Equations in Free Groups with One Variable: I

Transcription

1 Equations in Free Groups with One Variable: I I. M. Chiswell and V. N. Remeslennikov 1. Introduction. R.Lyndon [8] investigated one-variable systems of equations over free groups and proved that the set of solutions of a single equation can be defined by a finite system of parametric words. These parametric words were complicated, and the number of parameters on which they depended was restricted only by the type of each equation considered. Further progress in this direction was made by Appel [2] and Lorents [6], [7], who gave the exact form of the parametric words, and Lorents extended the results to systems of equations with one variable. Unfortunately Appel s published proof has a gap (see the review 38#1149 by Tekla Lewin in Mathematical Reviews, reprinted on p.87 of [3]) and Lorents announced his results without proof. The goal of this paper is two-fold; firstly to give a full argument, and secondly to suggest a new approach for describing the set of all solutions of a one-variable system in a free group. Instead of working directly with solutions, we work only with the coordinate groups of irreducible algebraic sets. This idea will be explained in the next section. It seems to us a simpler point of view than that in the original papers, despite the number of different cases to consider. We then discuss the other ideas used in the proof (ultrapowers and Lyndon length functions), before stating the main results. These are then proved by means of a series of lemmas, one for each case that can arise. 2. Algebraic geometry over groups. We remind readers of the main definitions concerning algebraic geometry over groups, following the preprint of G.Baumslag, A.Myasnikov and V.Remeslennikov [4]. Definition 2.1. Let G be a fixed group. A group H is called a G-group if G is a fixed subgroup of H. The class of G-groups forms a category in the obvious way, in particular a group homomorphism ϕ : H 1 H 2 between two G-groups is termed a G- homomorphism if ϕ(g) = g for all g G. Definition 2.2. Let X be a set and G a group. The group G F (X), where F (X) is the free group on X, is called the free G-group on X and is denoted by G[X]. With the obvious embedding of G, G[X] is a free object, in the usual sense, in the category of G-groups. A G-group H is called a free G-group if it is isomorphic as a G-group to G[X] for some set X, and X is then called a set of free G-generators for H. If X has a single element x, we denote G[X] = G x by G[x]. 1

2 Definition 2.3. The set G n = {(g 1,..., g n ) g i G} is called affine n-space over G; when n = 1, G 1 = G. Suppose X = {x 1,..., x n } is a finite set. Using the normal form theorem for free products and for F (X), an element f G[X] may be viewed as a word in the elements of G and X ±1, which we view as a monomial in the variables X ±1 with coefficients in G. Given p = (g 1,..., g n ) G n, we can substitute g ±1 i for x ±1 i in f to obtain an element f(p) G. (More formally, there is a homomorphism ϕ : G[X] G which is the identity mapping on G and sends x i to g i, and f(p) means ϕ(f)). If f(p) = 1, we think of p as a solution of the equation f = 1, and sometimes call a subset of G[X] a system of equations over G. The definitions which follow are made by analogy with the situation for affine algebraic sets over a field. Definition 2.4. Let S G[X], where X = {x 1,..., x n }. Then the set V G (S) = {p G n f(p) = 1 for all f S} is termed the algebraic set over G defined by S. Typical Examples. (1) Every singleton {g} is an algebraic set: V G (xg 1 ) = {g} (here n = 1 and X = {x}). (2) For any subset M G, the centraliser C G (M) is algebraic, being equal to V G ([x, m] m M). (3) More generally, for any g G, the coset C G (M)g = V G ([xg 1, m] m M) is algebraic. Definition 2.5. Let S G[X] be a system of equations over G, and put Y = V G (S). Then we define Rad(S) = Rad(Y ) = {f G[X] f(p) = 1 for all p Y }. Call Rad(S) the radical ideal defined by the system S (or by Y ). Clearly Rad(S) is always a normal subgroup of G[X]. Definition 2.6. The factor group G Y = G S = G[X]/Rad(S) is termed the coordinate group of the algebraic set Y. Examples. Let G = F be a non-abelian free group and X = {x}. (1) If Y = V F (xg 1 ) = {g}, then G Y = F [x]/ncl(xg 1 ) = F. (2) If Y = C G (g) = V F ([x, g]), then G Y = F [x]/ncl([x, g]) = F, t [g, t] = 1. The proof is left to the reader, using the Normal Form Theorem for HNN extensions. Such an HNN extension is called a free extension of centralisers of rank 1. Definition 2.7. Let G be a group and let Y G n, Z G p be algebraic sets. Then a map φ : Y Z is termed a morphism of the algebraic set Y to the 2

3 algebraic set Z if there exist f 1,..., f p a 1,... a n Y, G[x 1,... x n ] such that for any point φ(a 1,... a n ) = (f 1 (a 1,... a n ),... f p (a 1,... a n )) Z. Definition 2.8. Two algebraic sets Y, Z are said to be isomorphic if there exist morphisms φ : Y Z, θ : Z Y such that θφ = 1 Y and φθ = 1 Z. Let AS G be the category of all algebraic sets over G, with morphisms as defined above (so the use of the term isomorphic is consistent with that in category theory). Denote by AG G the category of all coordinate groups of the algebraic sets from AS G, where morphisms are G-homomorphisms. The following is Theorem 4 in [4]. Theorem 2.1 ([4]). The categories AS G and AG G are equivalent. From the theorem follows a very important fact: to classify algebraic sets up to isomorphism, it is enough to classify coordinate groups of the corresponding algebraic sets. Typical Examples. (1) Let V be an algebraic set in G n. Then for any two points p, q G n the set pv q is also an algebraic set which is isomorphic to V. Let p = (p 1,..., p n ), q = (q 1,..., q n ), and let S(x 1,..., x n ) be the system of equations that defines V. Then S(y 1,..., y n ), where y i = p 1 i x i q 1 i defines pv q and the map φ : V pv q, φ(a 1,..., a n ) = (p 1 a 1 q 1,..., p n a n q n ) defines the isomorphism between V and pv q. In particular, (1a) Any two points p 1, p 2 of G n are isomorphic to each other (as singleton algebraic sets). (1b) If C G (M) is the centralizer of a subset M and f, g G then C G (M) is isomorphic to fc G (M)g. (2) If V G 1 is an algebraic set and V is isomorphic to G 1, then V = G 1. For by Theorem 2.1, an isomorphism from V to G 1 induces a G-isomorphism φ : G[x] G[y] of two free G-groups of rank 1, where G[x] is the coordinate group of G and G[y] is the coordinate group of V. Then φ(x) = py e q, where e { 1, 1}, p, q G. Further, φ defines an isomorphism φ between G and V in the following way (see the proof of Theorem 2.1 given in [4; Theorem 4]): φ (g) = pgq. Therefore V = G. (3) Let F be a free group. If C F (f) is the centralizer of f in F, f 1 is not a proper power in F and V is an algebraic set in F 1 isomorphic to C F (f), then there exist p, q F such that V = pc G (f)q. For let T (V ) and T (C(f)) be coordinate groups for V and C F (f) respectively, then they are F -isomorphic (by Theorem 2.1). Let T (C(f)) = F, t [t, f] and T (V ) = F, y where t, y are the images of x under the canonical projections fron F [x], and let φ : T (V ) T (C(f)) be an F -isomorphism. It is not hard 3

