∆. In this case the X-component containing x is infinite; and we say x is of type (C).
Proof. (A) If x belongs to an infinite X-component of ψ (of types (1), (3) or (4) that is), then so does xΓ, a contradiction. As ψ is in semi-normal form with respect to X it follows that x is in a complete finite X-component. Let d be the smallest positive integer such that xψ d = x, let 1 ≤ i ≤ d − 1 and let z ∈ X and ∆ ∈ A * be such that xψ i = z∆. Then z is also in a complete finite X-component, so there exists y ∈ X and Γ ∈ A * such that zψ d−i = yΓ. Hence From Lemma 3.16, we have y = x and Γ = ∆ = ε, so xψ i = z ∈ X, as claimed.
(B) If x belongs to a finite X-component then, from (A), the X-component of xΓ consists of elements zΓ, where z ∈ X, contrary to the hypotheses of (B). Therefore x belongs to an infinite X-component of ψ. Without loss of generality we may assume that there is i > 0 such that xΓψ i = x∆. Suppose first that xψ k ∈ X A , for all k ≥ 0. Then xψ i = vΛ, for some v ∈ X and Λ ∈ A * , and thus x∆ = xΓψ i = vΛΓ; so v = x and ∆ = ΛΓ, and we obtain Note that if xψ k ∈ X A for all k, then x = xΛΛ ′ , which forces Λ = Λ ′ = ε, so Γ = ∆, a contradiction. Hence the final statement of (B) holds.
(C) In this case x must belong to an infinite X-component, as (A) does not hold. As X is finite there is z ∈ X such that zΓ and z∆ belong to the X-component of x, for distinct Γ and ∆ in A * ; and then z is of type (B), as required.
Definition 4.19. Let u ∈ V n,r and ψ ∈ G n,r . If uψ d = uΓ, for some d ∈ Z \ {0} and some Γ ∈ A * \ {1}, then u is a characteristic element for ψ. If u is a characteristic element for ψ then the characteristic of u is the pair (m, In this case Γ is called the characteristic multiplier and m is the characteristic power for u, with respect to ψ. From the definition, if ψ is in semi-normal form with respect to X then an element x ∈ X is of type (B) if and only if x is a characteristic element: in which case it follows from Lemma 4.24 below that the ψ-orbit of x is of type 4 or 5, in Corollary 4.14. On the other hand, if x ∈ X has type (C) then the ψ-orbit of x may be of types 3, 4, 5 or 6, in Corollary 4.14.
Example 4.20. In Example 4.13, the automorphism ψ is in semi-normal form with respect to an A-basis X. The elements xα 2 α 1 and xα 2 2 of X are of type (A). The element xα 2 1 ∈ X is of type (B) with characteristic (−1, α 1 ), while xα 1 α 2 ∈ X is of type (B) with characteristic (1, α 2 ); and both of these elements are extremal in the unique semi-infinite X-components contained in their ψ-orbits.
In Example 4.17 elements xα 2 α 1 and xα 2 2 of X are of type (C), are not characteristic and belong to complete infinite X-components. An example of an element of type (C) in a semi-infinite Xcomponent can be found at [D15]: follow the link to "Examples" and refer to the example named "semi inf c".
Lemma 4.21. If u ∈ V n,r is a characteristic element for ψ ∈ G n,r then 1. (m, Γ) is uniquely determined and 2. if v is in the same ψ-orbit as u then v is a characteristic element with the same characteristic as u.
Interchanging u and v we see also that whenever vψ k = v∆ then uψ k = u∆.
From Lemma 4.21, if a ψ-orbit has a characteristic element, then every X-component of this ψorbit contains a characteristic element, and all these elements have the same characteristic. Bearing this in mind we make the following definition. Theorem 4.23 ([Hig74, Theorem 9.4]). Let ψ ∈ G n,r be in semi-normal form with respect to X. Then ψ is of infinite order if and only if it has a characteristic element. Moreover, if ψ is of infinite order then this characteristic element may be taken to belong to X.
Proof. If u is a characteristic element for ψ with characteristic (m, Γ) then uψ m = uΓ, so uψ mq = uΓ q and, for sufficiently large q, uψ mq ∈ X A . Then uψ mq is a characteristic element and uψ mq = x∆, for some x ∈ X and ∆ ∈ A * . Now x∆Γ = uψ mq Γ = uψ m(q+1) = x∆ψ m , so from Lemma 4.18, x has type (B). Thus we may assume u ∈ X. Now, for j ∈ N. Since Γ is a characteristic multiplier, the elements uΓ j are all different for j ∈ N, so ψ has infinite order.
Conversely, if ψ has no characteristic element, then certainly there are none in X, so X has no elements of type (B) nor type (C). Thus all elements of X are of type (A), as ψ is in semi-normal form; whence ψ is a permutation of X and has finite order.
Lemma 4.24. Let ψ be in semi-normal form with respect to an A-basis X and let u ∈ V n,r . If u has characteristic (m, Γ) then the ψ-orbit of u has precisely one X-component, which is semi-infinite (right semi-infinite if m > 0 and left semi-infinite if m < 0) and consists of elements of the form xΛ, where x ∈ X, x is of type (B), and Λ ∈ A * . Furthermore, if xΛ belongs to the X-component of the ψ-orbit of u, where x ∈ X and Λ ∈ A * , then x has characteristic (m, Γ 1 Γ 0 ), where Γ = Γ 0 Γ 1 , Λ = (Γ 1 Γ 0 ) p Γ 1 = Γ 1 Γ p , p ≥ 0, and Γ 0 is non-trivial.
Proof. As uψ m = uΓ we have uψ mq = uΓ q , for all integers q, and choosing q sufficiently large uΓ q ∈ X A . Thus we may assume that u ∈ X A . Let u = xΛ, where x ∈ X and Λ ∈ A * and assume first that m > 0. As u has characteristic (m, Γ), both xΛ and xΛΓ belong to S, so from Lemma 4.18, x is of type (B). Suppose there is an integer K ≥ 0 such that uψ −k ∈ X A , for all k ≥ K. Let Λ = Λ 0 Γ t , where Λ 0 has no terminal segment equal to Γ. Then, for j such that m(j + 1) ≥ K and j ≥ t, uψ −m(j+1) ∈ X A , so for some z ∈ X and Ξ ∈ A * , uψ −m(j+1) = zΞ and, from Lemma 4.21, zΞ has characteristic (m, Γ). Hence which implies z = x and ΞΓ j−t+1 = Λ 0 , a contradiction. As ψ is in semi-normal form with respect to X and u is not in a complete X-component, the X-component C of u must be in a right semi-infinite. Moreover, as we have just shown the ψ-orbit of u contains no left semi-infinite X-component, so C is the unique X A -component of this ψ-orbit. Suppose x has characteristic (k, Ω). If the X-component of x is left semi-infinite then xΛψ −j ∈ X A , for all k ≥ 0, so C is not right semi-infinite. Hence x is in a right semi-infinite X-component and k > 0. If Λ = Ω j Λ 1 then xΛ 1 ψ kj = xΩ j Λ 1 = u and so C contains xΛ 1 ; and it suffices to prove the Lemma under the assumption that that Λ has no initial segment equal to Ω. Suppose that m = kp + r, where 0 ≤ r < k. Then xΛψ kp = xΩ p Λ and xΩ p Λψ r = xΛψ kp+r = xΛψ m = xΛΓ.
In the case when m < 0 the result follows from the above on replacing ψ by ψ −1 .
Lemma 4.25. Let θ ∈ G n,r and u ∈ V n,r such that uθ k = u∆, where ∆ = ε. Then u has characteristic (m, Γ) with respect to θ, where k = mq and ∆ = Γ q , for some positive integer q.
Proof. Let θ be in semi-normal form with respect to X. Suppose first that k > 0. Let (m, Γ) be the characteristic of u. As in the proof of Lemma 4.24, we may assume that u ∈ X A , the X-component of u is right semi-infinite and that there exist x ∈ X and Γ 1 ∈ A * such that u = xΓ 1 and x has characteristic power m. Then k ≥ m, say k = mq + s, where 0 ≤ s < m and q ≥ 1. Let xθ s = yΛ ′ , where y ∈ X and Λ ′ ∈ A * . Now xΓ 1 ∆ = u∆ = uθ k = uθ mq+s = uΓ q θ s = xθ s Γ 1 Γ q = yΛ ′ Γ 1 Γ q . Hence x = y and so s = 0 and m = kq. Moreover xΛ∆ = u∆ = uθ k = uθ mq = uΓ q = xΛΓ q , so Λ∆ = ΛΓ q , from which ∆ = Γ q , as required.
Corollary 4.26. Let ψ be in semi-normal form with respect to an A-basis X and let u ∈ V n,r . Then there exists an element Λ ∈ A * such that uΛ belongs to a complete X-component of ψ.
Proof. Multiplying by a sufficiently long element of A * we may, as usual, assume that u ∈ X A , so u belongs to either a complete or a semi-infinite X-component of ψ. There are finitely many semi-infinite X-components. If S is a characteristic semi-infinite X-component with characteristic (m, Γ) then, from Lemma 4.24, elements of S have the form xΛ where x ∈ X, Λ ∈ A * and, for all but finitely many elements of S, Λ is periodic of period m. Let F S be the finite subset of elements of A * such that Λ ∈ F S only if xΛ ∈ S and Λ is not periodic of period m. Let F 0 be the union of the F S over all characteristic semi-infinite X-components. If S is non-characteristic then, from Lemma 4.18, S contains an element z∆, where z ∈ X of type (B), with characteristic (m ′ , Γ ′ ), say. It follows, from Lemma 4.24 again, that all but finitely many elements of S have the form xΛ∆ where x ∈ X, Λ ∈ A * and Λ is periodic of period m ′ . This time, let F S be the finite subset of elements of A * such that Λ∆ ∈ F S only if xΛ∆ ∈ S and Λ is not periodic of period m ′ . Let F 1 be the union of the F S over all non-characteristic semi-infinite X-components. Let M be the maximum of lengths of elements of F 0 ∪ F 1 and assume u = xΓ, where x ∈ X, Γ ∈ A * . Choose element Ξ of A * such that ΓΞ has length greater than M , is not periodic and does not factor as Λ∆, where Λ is periodic and ∆ ∈ F 1 . Then uΞ = xΓΞ cannot belong to a semi-infinite X-component, so must belong to a complete X-component.

Quasi-normal forms
Quasi-normal forms, introduced in [Hig74, Section 9], are particular semi-normal forms which give representations of automorphisms minimising the number of elements in pond orbits. In [Hig74] it is claimed that if an automorphism is given with respect to a quasi-normal form, then it has no pond orbits. In this section we shall see that this is not the case.  Hig74, Section 9]). An element ψ of G n,r is in quasi-normal form with respect to the A-basis X if it is in semi-normal form with respect to X, but not with respect to any proper contraction of X.
It follows from Lemma 4.3 that for ψ ∈ G n,r there exists an A-basis X with respect to which ψ is in quasi-normal form. For instance, the automorphisms ψ in Examples 4.11, 4.13 and 4.17 are in quasi-normal form with respect to the bases X in those examples. Additionally, the automorphism ψ of Example 4.12 is in quasi-normal form with respect to the basis X 3 .
Lemma 4.28. Given an element ψ ∈ G n,r there exists a unique A-basis X ψ with respect to which ψ is in quasi-normal form and which can be effectively constructed.
Proof. Assume ψ is given by listing the images of elements of X, where X is an A-basis of V n,r . We modify X to find an A-basis X ′ with respect to which ψ is in semi-normal form. For each y ∈ X we list elements of the ψ-orbit of y: . . . , yψ −3 , yψ −2 , yψ −1 , y, yψ, yψ 2 , yψ 3 , . . . .
We begin with y and go forward in the sequence yψ i , for i > 1, until we reach i = m ≥ 0 such that, (1F) either yψ m ∈ X A with yψ m+1 ∈ X A or, (2F) for some l with 0 ≤ l < m and for someŷ ∈ X and Γ, ∆ ∈ A * , yϕ l =ŷΓ and yϕ m =ŷ∆.
Given y ∈ X, the forward part of the process above produces a sequence of elements of X A , until it halts. As X is finite, if it does not halt at step (1F) then it halts at step (2F); so always halts. Similarly, the backward part of the process always halts.
If some y satisfies (1F) and (1B), then ψ is not in semi-normal form with respect to X. In this case we take a simple expansion X ′ of X at the element y and start again, replacing X with X ′ . That is, we implement the process described in the proof of Lemma 4.9, to find a basis with respect to which ψ is in semi-normal form. It follows from that proof that, eventually we shall find X such that no y ∈ X satisfies both (1F) and (1B), and then ψ is in semi-normal form with respect to X, by Lemma 4.18. We can now assume ψ is in semi-normal form with respect to X. We can thus test all the contractions of the A-basis X to find an expansion of x with respect to which ψ is in a quasi-normal form.
For uniqueness, we will argue by contradiction. Let ψ be in quasi-normal form with respect to X 1 and X 2 , with X 1 = X 2 . Since X 1 = X 2 and X 1 , X 2 are expansions of x, (without loss of generality) there exists a contraction X ′ 1 of X 1 which contains an element y of X 2 \ X 1 . Then X ′ 1 A = X 1 A ∪ {y} A and, as ψ is in semi-normal form with respect to X 2 , it is also in seminormal form with respect to X ′ 1 , contrary to the definition of quasi-normal form. Remark 4.29. Let ψ ∈ G n,r be in quasi-normal form with respect to X. The proof of this lemma illustrates that if ψ is in semi-normal form with respect to X ′ , then X ′ is an expansion of X. The converse is false: it is not true in general that ψ is in semi-normal form with respect to all expansions of X.
Lemma 4.30. Let ψ ∈ G n,r be in semi-normal form with respect to an A-basis X and let u, v ∈ X A . Then we can effectively decide whether or not u, v are in the same X-component, and if so, find the integers m for which uψ m = v.
Proof. As u ∈ X A , we have u = yΛ, where y ∈ X and Λ ∈ A * . We now run the process of Lemma 4.28 on y. If the process halts with yψ m = y, for some m then we may list the elements uψ i = yψ i Λ, i = 0, . . . , m − 1, of the (complete finite) ψ-orbit of u. In this case v is in the same ψ-orbit as u if and only if it appears in the list, so we are done. Otherwise the process halts at (1F) and (2B), at (2F) and (1B) or at (2F) and (2B). In all cases we obtain y ∈ X such that, for some k = l and Λ 1 = Λ 2 ∈ A * , we have yϕ k = yΛ 1 and yϕ l = yΛ 2 . It follows from Lemma 4.18 that y is of type (B). As uϕ k = yΛϕ k = yΛ 1 Λ we may replace u = yΛ with u = yΛ 1 Λ. Therefore we now assume that u = yΛ, where y is of type (B). Now, when we run the process of Lemma 4.28 on y it halts at (2F) and (1B) or at (1F) and (2B). Suppose first the forward part halts at (2F). Then y is in a right semi-infinite X-component and there is a minimal positive integer m such that yψ m = yΓ, with Γ = 1. That is y has characteristic (m, Γ), with m > 0.
If Λ = Γ i Λ 0 , where Λ 0 has no initial segment Γ, and we set u 0 = yΛ 0 then, so u 0 is in the same ψ-orbit as u. Hence we may replace u = yΛ by u 0 = yΛ 0 . Once we have done this we may suppose Λ has no initial segment equal to the characteristic multiplier Γ of y.
Next we run the process of Lemma 4.28 on u instead of y. As y is in a right semi-infinite X-component the forward part of the process halts at (2F). We obtain a list of elements of the X-component of u of the form where y j , z j ∈ X, Γ ′ j , Φ j ∈ A * , z j Φ j = uψ −j , for 1 ≤ j ≤ r and for some r ≥ 0, and yψ s = y s Γ ′ s , for 0 < s < m. (The y i 's must be distinct otherwise u would have characteristic power less than m.) If the backward part of the process halts at (1B) then z r Φ r ψ −1 = uψ −r−1 / ∈ X A . In this case, the entire X-component of u consists of the elements on this list together with elements where we set y 0 = y, Γ ′ 0 = Γ. As v ∈ X A we also have z ∈ X and ∆ in A * such that v = z∆. If z is in a finite X-component then v cannot belong to the same X-component as u, so we assume z is in an infinite X-component. As in the case of u, we may adjust v so that z is of type (B). As before we find a characteristic multiplier Φ for z and, replacing ∆ with a shorter element if necessary, we may assume that ∆ has no initial segment equal to Φ. If for some q ≥ 0 and i with 0 ≤ i < m. In this case, z = y i and by Lemma 4.24 and our assumption on v we have q = 0, so v = y i Γ ′ i Λ, which appears on list (6). Assume then that v = uψ d , where d < 0. As the backward part of the enumeration of the ψ-orbit of u halts at (1B), the X-component of u has initial element z r Φ r , and v must appear on list (6).
On the other hand, if the backward part of the process stops at (2B) then u is in a complete infinite X-component and, for some s with 0 ≤ s ≤ r, we have z r = z s (and r is minimal with this property). It follows that z s is of type (B) and in a left semi-infinite X-component. Again, we may assume that v = z∆, where ∆ ∈ A * , z ∈ X is of type (B) and has characteristic multiplier Φ, such that ∆ has no initial segment equal to Φ. As before if v = uψ d , where d ≥ 0, then v appears on list (6). Assume then that v = uψ d , where d < 0. Repeating the argument above, using the left semi-infinite X-component of z s instead of the right semi-infinite X-component of y, it follows again that v appears on list (6).
Therefore, in the case where y is in a right semi-infinite X-component we have v in the Xcomponent of u if and only if v lies on the list (6); and we may compute m such that uψ m = v, if this is the case. Finally, if the enumeration of the X-component of y halts at steps (1F) and (2B) then the process is essentially the same, except that we deal with a left, rather than a right, semi-infinite X-component of y.
This procedure allows us to decide if two given words belong to the same X-component so, if there are no pond orbits, we may decide if two words belong to the same ψ-orbit. On the other hand, as the enumeration of components always stops once we fall outside X A , we cannot detect when a pair of elements lie in the same ψ-orbit but on opposite sides of a pond. We demonstrate below that in some cases there are no semi-normal forms which are free of ponds; and therefore we require a strategy to deal with ponds.