4 to check that φ(y) = pt e q, where e { 1, 1}, p, q F. Then (see the proof of Theorem 2.1 in [4; Theorem 4]) the map φ : f k pf ek q, k Z defines an isomorphism from C F (f) to V. Therefore V = pc F (f)q. Definition 2.9. A group G is said to be equationally Noetherian if, for every n > 0 and any subset S of G[x 1,... x n ], there exists a finite subset S 0 of S such that V G (S) = V G (S 0 ). Remark. (see [4]) Every linear group is equationally Noetherian, in particular, every free group is equationally Noetherian. If G = F is a free group, it is possible to introduce on F n the so-called Zariski topology, by taking the set of all algebraic sets as the set of closed subsets. One can therefore define the notion of irreducible algebraic set. (A non-empty subset Y of a topological space X is called irreducible if it cannot be expressed as a union Y = Y 1 Y 2 of proper subsets, each of which is relatively closed in Y ). Theorem 2.2 ([4]). Every algebraic set V over an equationally Noetherian group H is a finite union of irreducible algebraic sets, which are uniquely determined by V, up to order. (They are called the irreducible components of V ). 3. Ultraproducts and coordinate groups of irreducible algebraic sets. Let I be a set and let P(I) be the Boolean algebra of all subsets of I. An ultrafilter over I is a subset D of P(I) such that (i) A D and A B I implies B D (ii) A, B D implies A B D (iii) for all A P(I), exactly one of A, I \ A belongs to D. Let {X i i I} be a family of sets indexed by I, and let D be an ultrafilter over I. The ultraproduct i I X i/d is defined to be the quotient set i I X i/, where (x i ) i I (y i ) i I if and only if {i I x i = y i } D. This is easily checked to be an equivalence relation. We shall denote the equivalence class of an element (x i ) i I i I X i by x i i I, and usually abbreviate this to x i. If all X i = X, a single set, then the ultraproduct is X I /D, which is called an ultrapower of X. Note that X embeds canonically in X I /D, by mapping x to x i, where x i = x for all i I. Denoting the ultrapower X I /D by X, given a function f : X n Y of n variables, we obtain an extension to a function f : ( X) n Y by defining f(x (1),..., x (n) ) = f(x (1) i,..., x (n) i ), where x 1 = x 1 i, etc. Similarly any finitary relations on X can be extended to X. Thus if X is a group, then we can define a multiplication in X by x i y i = x i y i, and this makes X into a group. If X is abelian, so is X I /D, and if X is an ordered abelian group, then X I /D can be made into an ordered abelian group by defining x i y i if and only if {i I x i y i } D. This can be proved directly, but follows from Loš s Theorem (see [14]). Further, the canonical embedding of X into X I /D is clearly structure-preserving (if X is a group, it is a group homomorphism, etc.) 4

5 An ultrafilter D over I is called principal if some finite subset of I belongs to D. This is equivalent to saying there is an element j I such that j A for all A D. In this case, the canonical embedding X X I /D is surjective and so identifies X and X I /D. On the other hand, if I is countable and D is non-principal, the canonical embedding is not surjective. Now suppose G is an equationally Noetherian group. Note that any ultrapower G I /D is a G-group under the canonical embedding of G into G I /D. The following result says that certain ultrapowers of G are, in a sense, almost universal for the category of all coordinate groups of irreducible algebraic sets (a subcategory of AG G ). Theorem 3.1 ([4]). Let G be an equationally Noetherian group. Then for every countable I and any non-principal ultrafilter D over I, the ultrapower G I /D has the following properties: (1) every coordinate group of an irreducible algebraic set over G is embeddable in G I /D; (2) every finitely generated G-subgroup of G I /D is isomorphic to the coordinate group of some algebraic set Y over G. 4. Length Functions. Let G be a group and let Λ be an ordered abelian group. A mapping L : G Λ is called a Lyndon length function if (1) L(1) = 0 (2) for all g G, L(g) = L(g 1 ) (3) for all g, h, k G, c(g, h) min{c(h, k), c(k, g)}, where c(g, h) is defined to be 1 2 (L(g) + L(h) L(g 1 h)). (Thus c(g, h) is an element of the ordered abelian group 1 2Λ, which in turn is a subgroup of the ordered abelian group Q Z Λ). Axiom (3) means that at least two of c(g, h), c(h, k), c(k, g) are equal, and not greater than the third, for all g, h, k G. We leave it as an exercise to verify that these axioms imply: (4) for all g, h G, 0 c(g, h) L(g). In particular, since c(g 1, h) 0, we obtain the triangle inequality: (5) for all g, h G, L(gh) L(g) + L(h). Example. Take a free group F with basis X, and define L : F Z by L(g) = the length of the reduced word on X ±1 representing g. Then L is a Lyndon length function, and c(g, h) is the length of the largest common initial segment of the reduced words for g and h (this is the reason for the factor 1 2 in the definition of c). Call this the canonical length function defined by X. Definition 4.1. A Lyndon length function L : G Λ is called free if (6) for all g G, g 1, L(g 2 ) > L(g). 5

6 The terminology comes from the connection between Lyndon length functions and actions by isometries on Λ-trees [1; 5], because (6) is the condition for the action to be free and without inversions [1; 7.2]. Lyndon used the term Archimedean to describe length functions with property (6). The example above of the canonical length function on a free group corresponding to a basis is free, reflecting the fact that the group acts freely on the corresponding Cayley graph. The idea of a product of elements of the free group being reduced as written (having no cancellation) can be introduced for arbitrary length functions. Definition 4.2. Let L be a Lyndon length function on a group G. A sequence (g 1,..., g n ), of elements of G is reduced relative to L if L(g) = L(g 1 ) +... L(g n ). We write g = g 1... g n to mean g = g 1... g n and (g 1,..., g n ) is reduced. Note that uv = u v if and only if c(u 1, v) = 0. Also, uvw = u v w if and only if uv = u v and uvw = (uv) w, if and only if vw = v w and uvw = u (vw). This can be interpreted as a kind of associativity : (u v) w = u (v w). We shall use this without comment; a more general statement can be found in Lemma 2.2 of [13]. Lemma 4.1. For any length function L, (1) If u = v w and L(z) L(v) then zu = (zv) w; (2) If u = u 1 v 1 = u 2 v 2 and L(u 1 ) L(u 2 ) then u 2 = u 1 u 3 for some u 3 ; (3) If uv = u v, vw = v w and L(v) > 0, then uvw = u v w. Proof. The proof of (1) is essentially contained in [5; Lemma 3]. We need to prove that c((zv) 1, w) = 0. This follows from observation (4) if L(zv) = 0, Otherwise, c((zv) 1, v 1 ) 1 2 L(zv) > 0, and c(v 1, w) = 0, so by Axiom (3) c((zv) 1, w) = 0. To prove (2), put u 3 = u 1 1 u 2. By (1), v 1 = u 1 1 u = u 3 v 2. Hence L(v 1 ) = L(u) L(u 1 ) = L(u 3 ) + L(v 2 ) = L(u 3 ) + L(u) L(u 2 ) so L(u 3 ) = L(u 2 ) L(u 1 ). Therefore, c(u 1 1, u 3) = 1 2 (L(u 1) + L(u 3 ) L(u 2 )) = 0, and so u 2 = u 1 u 3. In (3), we are assuming c(u 1, v) = 0 = c(v 1, w) and have to show that c((uv) 1, w) = 0. Using c(u 1, v) = 0 we calculate that c((uv) 1, v 1 ) = L(v) > 0. By Axiom (3) for a length function, c((uv) 1, w) = 0, as required. Let F be a free group with basis X and let L be the canonical length function defined by X. Let I be an index set, let D be an ultrafilter in P(I) and denote the ultrapower F I /D by F. Then as noted in 3, F (resp. Z) is canonically embedded in F (resp Z = Z I /D), and L extends to a mapping L : F Z, given by L( u i ) = L(u i ), and L is a free Lyndon length function. This can be checked directly, but follows from the version of Loš s Theorem in (vdw). 6