Lemma 4.31. Let ψ ∈ G n,r be in semi-normal form with respect to X, and suppose that some ψorbit O contains a pond with respect to X. If ψ is in semi-normal form with respect to an expansion X ′ of X, then O is also a pond-orbit with respect to X ′ .
where l, r ∈ X A are endpoints of semi-infinite X-components and the p i / ∈ X A form a pond of length m. To begin with we claim that, for sufficiently large s ≥ 0, we have rψ s ∈ X ′ A . Indeed, because r belongs to a semi-infinite X-component, Lemma 4.18 implies that there is some s ′ ≥ 0 for which rψ s ′ = r ′ ∆, where ∆ ∈ A * and r ′ ∈ X has characteristic (m, Γ). Therefore, for all q ≥ 0, rψ s ′ +mq = rψ s ′ ψ mq = r ′ ∆ψ mq = r ′ ψ mq ∆ = r ′ Γ q ∆.
By taking q sufficiently large, we can ensure that rψ s ′ +mq ∈ X ′ A , since the difference X A \X ′ A is finite. That is, we can find s ≥ 0 such that rψ s ∈ X ′ A . Similarly, there is some t ≥ 0 for which lψ −t ∈ X ′ A .
Since X ′ A ⊂ X A , it follows that each p i / ∈ X ′ A . Appealing to Corollary 4.14, the only possibility is that O is a pond-orbit with respect to X ′ . This proof demonstrates that the pond width with respect to X ′ is at least the previous width k with respect to X. Additionally, if ψ was in quasi-normal form with respect to X, this lemma shows that every semi-normal form X ′ for ψ contains the pond given above.
Lemma 4.32. Given an element ψ ∈ G n,r in semi-normal form with respect to an A-basis X we may effectively construct a list of all initial and terminal elements of semi-infinite X-components of ψ and the set P (ψ) of the pairs (l, r) such that l and r are initial and terminal elements of X-components of a pond orbit, and for each such pair the integer k such that r = lψ k .
Proof. Let Y be the minimal basis for ψ and Z = Y ψ. Then the set of initial elements of right semi-infinite X-components is R = X A \Z A , which we may enumerate effectively. An analogous statement applies to the set L = X A \Z A of terminal elements of left semi-infinite Xcomponents.
To enumerate the required set P (ψ), for each (l, r) ∈ L × R note that (l, r) is in P (ψ) only if r = lψ k , for some k; in which case rΓ = lΓψ k , for all Γ ∈ A * . With this in mind, first find Γ ∈ A * such that lΓ is in a complete orbit. For a given Γ this may be done using the process of Lemma 4.28: we iterate this until we find Γ such that, on input lΓ, the process halts at (2F) and (2B). Now check, using Lemma 4.30, whether rΓ and lΓ are in the same X-component. If not, then (l, r) cannot be in P (ψ).
Assume then that rΓ = lΓψ m , for some m. If (l, r) ∈ P (ψ), with r = lψ k , then lψ k Γ = rΓ = lΓψ m , so lΓψ k−m = lΓ, and as lΓ belongs to a complete infinite X-component, this implies k = m. Hence, once we have found m such that rΓ = lΓψ m , we conclude that (l, r) is in P (ψ) if and only if r = lψ m .
In practice, in enumerating the sets L and R in the proof above, we need consider only noncharacteristic elements, as Lemma 4.24 implies that no characteristic element belongs to a pond orbit.
Example 4.33. Let ψ and X be the automorphism and basis described in Example 4.17. As noted above ψ is in quasi-normal form with respect to X and we have seen that ψ-orbit of xα 2 1 α 2 is a pond-orbit. We claim that this ψ-orbit is the only pond orbit with respect to X.
The endpoints of semi-infinite X-components are precisely The four endpoints xα 2 1 , xα 3 1 , xα 1 α 2 and xα 1 α 2 α 1 have characteristics (−1, α 2 1 ), (−1, α 2 1 ), (1, α 1 α 2 ) and (1, α 2 α 1 ) respectively. The two remaining endpoints are separated by the pond we identified in Example 4.17. Hence P (ψ) = {(xα 2 1 a 2 , xα 1 α 2 2 )}. for all Γ ∈ A * of length s (using Lemma 3.20). Now, suppose that we have an algorithm A to decide whether v ′ = u ′ ψ m , for some m, for elements u ′ , v ′ of X A (and to return m, if so). Then if u, v are arbitrary elements of V n,r we may choose s such that uΓ and vΓ belong to X A , for all Γ ∈ A * of length s, and input all these elements to the algorithm A in turn. In the light of the previous remark, this allows us to determine whether or not u and v belong to the same ψ-orbit (and to return appropriate m, if so). Hence we may assume u, v ∈ X A . By Corollary 4.14, u and v belong to the same ψ-orbit if and only if either they belong to the same X A -component of a ψ-orbit, or they belong to different X-components of a single pond orbit. We may use Lemma 4.30 to decide whether or not u and v both belong to the same Xcomponent. If so we are finished. If not, and both belong to semi-infinite orbits, then for each pair (l, r) in P (ψ) we check whether u and l or u and r belong to the same X-component. If so, say u and l belong to an X-component, then we run the same check on v and r. This allows us to check whether or not u and v belong to the same pond orbit; and as ponds have finite width we may compute m such that uψ m = v, if this is the case.
We now need to check if u and v ′ are separated by a pond. In Example 4.33 we showed that ψ has only one pond-orbit, and referring to the enumeration given in Example 4.17 we see that neither u nor v ′ belong to this orbit. Hence u and v ′ do not share a ψ-orbit. Now let us test if u and w = xα 4 1 α 2 α 2 1 α 2 = q 1 α 2 1 α 2 α 2 1 α 2 share a ψ-orbit. We remove the characteristic multiplier α 2 1 of q 1 from w, obtaining w ′ = q 1 α 2 α 2 1 α 2 where w ′ ψ −1 = w. From the list (6') we notice that uψ −2 = w ′ , so uψ −3 = w.

The Conjugacy problem
For a group with presentation X | R , the conjugacy problem is to determine, given words g, h ∈ F(X) whether or not g is conjugate to h in G; denoted g ∼ h. The strong form, which we consider here, requires in addition that, if g is conjugate to h, then an element c ∈ F(X) is found, such that c −1 gc = G h. We say the conjugacy problem is decidable if there is an algorithm which, given g and h outputs "yes" if they're conjugate and "no" otherwise. The stronger form entails the obvious rejoinder. The word problem is the special case of the conjugacy problem where h = 1.
As pointed out at the beginning of Section 4, an element ψ of G n,r may be uniquely represented by the triple (Y, Z, ψ 0 ), where Y is the minimal expansion of ψ, Z = Y ψ and ψ 0 is a bijection between Y and Z, namely ψ 0 = ψ| Y . This triple is called a symbol for ψ. In [Hig74, Section 4] a finite presentation of G n,r is given, with generators the symbols (Y, Z, ψ 0 ) such that Y is a d-fold expansion of x, for d ≤ 3. As we may effectively enumerate symbols and effectively construct the symbol for ψ 1 ψ 2 , from the symbols for ψ 1 and ψ 2 , words in Higman's generators effectively determine symbols and vice-versa. Therefore when we consider algorithmic problems in G n,r we may work with symbols for automorphisms, and leave the presentation in the background. That is, we always assume that automorphisms are given as maps between bases of V n,r (from which a symbol may be computed). As minimal expansions are unique it follows immediately that the word problem is solvable in G n,r . In this section we give an algorithm for the conjugacy problem in G n,r , based on (a complete version of) Higman's solution.

Higman's ψ-invariant subalgebras V P and V RI
Let ψ be an element of G n,r . Higman defined two subalgebras of V n,r , determined by ψ; namely • the subalgebra V P,ψ generated by the set of elements of V n,r which belong to finite ψ-orbits.
• the subalgebra V RI,ψ generated by the set of characteristic elements for ψ.
Where there is no ambiguity, we will write V P for V P,ψ and V RI for V RI,ψ .
If u ∈ V n,r then the ψ-orbit of u is identical to the ψ-orbit of uψ; so u is in a finite ψ-orbit if and only if uψ is in a finite ψ-orbit. From Lemma 4.21, an element u is a characteristic element for ψ if and only if uψ is a characteristic element for ψ. Therefore V P,ψ and V RI,ψ are ψ-invariant subalgebras of V n,r . (A subalgebra S is ψ-invariant if Sψ = S.) Hence ψ P = ψ| V P,ψ is an automorphism of V P,ψ and ψ RI = ψ| V RI,ψ is an automorphism of V RI,ψ .
If ψ and ϕ are conjugate elements of G n,r and ρ −1 ψρ = ϕ, for some ρ ∈ G n,r , then, for all Γ ∈ A * we have uϕ m = uΓ if and only if uρ −1 ψ m ρ = uΓ if and only if (uρ −1 )ψ m = (uρ −1 )Γ. Thus u is in a finite ϕ-orbit if and only if uρ −1 is in a finite ψ-orbit (taking Γ = ε) and u is a characteristic element for ϕ if and only if uρ −1 is a characteristic element for ψ (Γ = ε). It follows that the restriction ρ| V P,ψ of ρ to V P,ψ maps V P,ψ isomorphically to V P,ϕ and similarly, ρ| V RI,ψ is an isomorphism from V RI,ψ to V RI,ϕ . Now suppose that ψ is in semi-normal form with respect to an A-basis X. Partition X into X P = X P,ψ = {y ∈ X| y is of type (A)}, and X RI = X RI,ψ = {y ∈ X| y is of type (B) or (C)}.
Theorem 5.1 ([Hig74, Theorem 9.5]). Let ψ be an element of G n,r , in semi-normal form with respect to A-basis X. Then, with the notation above, the following hold.
1. V n,r = V P * V RI , the free product of the ψ-invariant subalgebras V P and V RI .
Proof. Let W P = X P A λ and W RI = X RI A λ . As X is the disjoint union of X P and X RI , we have V n,r = W P * W RI , using Lemma 3.11. We shall show that V P = W P and V RI = W RI . By definition, W P ⊆ V P . If x ∈ X RI is of type (B) then x ∈ V RI , by definition. If x ∈ X RI is of type (C) then there exists z ∈ X RI , of type (B), and ∆ ∈ A * , such that xψ i = z∆. As z ∈ V RI , so is z∆, and as To see that V P ⊆ W P , let u ∈ V n,r have a finite ψ-orbit. Choose d ∈ N such that, uΓ ∈ X A , for all Γ ∈ A * of length d. Let Γ be a length d element of A * , so uΓ = x∆, for some x ∈ X, ∆ ∈ A * . As u is in a finite ψ-orbit so is uΓ, so x ∈ X P and thence uΓ = x∆ ∈ W P . As this holds for all Γ in A * of length d, we have u ∈ W P , by Lemma 3.11. Hence V P ⊆ W P .
To see that V RI ⊆ W RI , we first show that W RI is ψ-invariant. Let Y be the minimal expansion of X associated to ψ and let x ∈ X RI . Then choose d such that xΓ ∈ Y A , for all Γ ∈ A * of length d. Given Γ ∈ A * of length d, let y ∈ Y and ∆ ∈ A * such that xΓ = y∆. Then xΓψ = yψ∆ ∈ X A , so xΓψ = zΛ, for some z ∈ X and Λ ∈ A * . Moreover, z must have type (B) or (C), as x does, so xΓψ ∈ X RI A ⊆ W RI . This holds for all Γ of length d, so again xψ ∈ W RI . It follows that W RI ψ ⊆ W RI . Repeating the same argument, using Z = Y ψ instead of Y and ψ −1 instead of ψ gives W RI ψ −1 ⊆ W RI ; so W RI is ψ-invariant as claimed. Now let u ∈ V n,r be a characteristic element for ψ. Then, from Lemma 4.24, we have uψ i = xΛ, for some integer i, x ∈ X RI and Λ ∈ A * . Thus u = xΛψ −i ∈ W RI , as W RI is ψ-invariant; and we have V RI ⊆ W RI . This proves 1 and 2 of the Theorem, and 3 then follows from the discussion preceding the statement of the Theorem.
Note that in the case that ρ −1 ψρ = ϕ in the theorem above we have ρ = ρ P * ρ RI an isomorphism from V P,ψ * V RI,ψ to V P,ϕ * V RI,ϕ , both of which are isomorphic to V n,r .
Theorem 5.1 allows us to decompose the conjugacy problem for ψ and ϕ into conjugacy problems for ψ P and ϕ P and for ψ RI and ϕ RI . Indeed, if |X P | = d p and |X RI | = d RI then V P ∼ = V n,dP and V RI ∼ = V n,dRI and we regard ψ P and ψ RI as automorphisms of V n,dP and V n,dRI , respectively. It turns out that ψ P and ψ RI are each of particularly simple types; so if we can solve the conjugacy problem for these simple types of automorphism, then we can solve it in general. In the remainder of this subsection we describe in detail how this decomposition works.
First consider a single automorphism ψ ∈ G n,r , where ψ is in semi-normal form with respect to an A-basis X. Here we assume that V n,r is the free V n algebra on a set x of size r, and that X is an expansion of x. Let X P and X RI be defined as above, let Y be the minimal expansion of X associated to ψ and let Z = Y ψ. As Y is an expansion of X, for all x ∈ X the set Y x = Y ∩ {x} A is an expansion of {x}, by Lemma 3.16. Therefore Y P = Y ∩ X P A is an expansion of X P , and Y RI = Y ∩ X RI A is an expansion of X RI . Similarly, Z P = Z ∩ X P A and Z RI = Z ∩ X RI A are expansions of X P and X RI , respectively. In fact, as ψ permutes the elements of X of type (A), ψ P permutes the elements of X P , so X P = Y P = Z P . Therefore ψ P is an automorphism of V P = X P A λ , which permutes the elements of X P . For all y ∈ Y RI we have yψ = z ∈ Z, and z is in where V RI is freely generated by X RI , and Y RI is the minimal expansion of X RI associated to ψ RI (as Y is the minimal expansion of X associated to ψ). Furthermore Y RI ψ RI = Z RI and if u is an element of X RI A such that uψ ∈ X A then uψ ∈ X A ∩ V RI = X RI A ; so no element of X RI A is in an incomplete finite X RI -component of ψ RI . Now let |X P | = a and |X RI | = b and let X P = {x 1 , . . . , x a } and X RI = {x a+1 , . . . , x a+b }, where x i ∈ x A . Then, regarding the x i as new symbols, we may view V P as V n,a , the free V n algebra on {x 1 , . . . , x a }, and V RI as V n,b , the free V n algebra on {x a+1 , . . . , x a+b }. We may thus regard ψ P and ψ RI as elements of G n,a and G n,b , respectively. In this case, ψ P is in quasi-normal form with respect to the A-basis X P = {x 1 , . . . , x a } and ψ RI is in quasi-normal form with respect to X RI = {x a+1 , . . . , x a+b }. Moreover, the minimal expansion of X RI associated to ψ RI is Y RI . (Here we write all elements of Y and Z in terms of the x i , rather than as expansions of elements of x.) Example 5.3. Let n = 2, r = 1 and V 2,1 be free on {x}. Let Y = {xα 4 1 , xα 3 1 α 2 , xα 2 1 α 2 , xα 1 α 2 α 1 , xα 1 α 2 2 , xα 2 α 1 , xα 2 2 α 1 , xα 3 2 } and Z = {xα 3 1 , xα 2 1 α 2 α 1 , xα 2 1 α 2 2 , xα 1 α 2 α 1 , xα 1 α 2 2 , xα 2 α 2 1 , xα 2 α 1 α 2 , xα 2 2 } and let ψ be the element of G n,r determined by the bijection from Y to Z given by xα 4 1 ψ = xα 3 1 , xα 3 1 α 2 ψ = xα 2 1 α 2 α 1 , xα 2 1 α 2 ψ = xα 2 1 α 2 2 , xα 1 α 2 α 1 ψ = xα 1 α 2 2 , xα 1 α 2 2 ψ = xα 1 α 2 α 1 , xα 2 α 1 ψ = xα 2 α 2 1 , xα 2 2 α 1 ψ = xα 2 α 1 α 2 and xα 3 2 ψ = xα 2 2 .
Definition 5.4. Let ψ be an element of G n,r . Then ψ is called periodic if V RI = ∅ and ψ is called regular infinite if V P = ∅.
Lemma 5.5. Let ψ be an element of G n,r in semi-normal form with respect to an A-basis X.
1. ψ is periodic if and only if ψ permutes the elements of X.

ψ is regular infinite if and only if no element of X is of type (A).
Proof.
1. If ψ permutes the elements of X then X contains no element of type (B) or (C); so X = X P and V n,r = V P , by Theorem 5.1. As V n,r is the free product of V P and V RI it follows that V RI = ∅ so ψ is periodic.
If ψ is periodic then X RI ⊆ V RI = ∅, so X = X P . Thus X consists of elements of type (A), which are permuted by ψ, by Lemma 4.18.
2. If ψ is regular infinite then V P = ∅, so X p = ∅; i.e. no element of X is of type (A). If X contains no element of type (A) then X P = ∅ and therefore V P = ∅, from Theorem 5.1, and ψ is regular infinite. It follows that, in the notation established above Example 5.3, the automorphism ψ P ∈ G n,a is periodic and ψ RI ∈ G n,b is regular infinite. Thus, the decomposition of Theorem 5.1 may be viewed as factoring ψ into a product of a periodic and a regular infinite automorphism. It remains to see how to regard a pair of automorphisms in this way, simultaneously in the same algebra.
To this end suppose that ψ i ∈ G n,ai is in semi-normal form with respect to an A-basis X i , where |X i | = a i , for i = 1, 2. If there exists an isomorphism ρ : V n,a1 → V n,a2 with the property that ρ −1 ψ 1 ρ = ψ 2 then, from Corollary 3.14, a 1 ≡ a 2 mod n − 1. Also, if a 1 ≡ a 2 mod n − 1 then V n,ai is isomorphic to V n,s where 1 ≤ s ≤ n− 1 and s ≡ a i . If this is the case then we may take an A-basis x s of s elements of V n,s and choose expansions X ′ 1 and X ′ 2 of x s of a 1 and a 2 elements respectively. Now let f i be the map taking X i to X ′ i . Then there exists an isomorphism ρ : V n,a1 → V n,a2 such that ρ −1 ψ 1 ρ = ψ 2 if and only if a 1 ≡ a 2 mod n − 1 and, setting Figure 5.1.1.) Combining this with Theorem 5.1.1 gives a decomposition of the conjugacy problem into the conjugacy problem for periodic and for regular infinite elements, separately. Let ψ and ϕ be elements of G n,r , write V n,a1 = V RI,ψ , ψ 1 = ψ RI , V n,a2 = V RI,ϕ and ψ 2 = ϕ RI . Using the procedure above, if ρ RI exists (in the notation of Theorem 5.1) then we may regard ψ i , i = 1, 2, as a regular infinite element of G n,s , namely ψ i , for appropriate s. Similarly, we may regard ψ P and ϕ P as periodic automorphisms of a single algebra.