7 We make some observations on the length function L which will be used. The basic idea is that one can to some extent carry out arguments analogous to cancellation arguments one uses in F, using the length function L. (Lyndon introduced length functions to formalise cancellation arguments used in free groups and free products). We denote 1 2 ( L(g) + L(h) L(g 1 h)) by c(g, h). Lemma 4.2. (1) Let u F and suppose 0 m n = L(u), where m, n Z. Then there are unique v, w F such that u = v w and L(v) = n. (v is called an initial segment, and w a terminal segment, of u). (2) If u, v F, there are unique u, v and t in F such that u = u t, v = t 1 v uv = u v, and L(t) = c(u 1, v). Proof. (1) Let u = u i, m = m i and n = n i. Then m i n i and L(u i ) = n i for almost all i, i.e. for all i A, where A D, and for such i, u i = v i w i, where L(v i ) = m i. Put v = v i, w = w i (v i and w i may be taken to be any elements of F, e.g. 1, for i A). This shows existence. If also u = v w, where L(v ) = m, put v = v i. For almost all i, L(v i) = L(v i ) = m i, and u i = v i w i = v i w i, and for such i, v i = v i, hence v = v, and so w = w. (2) This is proved by a similar but more elaborate argument to that in (1), by expressing elements in terms of coordinates. Definition 4.3. A pair (u, v) of elements of F is said to have small cancellation if the element t in Lemma 4.2(2) belongs to F, otherwise (u, v) is said to have big cancellation. Definition 4.4. An element t F is called cyclically reduced if it cannot be written as t = f 1 t 1 f, where t 1, f F and f 1. It is easy to see, by taking coordinates, that any element is conjugate in F to a cyclically reduced element. Let a F ; the mapping Z F, n a n, has an extension to a mapping Z F, denoted by α a α, as explained in 3, and this makes F into an exponential Z-group, in the sense of Lyndon [9]. Let 1 a F, and let b be the root of a (so a = b m for some m Z with m > 0 and b is not a proper power in F ). Thus the centraliser C F (a) = b, and it follows that C F (a) = {b α α Z}. We call a a root element if a is its own root (i.e. a is not a proper power). Definition 4.5. An element h F is called an f-power if h = f α, where f F and α Z. If α Z, call h a big f-power. It follows from Proposition 1 in [15] that for any f F and α Z, L(f α ) = L(f) + ( α 1)( L(f 2 ) L(f)), where α = α if α 0 and α = α if α < 0. In particular, if f α is a big f-power, then L(f α ) Z\Z. If f is a cyclically reduced element of F, the initial segments 7

8 of f α are the elements f β g, where β Z, 0 β α and g is an initial segment of f, and if β = α then g = 1. This can again be seen by taking coordinates. Similarly, the terminal segments of f α are f β g, where β Z, 0 β α and g is a terminal segment of f, with g = 1 if β = α. Lemma 4.3. (1) If u F and u = v w, where v, w F, then v, w F. (2) If u = u 1 v 1 = u 2 v 2, where u 2 F, u 1 F \ F, then u 1 = u 2 w for some w. (3) Let f F, t F, α Z, α > 0. Then f α t = f α t if and only if ft = f t, and tf α = t f α if and only if tf = t f. Proof. (1) Let v = v i and w = w i. Then u = v i w i for almost all i, say for i A, where A D. Put n = L(u) and for 0 j L(u), let A j = {i A L(v i ) = j}. Then A = A 0... A n, hence A k D for some k. For otherwise I \ A j D for all j, so n j=0 (I \ A j) D, i.e. I \ A D, contradicting A D. Thus v i = u for all i A k, where u is the initial segment of u of length k, hence v = u F, and w = uv 1 F. (2) Under the assumptions of (2), if u 1 cannot be written as claimed, then we can write u 2 = u 1 w for some w, by Lemma 4.1(2), and by part (1), u 1 F, a contradiction. (3) For the first part, we have to show that c(f α, t) = 0 if and only if c(f 1, t) = 0. This is clear if f = 1 (property (4) of a length function) so assume f 1. From the formula for L(f α ) above, we calculate that c(f 1, f α ) = 1 2 L(f 2 ) if α 1, and c(f 1, f 1 ) = L(f), so in any case c(f 1, f α ) > 0. By Axiom (3) for a length function, c(f 1, t) = 0 if and only if c(f α, t) = 0. The last part follows by applying what has been proved to f 1 and t 1 in place of f, t. It is useful to define an equivalence relation on F. Definition 4.6. Let f 1, f 2 F. Call then F -equivalent (written f 1 F f 2 ) if there exist g, h F with f 1 = gf 2 h. (That is, f 1, f 2 are in the same (F, F ) double coset in F ). The equivalence class of f F under this relation is denoted by [f]. Definition 4.7. An element t F will be called a big element if t F. Let t be a big element, f F \ {1}. We say f left divides t if t = f α t 1 for some f F, α Z \ Z, α > 0 and t 1 F. By considering coordinates, we see that there is then an α, with α as large as possible, for which this happens, and for this α, t 1 cannot be written in the form t 1 = f t 2. Symmetrically, f right divides t if f 1 left divides t 1. In this case, we can write t = t 1 f α, where α Z \ Z, α > 0 and t 1 cannot be written as t 1 = t 2 f. 8

9 Lemma 4.4. Let t be a big element, u F. (1) If the pair (t, ut) has big cancellation then we can write u = u 1 u 2, t = u 1 2 t u 1 1 and t = f 1 t f, where f is a big element. (2) Assume u 1, and write u = v ũ v 1, where ũ is cyclically reduced. If the pair (t 1, ut) has big cancellation then we can write t = v ũ α t, where α Z \ Z, t F. Proof. (1) From Lemma 4.2(1) and 4.2(2) we can write u = u 1 u 2, t = u 1 2 t 1 and ut = u 1 t 1 for some u 1, u 2 F and t 1 F. Also from Lemma 4.2(2), we can write t = t 2 v, ut = v 1 t 3, tut = t 2 t 3 and by assumption v is a big element. By Lemma 4.3(2), v 1 = u 1 v 1 for some v 1, which is also a big element. Thus t = t 2 v1 1 u 1 1 = t 4 u 1 1, where t 4 is big. Again by Lemma 4.3(2), t 4 = u 1 2 t 5 for some t 5, and t 5 is big. Hence t = u 1 2 t 5 u 1 1, so t 1 = t 5 u 1 1 and ut = u 1 t 1 = u 1 t 5 u 1 1. Since t = t 2 v1 1 u 1 1 = u 1 2 t 5 u 1 1, we have t 2 v1 1 = u 1 2 t 5. If L(v 1 ) L(t 5 ), then v1 1 = v 2 t 5 for some v 2, by Lemma 4.1(2). But then ut = v 1 t 3 = u 1 v 1 t 3 = u 1 t 5 u 1 1, so v 1 t 3 = t 5 u 1 1, which implies t 5 = t 1 5 (they are initial segments of v 1 of the same length), hence t 5 = 1, contradicting that t 5 is big. Therefore L(v 1 ) < L(t 5 ) and by Lemma 4.1(2), t 5 = t 6 v1 1 for some t 6. We can therefore write v 1 t 3 = t 6 v1 1 u 1 1. If L(t 6 ) L(v 1 ) then v 1 = t 6 v 3 for some v 3. But then v 3 t 3 = v1 1 u 1 1 = v3 1 t 1 6 u 1 1, hence v 3 = v3 1, so v 3 = 1 and v 1 = t 6, again giving the contradiction t 5 = 1. Thus L(t 6 ) > L(v 1 ) and we can write t 6 = v 1 t 7 for some t 7, so t 5 = v 1 t 7 v 1 giving the required decomposition of t with f = v1 1. (2) Write u = u 1 u 1 2, t = u 2 t 1, ut = u 1 t 1 as in Lemma 4.2. If L(v) > L(u 2 ), then v = u 2 z for some z 1, by Lemma 4.1(2), and t 1 (ut) = t 1 (u 2 z ũ z 1 t 1 ) = t 1 1 (z ũ z 1 t 1 ) = t 1 1 z ũ z 1 t 1 using Lemma 4.1(3). Since u 2 F, the pair (t 1, ut) has small cancellation, a contradiction. Hence L(v) L(u 2 ) and u 2 = v z for some z, so t 1 ut = (t 1 1 z 1 )(ũ(z t 1 )), and the pair (t 1 1 z 1, ũ(z t 1 )) has big cancellation since v F. Since t = v (zt 1 ), it suffices to show that, if u is cyclically reduced, then t = u α t for some t, where α Z \ Z. Write t = u α s with α as large as possible. We need to show that α Z. Suppose α Z; then t 1 ut = s 1 us, and the pair (s 1, us) has big cancellation since u α F. Therefore it suffices to show t = u ±1 t for some t. For applying this to s, we contradict either the choice of α or the fact that (u α, s) is reduced. Recall that u = u 1 u 2, t = u 2 t 1, ut = u 1 t 1, and write t 1 = t 2 v 1, ut = v w, t 1 ut = t 2 w as in Lemma 4.2. By assumption v is big, so v = u 1 v 1 for some big v 1, by Lemma 4.3(2). Thus t = u 2 t 1 = v t 1 2 = u 1 v 1 t 1 2, so 9 1,

10 either u 1 is an initial segment of u 2 or u 2 is an initial segment of u 1, by Lemma 4.1(2). Since u is cyclically reduced, either u 1 = 1 or u 2 = 1. If u 1 = 1 then t = u 1 t 1, and if u 2 = 1 then t = u (v 1 t 1 2 ). This completes the proof. Lemma 4.5. (1) Let t be a big element and let u, v F. Then the pairs (u, t), (t, v) have small cancellation if and only if (u, t) and (ut, v) do, if and only if (u, t) and (u, tv) do. In this case, tv = t v if and only if utv = (ut) v, and ut = u t if and only if utv = u (tv). (2) If v 1,..., v n are big elements (n 2) and the pairs (v i, v i+1 ) have small cancellation for 1 i n 1, then (v 1... v n 1, v n ) has small cancellation and v 1... v n is a big element. Proof. (1) Assume (u, t) and (t, v) have small cancellation. We can write t = t 1 z, v = z 1 v 1 and tv = t 1 v 1, where z F, so t 1 is a big element. We can also write u = u 1 w 1, t = w t 2 and ut = u 1 t 2, where w F, so t 2 is big. By Lemma 4.3(2), we can write t 1 = w t 3 for some t 3, so t = w t 3 z, t 2 = t 3 z, and t 3 is big. Further, ut = (u 1 t 3 ) z, v = z 1 v 1 and (ut)v = u 1 t 3 v 1. Also, tv = t 1 v 1 = w t 3 v 1, and it follows from Lemma 4.1(3) that (ut)v = (u 1 t 3 ) v 1, showing that (ut, v) has small cancellation. Conversely, assume (u, t) and (ut, v) have small cancellation. Write u = u 1 w 1, t = w t 2, ut = u 1 t 2 as above, and ut = u 2 z 1, v = z 1 1 v 2, where utv = u 2 v 2, and z 1 F. Now z 1, t 2 are terminal segments of ut and t 2 is big, so by Lemma 4.3(2), z 1 is a terminal segment of t 2, not equal to t 2, hence we can write t = w t 3 z 1 with t 3 1. Thus utv = u 1 t 3 v 2, so t 3 v 2 = t 3 v 2 and tv = w t 3 v 2 by Lemma 4.1(3). Since v = z 1 1 v 2, this shows (t, v) has small cancellation. Also, it follows that z 1 = z and tv = t v if and only if z = 1, if and only if utv = (ut) v. The rest of (1) follows by applying this with v 1, t 1 and u 1 in place of u, t and v. (2) This follows easily from (1) using induction on n, using Lemma 4.3(1) to see that v 1... v n is big. The next result is a consequence of Proposition 4 in [12], concerning the stabilizing condition discussed there. Lemma 4.6. Let V be a subset of F such that: (i) for all v V, v is cyclically reduced and a root element; (ii) For all v, w V with v w, v is not conjugate to w ±1. Let α, β Z \ Z, with α, β > 0. Let g F, let v, w V ±1 {1} and assume that, if v = w ±1 then [g, v] 1, Then there exists a natural number n such that v α gw β = v α n v n gw n w β n. Further, (v α g, w β ) and (v α, gw β ) have small cancellation. Proof. The first part follows easily from [12; Prop. 4] by taking coordinates, provided v, w V {1}, since α, β > k for all k Z. The first part then follows 10

11 in general by replacing v by v 1, w by w 1 or both in V, as appropriate, unless v = w ±1. But the argument in [12] (part (b)) actually takes care of this case as well, again on taking coordinates. Supppose (v α g, w β ) has big cancellation. Using Lemma 4.2(2), write v α = v z, g = z 1 g 1 with v α g = v g 1, so g 1, z F since g F. Also, write v α g = v 1 u, w β = u 1 w 1 with v α gw β = v 1 w 1, so u is big. Now v α g = v g 1 = v 1 u, so u = u 1 g 1, where u 1 is a terminal segment of v, by Lemma 4.3(2) (applied to (v α g) 1 ). Hence u 1 z is a big terminal segment of v α, so has the form t v γ, where t is a terminal segment of v and γ Z \ Z (see remarks preceding Lemma 4.3). Therefore, L(u 1 ) = L(v γ ) + a, where a Z. Consequently, L(v α gw β ) = L(v α g) + L(w β ) 2 L(u) = L(v α ) + L(w β ) 2 L(u 1 ) + b, where b Z. (From above, L(v α g) = L(v ) + L(g 1 ) and L(v α ) = L(v ) + L(z).) Now from the first part, L(v α gw β ) = L(v α n )+ L(w β n )+L(v n gw n ) = L(v α )+ L(w β )+c, where c Z, using the formula preceding Lemma 4.3. We conclude that L(v γ ) Z, and again by the formula before Lemma 4.3, v = 1, so v = 1, hence u 1 = 1 and u is small, a contradiction. Similarly, (v α, gw β ) has small cancellation, completing the proof. Finally, the elements of F can be classified as follows. Lemma 4.7. Let t F \ F. Then one of the following occurs. (i) The equivalence class [t] contains no element which is left or right divisible by an element of F \ {1}. (ii) The equivalence class [t] contains an element of the form f α t 1, where t 1 is a big element, f F \ {1}, f is a root element and cyclically reduced, f α is a big f-power, and the equivalence class of t 1 contains no element of the form t 2 g β, where g F and g β is a big g-power. Further, t 1 cannot be written as f ±1 t. (iii) t 1 falls under Case (ii). (iv) The equivalence class [t] contains an element of the form f α t 1 g β, where f, g F \ {1}, f, g are root elements and cyclically reduced, f α, g β are big powers and g is not conjugate to f ±1. Further, t 1 cannot be written as f ±1 t or t g ±1. (v) The equivalence class [t] contains an element of the form f α t 1 f β, where f F \{1}, f is a root element and cyclically reduced, f α, f β are big powers, t 1 F \ {1} and t 1 cannot be written as f ±1 t or t f ±1. (vi) The equivalence class [t] contains an element of the form f α, where f F is cyclically reduced and a root element, and f α is a big power. Proof. Suppose (i) does not occur. Then [t] contains an element of the form f α t 1 or t 1 g β, wher f, g F \{1} and α, β Z\Z. In the first case, we can assume f is a root element, otherwise f = h n for some n Z, where h is the root of f, and we can replace f α by h nα. We can also assume f is cyclically reduced, otherwise f = u 1 v u, where u 1 and v is cyclically reduced, and t is F -equivalent to ut = v α (u t 1 ). Taking α as large as possible (applying the remarks preceding 11