We can now outline the algorithm for the conjugacy problem.

The conjugacy algorithm
Algorithm 5.6. Let ψ and ϕ be an elements of G n,r .
Step 1: Find A-bases X ψ and X ϕ such that ψ and ϕ are in quasi-normal form with respect to X ψ and X ϕ , respectively; as in Lemma 4.28. The sets X P,ψ , X RI,ψ , X P,ϕ and X RI,ϕ are obtained as part of this process.
Step 2: Find the minimal expansion Y ψ of X ψ associated to ψ and the minimal expansion Y ϕ of X ϕ associated to ϕ. (See Lemma 4.3.) Construct Y RI,ψ and Y RI,ϕ ; the sets elements of Y ψ and Y ϕ which are not in finite orbits (as in the the discussion following Theorem 5.1). Construct Z RI,ψ = Y RI,ψ ψ and Z RI,ϕ = Y RI,ϕ ϕ.
Step 3: For T = P and for T = RI carry out the following. Find the integer s T such that 1 ≤ s T ≤ n−1 and s T ≡ |X T,ψ |. Let x T be a set of s T elements, let V n,sT be free on x T and find expansions W T,ψ and W T ϕ of x T of sizes |X T,ψ | and |X T,ϕ |, respectively. Construct a map f T,ψ mapping X T,ψ bijectively to W T,ψ and f T,ϕ mapping X T,ϕ bijectively to W T,ϕ . Write ψ T and ϕ T as elements of G n,sT , using these maps.
Step 4: Input ψ P and ϕ P into Algorithm 5.13 for conjugacy of periodic elements of G n,r , below. If ψ P and ϕ P are not conjugate, return "No" and stop. Otherwise return a conjugating element ρ P .
Step 5: Input ψ RI and ϕ RI into Algorithm 5.27 for conjugacy of regular infinite elements of G n,sRI ; in Section 5.4 below. If ψ RI and ϕ RI are not conjugate, return "No" and stop. Otherwise return a conjugating element ρ RI .
Given this algorithm we have the following theorem.

Conjugacy of periodic elements
Let ψ ∈ G n,r be a periodic element. For u ∈ V n,r the size of the ψ-orbit of u is the least positive integer d such that uψ d = u.
Definition 5.8. Let ψ be a periodic element of G n,r in semi-normal form with respect to the A-basis X. The cycle type of ψ is the set For d ∈ N, define the ψ-multiplicity of d to be m ψ (d, X) = D/d, where D is the number of elements of X which belong to a ψ-orbit of size d.
Note that, as ψ is periodic and in semi-normal form with respect to X, all X-components of ψ are (ordered) ψ-orbits and all ψ-orbits of elements of X A are X-components (once ordered appropriately). Also, d ∈ T ψ (X) if and only if m ψ (d, X) = 0; the size of the set X is |X| = d∈T ψ dm ψ (d, X); if d ∈ T ψ then X contains m ψ (d, X) disjoint ψ-orbits of size d; and ψ is a torsion element of order equal to the least common multiple of elements of T ψ .
Lemma 5.10. Let ψ be a periodic element of G n,r in semi-normal form with respect to the Abasis X and the A-basis Z, where Z is a q-fold expansion of X. Then T ψ (X) = T ψ (Z) and Proof. Let z ∈ Z, say z = xΓ, where x ∈ X and Γ ∈ A * , and let x have ψ-orbit of size d. As ψ is periodic it permutes the elements of X, so the ψ-orbit of x is {x 0 , . . . , and thus the ψ-orbit of z has size d. It follows that T ψ (X) = T ψ (Z). As Z is an expansion of X, the set {x} A ∩ Z is an expansion of {x}, for all x ∈ X. Let E = {Γ ∈ A * : xΓ ∈ Z}, so {x} A ∩ Z = xE, and assume the ψ-orbit of x is O x = {x 0 , . . . , x d−1 }, as before. For Γ ∈ E we have shown that the ψ-orbit of z = xΓ is {x i Γ | 0 ≤ i ≤ d − 1} and as ψ is in semi-normal form with respect to Z it follows that x i Γ ∈ Z, for all i. This holds for all Γ ∈ E, The set O is precisely the set of elements of X with ψ-orbits of size d and there are dm ψ (d, X) such elements, that is |O| = dm ψ (d, X). Moreover, from the above, z ∈ Z has ψ-orbit size d if and only if z ∈ O j A ∩ Z, for some j, if and only if z ∈ O A ∩ Z. A dp-fold expansion of a set of size M has M + dp(n − 1) elements, so Z contains |O| + pd(n − 1) = dm ψ (d, X) + pd(n − 1) elements with ψ-orbits of size d. Therefore m ψ (d, Z) = m ψ (d, X)+p(n−1) ≡ m ψ (d, X) mod n−1, as required.
Note that it follows from this lemma that if ψ is in semi-normal form with respect to both X and X ′ then T ψ (X) = T ψ (X ′ ), since we may take a common expansion of both X and X ′ and then expand this to an A-basis Z with respect to which ψ is in semi-normal form. Hence, from now on, we refer to T ψ as the cycle type of ψ, without reference to the A-basis X.
Proposition 5.11. Let ψ and ϕ be periodic elements of G n,r in semi-normal form with respect to the A-bases X ψ and X ϕ , respectively. Then ψ is conjugate to ϕ if and only if Proof. Assume that ψ and ϕ are conjugate and let ρ ∈ G n,r be such that ρ −1 ψρ = ϕ. Let ρ be in semi-normal form with respect to X ρ , let Y be the minimal expansion of X ρ associated to ρ and let Z = Y ρ. Let W be a common expansion of X ψ and Y and let ψ be in semi-normal form with respect to an expansion X ′ ψ of W . (Such an expansion of W exists, by Lemma 4.9.) As ψ is periodic and in semi-normal form it permutes the elements of X ′ ψ , so for all x ∈ X ′ ψ we have x ′ ∈ X ′ ψ such that xρϕ = xψρ = x ′ ρ ∈ Xρ. Therefore ϕ permutes the elements of X ′ ψ ρ. Hence ϕ is in semi-normal form with respect to X ′ ϕ = X ′ ψ ρ. As X ′ ψ is an expansion of Y and Z = Y ρ it follows that X ′ ϕ is an expansion of Z. Now if x ∈ X ′ ψ and i ∈ Z then xρϕ i = xψ i ρ, so if xψ d = x then xρϕ d = xρ, and vice-versa, so x and xρ have orbits of equal size. This holds for all x ∈ X ′ ψ , so T ψ = T ϕ and both X ′ ψ and X ′ ϕ have the same number of elements with an orbit of size d. Therefore , for all d ∈ T ψ = T ϕ . Hence 2 follows, from Lemma 5.10 and the fact that X ′ ψ and X ′ ϕ are expansions of X ψ and X ϕ , respectively. Conversely, suppose that 1 and 2 hold. Let T ψ = T ϕ = {d 1 , . . . , d k } and write m j = m ψ (d j , X ψ ) and m ′ j = m ϕ (d j , X ϕ ). Fix j ∈ {1, . . . , k}. Assume first that m j > m ′ j . Then, by hypothesis, Let Y x be a q jfold expansion of {x} and, as in the proof of Lemma 5.10, let Moreover, as in the proof of Lemma 5.10, for given Γ ∈ E, the map ϕ cyclically permutes the elements of the set if O is expanded in this way then the resulting expansion of X ϕ has exactly m ′ j d j + d j q j (n − 1) elements with ϕ-orbits of size d j .
For each j such that m j > m ′ j apply this process to a single element of X ϕ with ϕ-orbit size d j . Now, for each j such that m ′ j > m j apply the process to an element of X ψ with ψ-orbit size d j , interchanging the roles of ϕ and ψ. The result is an expansion X ′ ψ of X ψ and an expansion X ′ ϕ of , where x i ψ = x i+1 and y i ϕ = y i+1 (subscripts modulo d), then we set x i ρ = y i , for 0 ≤ i ≤ d − 1. Then x i ρϕ = y i ϕ = y i+1 = x i+1 ρ = x i ψρ, for all x i , and it follows that xρϕ = xψρ, for all x ∈ X ′ ψ . Hence ρ −1 ψρ = ϕ, as required.
Algorithm 5.13. Let ψ and ϕ be periodic elements of G n,r .
Step 1: Construct A-bases X ψ and X ϕ with respect to which ψ and ϕ are in semi-normal form (Lemma 4.9).
Step 2: Compute the cycle types T ψ and T ϕ . If T ψ = T ϕ , output "No" and stop.
Step 4: Construct A-bases X ′ ψ and X ′ ϕ as described in the proof of Theorem 5.11.
Step 5: Choose a map ρ sending ψ-orbits of elements of X ′ ψ to ϕ-orbits of elements of X ′ ϕ , as in the proof of the theorem, and output ρ.