12 Lemma 4.4 to f or f 1, depending on the sign of α), we can assume that t 1 does not have f as an initial segment, so cannot be written as f ±1 t (ft 1 = f t 1 by Lemma 4.3(3)). If [t 1 ] contains no element of the form t 2 g β, where g F \ {1}, β Z\Z, then we are in Case (ii) if t 1 is big and Case (vi) if t 1 is small, otherwise we can take an element of this form, and as before assume g is cyclically reduced and a root element, and t 2 cannot be written as t 3 g ±1. Then t is F -equivalent to f α t 2 g β if t 2 1 (Lemma 4.1(3)), or to f α g β. Similarly, if t is F -equivalent to t 1 g β, we either arrive at Case (iii), Case (vi) or one of these two cases. Suppose t is F -equivalent to f α t 2 g β with t 2 1. If g is not conjugate to f ±1, we are in Case (iv). If it is, then since f, g are cyclically reduced, we can write g = (w f ±1 )w 1 for some w F, so t is F -equivalent to (f α t 2 ) (w f ±β ), by Lemma 4.3(3). This is of the form f α t f β, where β Z \ Z. Again we can maximise α and β so that t cannot be written as t f ±1 or f ±1 t ; if t = 1 we are in Case (vi), otherwise we are in Case (v). Suppose t is F -equivalent to f α g β. If g is not conjugate to f ±1, replacing f by f 1, g by g 1 as necessary, we can assume α, β > 0. By Lemma 4.6, we can then write f α g β = f α w g β for some w F and α, β Z \ Z. Again maximising α, β, we can assume w cannot be written as f ±1 w or w g ±1. This puts us in Case (iv). Suppose g = uf ±1 u 1 ; then t is F -equivalent to f α uf ±β. If u f, we are in Case (vi), otherwise by Lemma 4.6 we can write f α uf ±β = f α w f β for some w F and α, β Z \ Z. Again we can maximise α, β, putting us in Case (v). 5. Formulation of the Main Results. Let F be a free group and S any system of equations with one variable over F (i.e. S F x ), such that V F (S). By the remark preceding Theorem 2.2, there exists a finite subsystem S 0 S such that V F (S) = V F (S 0 ). Therefore it is enough to consider only finite systems of equations over the free group F. The full description of all algebraic sets from F 1 = F is given by the following theorems. Theorem 5.1. Any coordinate group F Y of an irreducible algebraic set Y F 1 satisfies one of the following. (a) F Y = F ; (b) F Y = F, t [u, t] = 1, where u is a root element of F ; (c) F Y = F x. By Theorem 3.1, Theorem 5.1 is a consequence of the following. Theorem 5.2. Let F be a non-abelian free group, D a non-principal ultrafilter over a countable set I, and Let F be the ultraproduct F I /D. Then any F - subgroup P of F with one F -generator satisfies one of the following: (a) P = F ; (b) P = F, t [u, t] = 1, where u is a root element of F and t F \ F ; 12

13 (c) P = F x, where x F \ F. Of, course, in Theorems 5.1 and 5.2, the isomorphisms are F -isomorphisms. Finally, we have the following. Theorem 5.3. Let F be a non-abelian free group and let V = V 1... V k be the decomposition of the algebraic set V from F 1 into its irreducible components. If V F 1, then every V i has one of the following forms: (a) V i is a point; (b) there exist elements f, g, h F such that V i = fc F (g)h. The last theorem follows directly from Theorem 5.1 together with Theorem 2.1 and the examples following it. Therefore it is enough to prove only Theorem Proof of the Main Theorem. Let F be a free group and let t be a big element of F. We shall prove Theorem 5.2 by a series of lemmas on the structure of F, t, corresponding to the possibilities for the element t given in Lemma 4.7. Lemma 6.1. Suppose t falls under Case (i) of Lemma 4.7. Then F, t is a free group, equal to F t for some t F -equivalent to t. Proof. Let h = u 1 t ε 1 u 2 t ε 2... u k t ε k u k+1, where ε i = ±1, u i F and if u i = 1 then ε i ε i+1. We need to show that h 1. Suppose first that no element of [t] has the form f 1 t 1 f where f is a big element. For 1 i n 1, the pair (t ε i, u i+1 t ε i+1 ) has small cancellation by Lemma 4.4, and the pair (u i, t ε i ) has small cancellation by Lemma 4.3(1). Therefore by Lemma 4.5, the pair (u i t ε i, u i+1 t ε i+1 ) has small cancellation for 1 i n 1 and (u 1 t ε 1 )(u 2 t ε 2 )... (u k t ε k ) is a big element, hence so is h, in particular, h 1. Suppose some element t of [t] does have the form f 1 t 1 f where f is a big element. Since F, t = F, t, we can assume t = f 1 t 1 f. We can also assume t 1 is cyclically reduced. (If t 1 = u 1 t 2 u, replace t 1 by t 2 and f by u f). We claim that h = g t ε k 1 fu k+1 for some g F. The proof is by induction on k, and the case k = 1 follows from Lemma 4.5(1) (since (f, u) has small cancellation for u F, by Lemma 4.3(1)). Assume that u 1 t ε 1 u 2 t ε 2... u k 1 t ε k 1 u k = g t ε k 1 1 fu k. Then h = ((g t ε k 1 1 ) fu k )(f 1 (t ε k 1 fu k+1 )), using Lemma 4.5(1). Suppose u k 1. If the pair (fu k, f 1 ) has big cancellation, then by Lemma 4.4, f = f 1 a α v for some a F, big a-power a α, f 1 F and v F. But then t is F -equivalent to (f 1 t ε k 1 f 1 ) a α, contrary to assumption on t. Thus (fu k, f 1 ) has small cancellation. By Lemma 4.5(1) and Lemma 4.1(3), we have h = g t ε k 1 1 fu k f 1 t ε k 1 fu k+1, which is of the required form with g = g t ε k 1 1 fu k f 1. Finally suppose u k = 1. Then by Lemma 4.5(1), h = (g t ε k 1 1 )(t ε k 1 fu k+1 ) and ε k 1, ε k have the same sign, so t ε k 1 1 t ε k 1 = t ε k 1 1 t ε k 1 since t 1 is cyclically reduced. By Lemma 4.1(3), h = (g t ε k 1 1 ) t ε k 1 fu k+1, which is of the required form. 13