Conjugacy of regular infinite elements
We begin with a necessary condition for two regular infinite elements to be conjugate. Let ψ be a regular infinite element of G n,r in quasi-normal form with respect to X. By Lemma 4.6, ψ has finitely many semi-infinite X-components, each of which has a unique characteristic (see Definition 4.22). Moreover, if ψ is in semi-normal form with respect to both X and Y and O is a semi-infinite X-component of ψ with characteristic (m, Γ) then O contains an element u of characteristic (m, Γ). From Lemma 4.24, the ψ-orbit of u has precisely one Y -component, which is again semi-infinite of characteristic (m, Γ). Therefore, the set of pairs (m, Γ) which are characteristics of semi-infinite X-components is independent of the choice of X (with respect to which ψ is in semi-normal form), and we may make the following definition.
Proof. Let ψ and ϕ be in quasi-normal form with respect to the A-bases X and Y respectively and let ρ be such that ρ −1 ψρ = ϕ. Thus, if u is an element of X A such that uψ m = uΓ, for some m and Γ, then uρϕ m = uψ m ρ = uΓρ = uρΓ.
The same argument can be applied starting with an element v ∈ Y A and interchanging ψ and ϕ. Hence if u belongs to a ψ-orbit of characteristic (m, Γ) then uρ belongs to an ϕ-orbit of characteristic (m, Γ). Thus, from Lemma 4.24, a ψ-orbit that contains a semi-infinite X-component of characteristic (m, Γ) is mapped by ρ to a ϕ-orbit which has a semi-infinite Y -component of the same characteristic.
Definition 5.17. Let ψ be in semi-normal form with respect to X. The equivalence relation ≡ on X, is that generated by the relation x ≡ x ′ , whenever xΓ and x ′ ∆ are in the same ψ-orbit, for some Γ, ∆ ∈ A * .
Proposition 5.19. Let ψ be a regular infinite element in quasi-normal form with respect to X. Let X = m i=1 X i where the X i are the equivalence classes of ≡ defined on X under the action of ψ. Then V n,r is the free product of the ψ-invariant subalegbras V 1 , . . . , V m , where V i is the subalgebra generated by X i .
Proof. As ψ is regular infinite, the sets X i partition X, so V n,r is the free product of the V i 's. To show that V i is ψ-invariant it suffices to show that if x ∈ X i then xψ and xψ −1 are in V i . To this end, choose d ≥ 0 such that xψΓ and xψ −1 Γ belong to X A , for all Γ ∈ A * of length d. Then, for Γ of length d, we have xψΓ = y∆ and xψ −1 Γ = zΛ, for some y, z ∈ X and ∆, Λ ∈ A * . By definition then y ≡ x ≡ z, so x, y, z ∈ X i . This implies that xψΓ = y∆ ∈ V i and xψ −1 Γ = zΛ ∈ V i . This holds for all Γ of length d, so from Lemma 3.20, xψ and xψ −1 belong to V i , as required. Hence V i is ψ invariant.
Lemma 5.20. Let ψ be a regular infinite element in quasi-normal form with respect to X and let X i , i = 1, . . . , m, be the equivalence classes of ≡ defined on X under the action of ψ. Then we may effectively construct the X i .
Proof. From Lemmas 4.28 and 4.3, we may effectively construct X, the minimal expansion Y of ψ with respect to X, and the basis Z = Y ψ. For each v ∈ X ∪ Y ∪ Z we may enumerate part of the X-component of v using the procedure of Lemma 4.30. Denote by O v the part of the ψ-orbit of v enumerated this way. Let ≡ 0 be the equivalence relation on X generated by y ≡ 0 z if yΓ and z∆ belong to O v , for some v ∈ X ∪ ∪Z and Γ, ∆ ∈ A * . We claim that ≡ 0 =≡.
Lemma 5.21. Let ψ be a regular infinite element in quasi-normal form with respect to X and let X i , i = 1, . . . , m, be the equivalence classes of ≡ defined on X under the action of ψ. Define for i = 1, . . . , m. Then θ i extends to an element of G n,r which commutes with ψ and with θ j , for all j ∈ {1, . . . , m}.
Lemma 5.22. Let ψ and ϕ be regular infinite elements of G n,r , in quasi-normal form with respect to the A-bases X and Y respectively and let X i , i = 1, . . . , m, be the equivalence classes of ≡ defined on X under the action of ψ. If ψ and ϕ are conjugate then, given x 1 , . . . , x m such that x i is an element of type (B) in X i , there exists a conjugator ρ such that x i ρ is a terminal or initial element in a semi-infinite Y -component of ϕ.
Proof. Since ψ and ϕ are conjugate, by Lemma 5.16 the sets M ψ and M ϕ , of characteristics for ψ and ϕ, coincide and there exists an element ρ ′ such that ρ ′−1 ψρ ′ = ϕ. Let x i be the given element of type (B) in X i . Then, from Lemma 5.16, x i ρ ′ belongs to a ϕ-orbit which contains a semi-infinite Y -component, with the same characteristic as x i . Let y i ∈ Y A be an initial or terminal element of this ϕ-orbit. Then there exists j i such that Thus For each equivalence class X i , define θ i as in Lemma 5.21 and ρ ∈ G n,r by Then θ = n i=1 θ −ji i commutes with ψ, so ρ −1 ψρ = ρ ′−1 θ −1 ψθρ ′ = ρ ′−1 ψρ ′ = ϕ and, for each chosen x i ∈ X i , Thus ρ is the required conjugator.
Definition 5.23. Let ψ and ϕ be regular infinite elements in quasi-normal form with respect to X and Y and let X i , i = 1, . . . , m be the equivalence classes of ≡ defined on X under the action of ψ. Let R i (ψ, ϕ) be the set of pairs (x, y), where x is of type (B) in X i and y is an initial or terminal element of a semi-infinite Y -component of ϕ with the same characteristic as x.
The set R i (ψ, ϕ) is finite since the number of elements of type (B) in X and the number of semi-infinite Y -components of ϕ is finite, so R(ψ, ϕ) is also finite.
Lemma 5.24. Given ρ 0 ∈ R(ψ, ϕ), there are finitely many ways of extending ρ 0 to an element ρ of G n,r such that ϕ = ρ −1 ψρ. Moreover the existence of such an extension ρ can be effectively determined, and if such ρ exists then the images yρ can be effectively determined, for all y ∈ X.
Proof. Throughout the proof, when we say ρ exists we mean that an extension ρ of ρ 0 to an element of G n,r exists and satisfies ϕ = ρ −1 ψρ. From Lemma 5.20, we may effectively construct the equivalence classes X i , and so also the sets R i (ψ, ϕ).
First consider a single equivalence class X i . We are given an element x i of type (B) and an element y i such that where y i is an initial or terminal element of a semi-infinite Y -component of ϕ with the same characteristic multiplier and power as x i . Let x ∈ X of type (B). Then, by definition of ≡, we have x ∈ X i if and only if there exist elements x i = u 0 , . . . , u m = x of X, elements Γ j , ∆ j ∈ A * and k j ∈ Z with u j+1 ∆ j+1 = u j Γ j ψ kj , for j = 0, . . . , m − 1. Before going any further we show that we may assume that u j is of type (B), for all j. Suppose not, say u j is of type (C). Then, by Lemma 4.18, there exist so we may replace u j by u ′ j . Continuing this way, eventually all u j will be of type (B). We show, by induction on m, that there are finitely many possible values of xρ, for an element ρ ∈ G n,r such that xϕ = xρ −1 ψρ (where x i ρ = x i ρ 0 = y i ) and describe an effective procedure to enumerate the set of all such elements. Suppose first that m = 1, so x = u 1 and we have Γ = Γ 0 , ∆ = ∆ 1 and k = k 0 such that x i Γψ k = x∆. Given that ρ exists, from Lemma 5.16, xρ belongs to a semi-infinite Y -component O of ϕ with the same characteristic as x. Therefore (if ρ exists) there exists an element (x, w) ∈ R i (ψ, ϕ) such that w is the initial or terminal element of O; and an integer l such that wϕ l = xρ. This implies that Lemma 4.34 gives an effective procedure to determine whether an integer l satisfying (7) exists, and if so find it. Given ρ 0 and x, the integer k and the elements Γ and ∆ are uniquely determined so, to decide whether an appropriate value xρ exists, we may check each pair (x, w) in the set R i (ψ, ϕ) to see if (7) holds for some l or not. For each such w there is at most one l such that (7) has a solution and, as R i (ψ, ϕ) is finite, we may effectively enumerate the values w∆ϕ l−k that could be assigned to xρ. Hence the result holds if m = 1. Now assume that m > 1 and the result holds for all x related to x i by a chain of length at most m − 1. Then u m−1 is of type (B) and by assumption u m−1 ρ may be given one of finitely many values, and we have a procedure to enumerate these values. Suppose then that u m−1 ρ = v. Now x = u m and we have Γ m−1 , ∆ m ∈ A * and k m−1 ∈ Z such that u m−1 Γ m−1 ψ km−1 = x∆ m . Applying the argument of the case m = 1 with u n−1 , Γ n−1 , ∆ n and v in place of x i , Γ, ∆ and y, we see that a finite set of possible values for xρ may be effectively determined. Therefore, by induction, the result holds for all x ∈ X i of type (B).
Finally, if x ∈ X i is of type (C), then by Lemma 4.18 there is a zΣ in the X-component of x, for some z of type (B) and Σ ∈ A * , i.e. xψ p = zΣ for some integer p. Since we have already determined the possible images of all the type (B) elements in X i , if ρ exists we have, for each choice of zρ, xρ = zΣψ −p ρ = zρΣϕ −p and this determines the image of the type (C) element under ρ (uniquely once we have made our initial choice for the image of zρ).
We carry out this process on each equivalence class in turn. An extension of ρ 0 exists only if the process results in a at least one possible value for each element of X. If the process returns a potential extension ρ of X then ρ is an extension of ρ 0 , of the required type, if Xρ is an A-basis of V n,r ; which may be verified effectively, using Lemma 3.16.
We are now able to state the main result of this section.
Proposition 5.25. Let ψ and ϕ be regular infinite elements of G n,r in quasi-normal form with respect to X and Y respectively.
Then, ψ is conjugate to ϕ if and only if there exists a map ρ 0 ∈ R(ψ; ϕ) such that ρ 0 extends to an element ρ of G n,r with ρ −1 ψρ = ϕ.
Proof. Obviously, if ρ 0 extends to an element of G n,r such that ρ −1 ψρ = ϕ, then ψ is conjugate to ϕ by ρ.
Assume that ψ is conjugate to ϕ. Lemma 5.22 tells us that there exists a conjugator ρ such that, for each equivalence class X i , there exists an element x i of type (B) in X i with y i = x i ρ an initial or terminal element of a semi-infinite Y -component of ϕ.
We define ρ 0 to be the map x 1 → y 1 , . . . , x m → y m , where y i = x i ρ for each i = 1, . . . , m. Thus, ρ 0 is an element of the finite set R(ψ; ϕ). Now ρ 0 is the restriction of ρ to {x 1 , . . . , x m }, so it certainly extends to ρ, as required.
Neither value of w results in ρ 1 which could be a conjugator.
Therefore, we find one conjugating element ρ 2 and we see that ψ and ϕ are conjugate.
The algorithm for the conjugacy of regular infinite elements of G n,r is as follows.
Algorithm 5.27. Let ψ and ϕ be regular infinite elements of G n,r .
Step 1: Construct A-bases X ψ and X ϕ with respect to which ψ and ϕ are in quasi-normal form (Lemma 4.28).
Step 3: Find the initial and terminal elements of semi-infinite X ϕ -components of ϕ, by finding the minimal expansion of X ϕ associated to ϕ (Lemma 4.9).
Step 5: For each equivalence class X i of ≡ on X ψ choose an element x i ∈ X i , of type (B).
Step 6: For each i and each pair (x i , y) of R i (ψ, ϕ), construct a map ρ i : X i → X ϕ , using equation (7), as in the proof of Lemma 5.24, if possible. In each case check that ρ i is an automorphism.

The power conjugacy problem
For a group with presentation X | R , the power conjugacy problem is to determine, given words g, h ∈ F(X) whether or not there exist non-zero integers a and b such that g a is conjugate to h b in G. We may in addition require that, if the answer to this question is "yes", then integers a and b, and an element c ∈ F(X), are found, such that c −1 g a c = G h b . We say the power conjugacy problem is decidable if there is an algorithm which, given g and h outputs "yes" if they're conjugate and "no" otherwise. Again, the stronger form entails the obvious extra requirements. Again, in G n,r we work entirely with symbols for automorphisms, ignoring the presentation. As in the case of the conjugacy problem, we break the power conjugacy problem down into two cases; one for periodic elements and one for regular infinite elements. Then, we construct an algorithm that combines the two parts.

The power conjugacy for periodic elements
Let ψ and ϕ be periodic elements of G n,r , of order k and m respectively, in quasi-normal form with respect to the A-bases X and Y Then, to test whether ψ a is conjugate to ϕ b for a, b ∈ Z, we can apply Proposition 5.11 to the pair ψ c , ϕ d for all c ∈ {1, . . . , k} and all d ∈ {1, . . . , m}.

Regular infinite elements
The first step is to compare the sets M ψ and M ψ a , a ∈ Z, |a| > 1, for a regular infinite automorphism ψ. First note that if ψ is in semi-normal form with respect to an A-basis X, then as X-components of ψ are super-sets of X-components of ψ a , we have ψ a in semi-normal form with respect to X.
Lemma 6.1. Let ψ be a regular infinite element of G n,r and let a be a non-negative integer. Then Proof. Let ψ be in semi-normal form with respect to X; so ψ a is also in semi-normal form with respect to X. First we show that the right hand side is contained in the left hand side. If (m, Γ) ∈ M ψ then there exists an element u of V n,r in a semi-infinite X-component for ψ of characteristic (m, Γ); and we may assume u ∈ X A . Suppose first that a > 0. If d = gcd(m, a), p = m/d, q = a/d and k = ma/d, then u(ψ a ) p = uψ mq = uΓ q , (as mq has the same sign as m). If a < 0 then, from the above, with d = gcd(m, −a), p = m/d, q = −a/d and k = −ma/d, we have uψ −ap = uΓ q . In all cases therefore u is a characteristic element of ψ a . Furthermore, if u(ψ a ) r = u∆, with ∆ = 1 then, from Lemma 4.25, m|ar, which we can rewrite as pd|qdr, so p|qr. As gcd(p, q) = 1, this implies p|r, so that |m/d| = |p| ≤ |r|. Hence u has characteristic (m/d, Γ q ), with respect to ψ a . As u belongs to a semi-infinite X-component for ψ a , it follows that (m/d, Γ q ) is in M ψ a and so we have M ψ a ⊇ {(m, Γ q ) | (md, Γ) ∈ M ψ , d > 0, gcd(m, q) = 1 and |a| = qd}.
On the other hand, suppose that (r, ∆) ∈ M ψ a . Assume first that a > 0. Then again, there exists u ∈ X A such that u is a characteristic element of ψ a , so uψ ar = u∆. Thus, from Lemma 4.25, u is a characteristic element for ψ, with characteristic (m, Γ) ∈ M ψ , such that m|ar and ∆ = Γ t , where ar = mt, t > 0. Let d = gcd(a, m), m = pd and a = qd. Then dqr = pdt, so qr = pt and gcd(p, q) = 1, so r = pr ′ and t = qt ′ , for some r ′ , t ′ . However, we have u(ψ a ) p = uψ dpq = uψ mq = uΓ q , and so, by definition of (r, ∆) ∈ M ψ a , we see that |p| ≥ |r|, so r ′ = ±1. Since a > 0, r ′ = 1. It now follows that r = p = m/d and ∆ = Γ q , and (r, ∆) belongs to the set on the right hand side of the equality in the lemma. That is M ψ a ⊆ {(m, Γ q ) | (md, Γ) ∈ M ψ , d > 0, gcd(m, q) = 1 and |a| = qd}.
If a < 0 then the lemma follows by applying the result above to M ψ −1(−a) , as for all θ ∈ G n,r we have (m, Γ) ∈ M θ if and only if (−m, Γ) ∈ M θ −1 .
Lemma 6.3. Let ψ and ϕ be regular infinite elements of G n,r and let c be an integer, such that c is coprime to m, for all m ∈ Z such that (m, Γ) ∈ M ψ ∪ M ϕ . Then ψ c ∼ ϕ c if and only if ψ ∼ ϕ.
If s < 0 then we have m(−s) + c(−t) = −1, with −s > 0 and the argument above implies instead that uψ −1 ρ = uρϕ −1 . In this case, let v = uψ, so v also has ψ-characteristic (m, Γ) and replacing u by v gives vψ −1 ρ = vρϕ −1 from which it follows that uψρ = uρϕ. This applies in particular to all elements of X of type (B). Let y ′ be an element of type (C); so there exists an element y ∈ X of type (B) such that y ′ ψ k = yΩ, for some k ∈ Z. Then y ′ = yΩψ −k , and yψ j has the same ψ-characteristic as y, for all j: and so is a characteristic element for ψ. From the above then yψ j ρ = (yρ)ϕ j , for all j. Now Therefore, yψρ = yρϕ, for all y ∈ X, so ψ ∼ ϕ.
Definition 6.4. Let ψ be a regular infinite element of G n,r and let a be a positive integer. Define a map ψ a : M ψ → M ψ a by ψ a (m, Γ) = (p, Γ α ), where d = gcd(m, a), p = m/d and α = a/d. Example 6.5. For ϕ in Example 6.2, with a = 2, the map Proof. Assume ψ a ∼ ϕ b , with a, b > 0, and that ρ −1 ψ a ρ = ϕ b . From Lemma 5.16, M ψ a and M ϕ b are equal, so M = N , and we may order M ψ a so that (p i , Γ αi i ) = (q i ∆ βi i ), so p i = q i and Γ αi i = ∆ βi i . With the notation for ( ψ a ) −1 (p i , Γ αi i ) and ( ϕ b ) −1 (q i , ∆ αi i ) given in the statement of the proposition, let d i,j = gcd(a, m i,j ) and e i,k = gcd(b, n i,k ), so by Definition 6.4, for i,k , by Proposition 3.9, there exist Λ i,j,k ∈ A * and positive integers s i,j,k , t i,j,k such that Γ i,j = Λ s i,j,k i,j,k and ∆ i,j = Λ t i,j,k i,j,k . Taking a power of Λ i,j,k if necessary, we may assume that gcd(s i,j,k , t i,j,k ) = 1. Then so s i,j,k α i,j = t i,j,k β i,k . As s i,j,k and t i,j,k are coprime this implies that α i,j /t i,j,k = β i,k /s i,j,k = c i,j,k ∈ Z, and α i,j = c i,j,k t i,j,k and β i,k = c i,j,k s i,j,k . Let Then there exist integers f i,j,k such that c i,j,k = gf i,j,k , for all i, j, k. From Lemma 6.1, M ψ a/g consists of elements (m/p, Γ α ), where (m, Γ) ∈ M ψ , p = gcd(m, a/g) and α = a/gp. Similarly, elements of M ϕ b/g are of the form (n/q, ∆ β ), where (n, ∆) ∈ M ϕ , q = gcd(n, b/g) and β = b/gq. Now g|c i,j,k and c i,j,k |α i,j and c i,j,k |β i,k . Therefore gcd(m i,j , a/g) = gcd(m i,j , a) = d i,j and similarly gcd(n i,k , b/g) = e i,k . Thus g is coprime to p i = m i,j gcd(m i,j , a/g) = n i,k gcd(n i,k , b/g) , for all i, j, k. From Lemma 6.3, it follows that ψ a/g ∼ ϕ b/g . Now a/g = α i,j d i,j /g = c i,j,k t i,j,k d i,j /g = f i,j,k t i,j,k d i,j and similarly b/g = f i,j,k s i,j,k e i,k , for all i, j, k. Also gcd({f i,j,k |1 ≤ i ≤ M, 1 ≤ j ≤ M i , 1 ≤ k ≤ N i }) = 1 so, for fixed i ′ , j ′ , k ′ , Corollary 6.7. The power conjugacy problem for regular infinite elements of G n,r is solvable.