14 Lemma 6.2. Suppose t falls under Case (ii) or (iii) of Lemma 4.7. Then F, t is a free group, equal to F t for some t F -equivalent to t. Proof. Since F, t = F, t 1 and F, t = F, t for any t F -equivalent to t, we can assume t is in Case (ii), and that t = f α t 1, where the conditions of Case (ii) are satisfied. We begin by remarking that if u F, the pair (t 1 u, f α ) has small cancellation. For write t 1 u = t 2 w 1, f α = w v and tuf α = t 2 v as in Lemma 4.2. Being an initial segment of f α, w = f β g, where β Z and g is an initial segment of f. Then t 1 u = (t 2 g 1 ) f β, and [t 1 ] = [t 1 u], so by the asssumptions on t 1, β Z, hence w F as claimed. A similar but more elaborate argument shows that (t 1, uf α ) has small cancellation, otherwise t 1 u has a terminal segment of the form f β g, where β Z \ Z and g is an initial segment of f, contradicting the conditions on t 1. Suppose t 1 = g t 2, where g is an initial segment of f, with L(g) as large as possible (so L(g) < L(f) by the conditions on t 1 in this case). We can replace t by g 1 t = g 1 (f α g t 2 ) = g 1 f α g t 2 = (g 1 fg) α t 2 (using Lemma 4.5) (g 1 fg is a cyclic conjugate of f in F, and t 2 is F -equivalent to t 1 ). By the choice of g, g 1 fg and t 2 have no common initial segment except 1. Thus we can assume f 1 t 1 = f 1 t 1, so in fact f β t 1 = f β t 1 for all β Z, by Lemma 4.3(3). Let h = u 1 t ε 1 u 2 t ε 2... u k t ε k u k+1, where ε i = ±1, u i F and if u i = 1 then ε i ε i+1. We need to show that h 1, and on conjugating by u k+1, we may assume u k+1 = 1. We shall do this by establishing an analogue of a well-known situation which occurs in the Nielsen method for free groups. In terms of the length function, this means the following. Let u, v, w F, Let ε i = ±1 for 1 i 3, and suppose v 1 if ε 1 + ε 2 = 0, and w 1 if ε 2 + ε 3 = 0. Then L(ut ε 1 vt ε 2 wt ε 3 ) > L(ut ε 1 ) L(vt ε 2 ) + L(wt ε 3 ). ( ) It then follows by Lemma 6.1 and the remarks following Cor. 6.2 in [10] that L(h) > 0, hence h 1. (Lyndon considered only Z-valued length functions, but the argument is valid for arbitrary Λ; alternatively, one can deduce the result for L by taking coordinates. One also needs to know that Lyndon s Axiom A5 is satisfied; see the proof of Prop. 2.6 in [10]). To prove ( ), we assume that ε 2 = 1, giving four cases to consider. The cases when ε 2 = 1 are similar and left to the reader. Case 1. ε 1 = ε 3 = 1. By repeated application of Lemmas 4.5(1) and 4.1(3), and using the observation at the beginning of the proof, we obtain utvtwt = u(f α t 1 )v(f α t 1 )w(f α t 1 ) = (uf α t 1 vf α t 1 )(wf α t 1 ) = uf α t 1 vf α t 1 wf α t 1. Now if a pair (r, s) has small cancellation, then L(rs) = L(r)+ L(s) 2 c(r 1, s), which by Lemma 4.2 is congruent mod Z to L(r) + L(s). Making repeated use 14

15 of this, we find that L(utvtwt) = 3 L(f α ) + 3 L(t 1 ) + a, where a Z. Using Lemma 4.5(1), L(ut) L(vt) + L(wt) = L(uf α t 1 ) L(vf α t 1 ) + L(wf α t 1 ) = L(f α ) + L(t 1 ) + b where b Z, so we need to show 2 L(f α ) + 2 L(t 1 ) > b a. Since L(t 1 ) > 0, it is enough to show 2 L(f α ) > b a, but this follows since L(f α ) Z \ Z and L(f α ) > 0. Case 2. ε 1 = 1, ε 3 = 1. Here we find, using Lemmas 4.5 and 4.1 again, that utvtwt 1 = uf α (t 1 vf α ) (t 1 wt 1 1 ) f α because the pair (t 1, wt 1 1 ) has small cancellation by Lemma 4.4(2). As in Case (1), we find that L(utvtwt 1 ) = 3 L(f α ) + 3 L(t 1 ) + a, where a Z, and the inequality ( ) follows as in Case 1, using the fact that L(f α ) is in Z \ Z. Case 3. ε 1 = 1, ε 3 = 1. We need to consider ut 1 vtwt = u(t 1 1 f α )v(f α t 1 )w(f α t 1 ); unless v = f n for some n Z, the pair (f α, vf α ) has small cancellation by Lemma 4.6, and as in previous cases we can write this as ut 1 1 f α vf α t 1 wf α t 1, and inequality ( ) follows once more from the fact that L(f α ) Z \ Z. Suppose that v = f n. Then ut 1 vtwt = u(t 1 1 f n t 1 )w(f α t 1 ) = u(t 1 1 f n t 1 )(wf α t 1 ) = (ut 1 1 f n t 1 )(wf α t 1 ) = ut 1 1 f n (t 1 wf α ) t 1 by the way t 1 has been chosen, and using Lemmas 4.5(1) and 4.1(3) again. Thus L(ut 1 vtwt) = L(ut 1 1 ) + L(f n ) + L(t 1 wf α ) + L(t 1 ). Further, L(ut 1 ) L(vt) + L(wt) = L(ut 1 1 ) + L(f α ) L(f n+α ) L(t 1 ) + L(wf α ) + L(t 1 ), so we have to show that L(f n )+ L(t 1 wf α ) > L(f α ) L(f n+α ) L(t 1 )+ L(wf α )+ L(t 1 )+ L(wf α ) and in view of the triangle inequality, it suffices to show that L(t 1 wf α ) > L(wf α ) L(t 1 ). Now the pair (t 1, wf α ) has small cancellation, so we can write t 1 = t 2 z, wf α = z 1 w 1, t 1 wf α = t 2 w 1, where z F. Consequently, z t 1 since t 1 is big, 15