Proof. Let ψ and ϕ be regular infinite elements of G n,r . Suppose that ψ a is conjugate to ϕ b , for some non-zero a, b. Replacing either ψ or ϕ or both by their inverse, we may assume that a, b > 0. Then, in the notation of the proposition above, we have ψ α ∼ ϕ β , where α = f i,j,k t i,j,k d i,j and β = f i,j,k s i,j,k e i,k . From the conclusion of the theorem it is clear that there are finitely many choices for f i,j,k , s i,j,k , t i,j,k , d i,j and e i,k . Hence there are finitely many possible α and β, and we may effectively construct a list of all possible pairs (α, β). Having constructed this list we may check whether or not ψ α ∼ ϕ β , using Algorithm 5.27. Hence we may decide whether or not there exist a, b such that ψ a ∼ ϕ b .
Remark 6.9. In Corollary 6.7 we supposed that the powers a and b were positive, giving us upper bounds a ≤â and b ≤b for the minimal powers which solve the power conjugacy problem. Now suppose that a < 0 and b > 0. We may write ψ a = (ψ −1 ) −a and then −a > 0. If we apply Corollary 6.7 to (ψ −1 , ϕ), we would obtain a second pair of bounds −a ≤ā and b ≤b. Observing that (m, Γ) ∈ M ψ if and only if (−m, Γ) ∈ M ψ −1 , we note that this replacement ψ → ψ −1 preserves the absolute value |m| of all characteristic multipliers. Thus each of the terms |m i,j , |n i,k |, |m| and |n| in equations (11-14) is unchanged. We conclude thatā =â andb =b. The same argument applies equally well to the remaining two cases a > 0, b < 0 and a < 0, b < 0. Thus, once we have obtainedâ andb, we need only to check the ranges 1 ≤ |a| ≤â and 1 ≤ |b| ≤b to find minimal conjugating powers.
Assume there exists positive integers a, b such that ψ a ∼ ϕ b ; with a = 1 and b ≤ 2. The map ϕ b : M ϕ → M ϕ b is given by Therefore there is no pair of positive integers a, b such that ψ a ∼ ϕ b . The same argument applies replacing ϕ or ψ by ϕ −1 or ψ −1 respectively, so no nontrivial power of ϕ is conjugate to a power of ψ.

The power conjugacy algorithm
We combine the algorithms of Sections 6.1 and 6.2 to give an algorithm for the power conjugacy problem in G n,r . In fact in Sections 6.1 and 6.2 we find a description of all solutions of the power conjugacy problem for periodic and regular infinite automorphisms, respectively: and the algorithm in this section does the same for arbitrary elements of G n,r . If we are only interested in the existence of a solution to the power conjugacy problem then we may essentially ignore the periodic part of automorphisms, as long as the regular infinite part is non-trivial. To see this, suppose ψ and ϕ are elements of G n,r and we have decompositions ψ = ψ P * ψ RI , ϕ = ϕ P * ϕ RI . Assume that we have found that V RI,ψ is non-trivial and ψ a RI is conjugate to ϕ b RI , a, b = 0. In this case, ψ P and ϕ P have finite orders, m and k say, and so we immediately have a solution ψ amk ∼ ϕ bmk , amk, bmk = 0, of the power conjugacy problem. The algorithm described below allows this type of solution but also tries to find a solution to the power conjugacy problem corresponding to each pair (c, d) such that ψ c P ∼ ϕ d P . Thus, in Theorem 6.14, we obtain a description of all solutions to the power conjugacy problem, for ψ and ϕ.
Algorithm 6.13. Let ψ and ϕ be elements of G n,r .
Step 3: If X RI,ψ is non-empty (that is, V RI,ψ is non-empty) and PC RI is empty, output "No" and stop.
Step 4: Compute the orders k and m of ψ P and ϕ P . Input ψ a P and ϕ b P to Algorithm 5.13, for all c, d such that 1 ≤ c ≤ k and 1 ≤ d ≤ m. Construct the set PC P of all triples (c, d, ρ) found such that ρ −1 ψ c ρ is conjugate to ϕ d . If X RI,ψ is non-empty, adjoin the triple (0, 0, θ 0 ) to PC P , where θ 0 is the identity map of the algebra V n,sP , of Step 3 of Algorithm 5.6.
Step 5: If PC P is empty, output "No" and stop. If PC P is non-empty and X RI,ψ is empty output PC P and stop.
Step 6: If this step is reached then both PC P and PC RI are non-empty. For all (α, β, ρ RI ) in PC RI and all pairs (c, d, ρ P ) in PC P consider the simultaneous congruences αx ≡ c mod k and βx ≡ d mod m, where k and m are the orders of ψ P and ϕ P found in Step 4. For each positive solution x = g (less than lcm(k, m)) add (αg, βg, g, ρ P * ρ RI ) to the set PC (which is empty at the start).
We verify that this algorithm solves the power conjugacy problem in the proof of the following theorem.
Theorem 6.14. The power conjugacy problem for the Higman-Thompson group G n,r is solvable. Furthermore, given elements ψ, ϕ ∈ G n,r , let ψ P have order k, let ϕ P have order m and let l = lcm(k, m). There is a finite subset PC ⊆ Z 3 × G n,r , which may be effectively constructed, such that ψ a ∼ ϕ b if and only if (ag/h, bg/h, g, ρ) ∈ PC, where ρ ∈ G n,r and g, h ∈ Z such that h ≡ g mod l, h|a and h|b. In this case ρ −1 ψ a ρ = ϕ b .
Proof. Apply Algorithm 6.13 to ψ and ϕ. If there exist a, b ∈ Z such that ψ a ∼ ϕ b then ψ a P ∼ ϕ b P and ψ a RI ∼ ϕ b RI . In this case let ψ P and ϕ P have orders k and m, respectively and let a 1 , b 1 ∈ Z be such that 1 ≤ a 1 < k and 1 ≤ b 1 < m and a 1 ≡ a mod k, b 1 ≡ b mod m. Then there exists ρ P such that (a 1 , b 1 , ρ P ) ∈ PC I . Furthermore, from Corollary 6.12, there exists (a 2 , b 2 , ρ RI ) ∈ PC RI and h ∈ Z such that a = a 2 h and b = b 2 h. Let g be such that 1 ≤ g < lcm(k, m), and g ≡ h mod lcm(k, m) so g ≡ h mod k and g ≡ h mod m. As h is a solution to the congruences a 2 x ≡ a 1 mod k and b 2 x ≡ b 1 mod m, it follows that g is also a solution to these congruences. Therefore (a 2 g, b 2 g, g, ρ P * ρ RI ) ∈ PC. As a 2 = a/h and b 2 = b/h, this is an element of Z 3 × G n,r of the required form.
Conversely, assume (u, v, g, ρ P * ρ RI ) ∈ PC, where u = ag/h and v = bg/h, for some a, h ∈ Z satisfying the hypotheses of the theorem. Then there exist (α, β, ρ RI ) in PC RI and (c, d, ρ P ) in PC P such that u = αg ≡ c mod k and v = βg ≡ d mod m. As g ≡ h mod l this implies that a = (u/g)h = αh ≡ c mod k and b = (v/g)h = βh ≡ d mod m. Therefore ψ a P = ψ c P ∼ ϕ d P = ϕ b P , by definition of PC P , and indeed ρ −1 P ψ a P ρ P = ϕ b P . Also, a = αh and b = βh implies ρ −1 RI ψ a RI ρ RI = ϕ b RI , by Corollary 6.12, so ψ a = (ψ P * ψ RI ) a = ψ a P * ψ a RI ∼ ϕ b P * ϕ b RI = (ϕ P * ϕ RI ) b = ϕ b and ρ P * ρ RI is a conjugating element.
Examples which illustrate how the algorithm works on automorphisms which are not necessarily periodic or regular infinite can be found at [D15]: follow the link to "Examples" and refer to the examples named "mixed pconj phi" and "mixed pconj psi".