16 so L(z) < L(t 1 ). Further, from these expressions we calculate that L(t 1 wf α ) = L(wf α ) + L(t 1 ) 2L(z) from which the required inequality follows. Case 4. ε 1 = 1, ε 3 = 1. This is similar to Case 3 and details will be left to the reader. If v is not a power of f, then ut 1 vtwt 1 = ut 1 1 f α vf α t 1 wt 1 1 f α, and inequality ( ) follows, as in previous cases, because L(f α ) Z. Otherwise, if v = f n, ut 1 vtwt 1 = ut 1 1 f n t 1 wt 1 1 f α, and by an argument similar to Case 3, it suffices to prove that L(wt 1 1 ) L(t 1 ) < L(t 1 wt 1 1 ), and this follows because the pair (t 1, wt 1 1 ) has small cancellation, by an argument similar to that in Case 3. This completes the proof. Lemma 6.3. Suppose t falls under Case (iv) of Lemma 4.7. Then F, t is a free group, equal to F t for some t F -equivalent to t. Proof. We can assume t = f α t 1 g β, where the conditions of Case (iv) are satisfied. Arguing as in Lemma 6.2, we can assume that f ±1 t 1 = f ±1 t 1. Similarly, we can assume t 1 g ±1 = t 1 g ±1. As in previous cases, we need to show that, if h = u 1 t ε 1 u 2 t ε 2... u k t ε k u k+1, where ε i = ±1, u i F and if u i = 1 then ε i ε i+1, then h 1. Assume first that t 1 1. Put h r = u 1 t ε 1 u 2 t ε 2... u r t ε r. We claim that, if ε r = 1, then h r has the form h g β, where h g ±1 = h g ±1, and if ε r = 1 then h r has the form h f α, where h f ±1 = h f ±1. This is proved by induction on r. Assume it is true for r; to prove it for r + 1, there are four cases. Case 1. ε r = 1 and ε r+1 = 1. Then h r+1 = (h g β )u r+1 (f α t 1 g β ) and (g β u r+1, f α ) has small cancellation by Lemma 4.6. Hence h r+1 = h g β u r+1 f α t 1 g β by Lemmas 4.5(1) and 4.1(3), that is, h r+1 = h g β, where h has the form a t 1 for some a F. This is of the required form, since h g ±1 = h g ±1 by the conditions we have imposed on t 1 and Lemma 4.1(3). Case 2. ε r = 1 and ε r+1 = 1. Then h r+1 = (h g β )u r+1 (g β t 1 1 f α ), u r+1 1 and (g β u r+1, g β ) has small cancellation by Lemma 4.6, unless u r+1 = g n for some n Z, n 1. Again using Lemmas 4.5(1) and 4.1(3), in this case h r+1 = h g β u r+1 g β t 1 1 f α, which is again of the required form by the conditions on t 1 and Lemma 4.1(3). If u r+1 = g n for some n Z, then h r+1 = h (g n )(t 1 1 f α ) = h (g n t 1 1 f α ) = h g n t 1 1 f α by the conditions on t 1, the induction hypothesis on h and Lemma 4.3(3), making repeated use of Lemma 4.1(3). This is again of the required form. The cases when ε r = 1 are similar and left to the reader (in effect, one replaces t by t 1 ). The case r = 1 is established by similar arguments and is also left to the reader. Thus h = h k u k+1, where h k = h z, and z is a big element. By Lemma 4.3(1) h k is big, hence h is big, in particular, h 1. 16

17 Now assume t 1 = 1. We claim that h r, as defined earlier, can be written in the form h r = v 1 s 1... v m s m, where m = m(r) 1, each s i is either f ±α or g ±β, all v i F and the following conditions are satisfied. (1) For i > 1, if s i 1 = f ±α and s i = f α, then v i f ; also, if s i 1 = g ±β and s i = g β, then v i g. (2) If ε r = 1 then s m = g β, and if ε r = 1 then s m = f α. (3) For i > 1, except in the cases when s i 1 = f α or g β and s i = f α or g β, v i is an alternating product of non-trivial small powers of f and g, beginning with a power of f if s i 1 = g ±β and a power of g if s i 1 = f ±α, and likewise ending in a power of f if s i = g ±β and a power of g if s i = f ±α. In the two cases s i 1 = f α and s i = g β, s i 1 = g β and s i = f α, we allow v i = 1, but not otherwise. (Note that f, g do not commute, so they freely generate a subgroup of F, hence v i 1 except in the specified cases. Also, in some cases v i may be a single power of f or g.) Again the proof is by induction on r. It is clearly true when r = 1. Assume it is true for r and consider h r+1 = (v 1 s 1... v m s m )u r+1 t ε r+1. Put u = u r+1 to simplify notation. Case 1. ε r+1 = 1. Then h r+1 = (v 1 s 1... v m s m )uf α g β. If ε r = 1, then s m = g β and we put v m+1 = u, s m+1 = f α, v m+2 = 1, s m+2 = g β, to express h r+1 as v 1 s 1... v m+2 s m+2, which is of the desired form. If ε r = 1, then s m = f α and we can proceed in a similar manner unless u = f n for some n Z, in which case n 0 by the conditions on h, and h r+1 = v 1 s 1... v m f n g β. Further, either v m is an alternating product of powers of f and g ending in a power of g, or v m = 1. Then h r+1 = v 1 s 1... v m 1 s m 1 v ms m, where v m = v m f n, s m = g β, is an expression of h r+1 in the required form. Case 2. ε r+1 = 1. This is similar to Case (1), in effect replacing t by t 1, and is left to the reader. Now (s i 1, v i s i ) has small cancellation by Lemma 4.6 and Condition (1), hence (v i 1 s i 1, v i s i ) has small cancellation by Lemma 4.5(1), so h r is a big element for r 1, by Lemma 4.5(2). Thus h = h k u k+1 is big, in particular h 1. This completes the proof. Before proceeding to the next case, a remark is needed. Remark. If x, y F, x is big and a is a positive integer, then (x, y) has small cancellation if and only if (x a, y) does, and xy = x y if and only if x a y = x a y. Also, (y, x) has small cancellation if and only if (y, x a ) does, and yx = y x if and only if yx a = y x a. Proof. Write x = u v u 1, where v is cyclically reduced. Then x a = u v a u 1 = (u v a 1 ) (v u 1 ) (using Lemma 4.1(3); note that v a = } v v {{... v } ). Now a times v u 1 is big, otherwise u and v would be small (Lemma 4.3(1)), so x would be 17

18 small. By Lemma 4.5(1), (x, y) has small cancellation if and only if (v u 1, y) does, if and only if (x a, y) does, and xy = x y if and only if (v u 1 )y = (v u 1 ) y, if and only if x a y = x a y. The last part of the remark is proved similarly. Lemma 6.4. Suppose t falls under Case (v) of Lemma 4.7. Then F, t is a free group, equal to F t for some t F -equivalent to t. Proof. We can assume t = f α t 1 f β where the conditions of Case (v) are satisfied. Again we need to show that, if h = u 1 t ε 1 u 2 t ε 2... u k t ε k u k+1, where ε i = ±1, u i F and if u i = 1 then ε i ε i+1, then h 1. Put h r = u 1 t ε 1 u 2 t ε 2... u r t ε r. There are two cases. Assume first that α and β have the same sign. We claim that h r can be written in the form h r = v 1 s 1... v m s m, where m = m(r) 1, v i F and (1) s i = x i w i y i, where x i = f α or f β, y i = f α or f β and w i = t η 1 1 f n 1 t η f n p 1 t η p 1 where p = p(i) 1, η j = η j (i) = ±1, n j = n j (i) Z, n j 0 and η 1 = 1 if x i = f α, η 1 = 1 if x i = f β η r = 1 if y i = f β, η r = 1 if y i = f α. Further, η j 1 η j = 1 for 2 j p. (2) For i > 1, if y i 1 = f α and x i = f α, then v i f ; also, if y i 1 = f β and x i = f β, then v i f. (3) If ε r = 1 then y m = f β, and if ε r = 1 then y m = f α. The proof is by induction on r. It is clearly true when r = 1. Assume it is true for r and consider h r+1 = (v 1 s 1... v m s m )u r+1 t ε r+1. Put u = u r+1. Case 1. ε r+1 = 1. Then h r+1 = (v 1 s 1... v m s m )uf α t 1 f β. If ε r = 1, then y m = f β and we put v m+1 = u, x m+1 = f α, w m+1 = t 1, y m+1 = f β, to express h r+1 as v 1 s 1... v m+1 s m+1, which is of the desired form. If ε r = 1, then y m = f α and we can proceed in a similar manner unless u = f n for some n Z, in which case n 0 by the conditions on h, and h r+1 = v 1 s 1... x m w m f n t 1 f β. Put w m = w m f n t 1, y m = f β and s m = x m w my m. Then h r+1 == v 1 s 1... v m 1 s m 1 v m s m is an expression of h r+1 in the required form. Case 2. ε r+1 = 1. This is similar to Case (1) and is left to the reader. We claim that (x i w i, y i ) and (x i, w i ) have small cancellation for all relevant i, so s i is big. Assuming this, (v i x i w i, y i ) has small cancellation by Lemma 4.5(1). By Condition (2) and Lemma 4.6, (y i 1, v i x i ) has small cancellation for i > 1. Also by repeated use of Lemma 4.5(1), (v i x i, w i ) has small cancellation and so does (y i 1, v i x i w i ), and therefore so does (y i 1, v i x i w i y i ), i.e. (y i 1, v i s i ) has 18

19 small cancellation. By further application of Lemma 4.5(1), (s i 1, v i s i ) has small cancellation and so does (v i 1 s i 1, v i s i ). By Lemma 4.5(2), h r is a big element, so h = h k u k+1 is big, hence h 1 as required. To establish the claim, it suffices by Lemma 4.6 to show that w i f. Suppose first that t 1 is big. By the conditions imposed on t 1 in this case and Lemma 4.4, the pairs (t 1 f n, t 1 1 ) and (t 1 1 f n, t 1 ) have small cancellation, for n Z, n 0, hence so do (t 1 f n, t 1 1 f m ) and (t 1 1 f n, t 1 f m ) for m Z by Lemma 4.5(1). Now put z j = t η j 1 f n j for 1 j p (where n p is defined to be 0). Then w i = z 1... z p is a big element by Lemma 4.5(2), so certainly is not in f. Suppose t 1 F. By the conditions on t 1, [t 1, f] 1, so t 1, f freely generate a subgroup of F, hence again w i f. Next, assume α, β have opposite signs. We claim that h r can be written in the form h r = v 1 s 1... v m s m, where m = m(r) 1, v i F and (1) s i = x i w i y i, where x i = f α and y i = f β, or x i = f β and y i = f α, and w i = t 1 f n1+α+β t 1... f np 1+α+β t 1 if x i = f α, y i = f β w i = t 1 1 f n1 α β t f np 1 α β t 1 1 if x i = f β, y i = f α where p = p(i) 1 and n j = n j (i) Z. (2) For i > 1, if y i 1 = f α and x i = f α, then v i f ; also, if y i 1 = f β and x i = f β, then v i f. (3) If ε r = 1 then y m = f β, and if ε r = 1 then y m = f α. The proof is by induction on r and is left to the reader, being similar to the previous case. As in the previous case, it suffices to show that (x i, w i ) and (x i w i, y i ) have small cancellation, and by Lemma 4.6, it is enough to show that w i f. Replacing w i by w 1 i if necessary, we can assume for this purpose that w i = t 1 f n1+α+β t 1... f np 1+α+β t 1. Assume first that α + β Z \ Z. Suppose γ, δ Z \ Z and have the same sign. Then the pair (f γ, t 1 ) has small cancellation. (Otherwise, a big initial segment of f γ has f as an initial segment by Lemma 4.3(2), so t 1 would have f 1 as an initial segment, contrary to the conditions on t 1.) Similarly, (t 1, f γ ) has small cancellation. Hence (f γ, t 1 f δ ) has small cancellation, by Lemma 4.5(1) if t 1 is big, and by Lemma 4.6 if t 1 is small. Further, (t 1 f γ, t 1 f δ ) and (t 1 f γ, t 1 ) have small cancellation by Lemma 4.5(1). Now put z j = t 1 f n j+α+β for 1 j p 1, and z p = t 1. Then z j is big for j < p, and z 1... z p 1 is big by Lemma 4.5(2). Hence w i = z 1... z p is big; this is obvious if t 1 is small, and otherwise follows from Lemma 4.5(2). Thus w i f. Now suppose α + β Z. We need to prove that a product t 1 f a 1... f a p 1 t 1 f, where all a j Z and p 1. Suppose t 1 is big. This product can be rewritten as t b 1 1 f c 1... f c q 1 t b q 1, where the b j are positive integers, the c j are non-zero integers and q 1. Replacing f by f 1 if necessary, we can assume α > 0, β < 0. We claim that if a > 0, b 0 are integers, then (t a 1f b, t a 1) has small cancellation. For otherwise, by Lemma 4.1 we can write t a 1 = u 1 2 t 2 u 1 1 and f b = u 1 u 2. Since t = f α t 1 f β, we have ft 1 = f t 1 and t 1 f 1 = t 1 f 1 by Lemma 4.3(3). 19

20 Suppose b > 0; then f b t 1 = f b t 1 by Lemma 4.3(3), hence f b t a 1 = f b t a 1 by the remark preceding the lemma, so u 2 = 1 and t a 1 = t 2 f b. Write t 1 = u v u 1 where v is cyclically reduced. Then t a 1 = (u v a 1 ) (v u 1 ) and v u 1 is big (see the proof of the remark preceding the lemma). Therefore, f b is a terminal segment of v u 1 (Lemma 4.3(2) applied to t a 1 ), so of t 1, hence f 1 is a terminal segment of t 1 (since f is cyclically reduced) contradicting the conditions on t 1. We similarly obtain a contradiction if b < 0, when t 1 f b = t 1 f b, and the claim is established. By the remark preceding the lemma, (t a 1f b, t c 1) has small cancellation for all integers c > 0, and for any d Z, t a 1f b, t c 1f d are big, and (t a 1f b, t c 1f d ) has small cancellation by Lemma 4.5(1). Now applying Lemma 4.5(2) to w i = (t b 1 1 f c 1 )... (t b q 1 1 f c q 1 )(t b q 1 ), w i is big, so w i f. Finally, if t 1 F, we have already observed that t 1 and f freely generate a subgroup of F, so w i f. Lemma 6.5. Suppose t falls under Case (vi) of Lemma 4.7, so t is F -equivalent to an element t = f α, where f is a cyclically reduced root element of F and α Z \ Z. Then F, t is the HNN-extension F, t [f, t ] = 1. Proof. For simplicity we can assume t = t. Clearly F, t is a homomorphic image of the HNN-extension, and to show the obvious epimomorphism is an isomorphism, it suffices by the normal form theorem for HNN-extensions to show that, if h = u 1 t ε 1 u 2 t ε 2... u k t ε k u k+1, where ε i = ±1, u i F and if u i f then ε i ε i+1, then h 1. We claim that (t ε i 1, u i t ε i ) has small cancellation for i > 1. This follows from Lemma 4.6 unless u i = f n for some integer n, when by assumption ε i 1 = ε i. Since f is cyclically reduced, f 2 = f f and f 1 = f 1 f 1, so by Lemma 4.3(3), f α f α+n = f α f α+n and f α f α+n = f α f α+n, since adding n to ±α does not change the sign. This establishes the claim. By Lemma 4.5(1), (u i 1 t ε i 1, u i t ε i ) has small cancellation for 1 < i k. By Lemma 4.5(2), g = (u 1 t ε 1 )(u 2 t ε 2 )... (u k t ε k ) is a big element, hence h = gu k+1 is big and h 1. References [1] R.C. Alperin and H. Bass, Length functions of group actions on Λ- trees. In: Combinatorial group theory and topology (ed. S.M.Gersten and J.R.Stallings), Annals of Mathematics Studies 111, pp Princeton: University Press [2] K.I. Appel, One-variable equations in free groups, Proc. Amer. Math. Soc. 19 (1968), [3] G. Baumslag, Reviews on infinite groups as printed in Mathematical Reviews 1940 through 1970, volumes 1-40 inclusive, Part 1. Providence, R.I.: American Mathematical Society [4] G. Baumslag, A.G. Myasnikov and V.N. Remeslennikov, Algebraic geometry over groups, preprint [5] I.M. Chiswell, Embedding theorems for groups with an integer-valued length function, Math. Proc. Cambridge Philos. Soc.85 (1979),