On Certain Elements of Free Groups

204, 394]405 Ž1998. JA977354

JOURNAL OF ALGEBRA ARTICLE NO.

On Certain Elements of Free Groups S. V. Ivanov* Department of Mathematics, Uni¨ ersity of Illinois at Urbana-Champaign, 1409 West Green Street, Urbana, Illinois 61801 E-mail: [email protected] Communicated by Alexander Lubotzky Received May 20, 1996

Let Fm be a free group of finite rank m. It is proven that for every n G 2 there is a non-trivial word wnŽ x 1 , . . . , x n . such that if values wnŽUn ., wnŽ Vn . of wnŽ x 1 , . . . , x n . on two n-tuples Un and Vn of elements of Fm are conjugate and non-trivial then these n-tuples themselves are conjugate. As a corollary, one has the existence of two elements in Fm whose images uniquely determine any monomorphism c : Fm ª Fm . Q 1998 Academic Press

INTRODUCTION It is well known due to Dehn, Magnus, and Nielsen Žsee wN, M, LS2x. that if f is an endomorphism of a free group F2 with a basis  x 1 , x 2 4 and y1 . f Žw x 1 , x 2 x. s w x 1 , x 2 x Žwhere w x 1 , x 2 x s x 1 x 2 xy1 , then f is an auto1 x2 morphism. A more general result was obtained by Zieschang wZ2x: If f : Fm ª Fm is an endomorphism of a free group Fm of rank m with basis  x 1 , . . . , x m 4 and f fixes the element w x 1 , x 2 xw x 3 , x 4 x ??? w x my1 , x m x Ž m is even. or f fixes x 1k x 2k ??? x mk with k ) 1, then f is an automorphism. Rips wRx generalized the Dehn]Magnus]Nielsen result in another direction: If w s w x 1 , . . . , x m x is a higher commutator Žthat is, w is a commutator of weight m involving all m letters x 1 , . . . , x m with arbitrary disposition of commutator brackets., then every endomorphism of Fm fixing w is an automorphism. * Supported in part by an Alfred P. Sloan Research Fellowship, a Beckman Fellowship, and NSF Grant DMS 95-01056. 394 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.

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395

These results gave rise to the following definition due to Shpilrain wSx: DEFINITION 1. An element w g Fm is called a test word if for any endomorphism f : Fm ª Fm the equality f Ž w . s w implies that f is an automorphism. Providing a solution to Shpilrain’s problem wSx on describing test words, Turner wTx proved that w is a test word if and only if w is not contained in a proper retract of Fm . In wTx new examples of test words are also given: If each k i is a multiple of k ) 1, then w s x 1k 1 x 2k 2 ??? x mk m is a test word in Fm . On the other hand, Zieschang wZ1x proved that the stabilizer in Aut Fm of w s x 1k 1 x 2k 2 ??? x mk m , where all k i are distinct and greater than 2, is as small as possible: it is generated by the inner automorphism tw defined by means of w. Putting together these two results, we see that if all k i are distinct and divisible by k ) 2, then the word w s x 1k 1 x 2k 2 ??? x mk m has the following interesting property: If f is an endomorphism of Fm and f Ž w . s w, then f g ²tw :, that is, f is a conjugation by a power of w. Let us consider two more definitions. DEFINITION 2. A non-trivial element w g Fm is an M-test word for Fm if for every endomorphism f and monomorphism c of Fm the equation f Ž w . s c Ž w . implies that f is also a monomorphism. DEFINITION 3. A non-trivial word w Ž x 1 , . . . , x n . is a C-test word in n letters for Fm if for any two n-tuples Ž A1 , . . . , A n ., Ž B1 , . . . , Bn . of elements of Fm the equality w Ž A1 , . . . , A n . s w Ž B1 , . . . , Bn . / 1 implies the existence of an element S g Fm such that Bi s SA i Sy1 for all i s 1, 2, . . . , n. Let us make several observations related to Definitions 2]3. First, any C-test word in n letters is both test and M-test one for Fn . In this sense Definition 3 is the most restrictive one. The commutator w x 1 , x 2 x is an M-test word Žfor F2 . but not a C-test word in 2 letters. If a C-test word w in n letters is not a proper power, then the stabilizer of w in Aut Fn is ²tw :. The word w s w x 1 , x 2 xw x 3 , x 4 x ??? w x 2 my1 , x 2 m x with m ) 1 is neither a C- nor M-test one. To see the latter, consider an even m and put A1 s x 1 , . . . , A m s x m , A mq1 s x 2 m x m xy1 A mq2 s x 2 m x my1 xy1 2m, 2m, y1 y1 . . . , A 2 my1 s x 2 m x 2 x 2 m , A 2 m s x 2 m x 1 x 2 m . Then rank² A1 , . . . , A 2 m : s 2 m but

w A1 , A 2 xw A 3 , A 4 x ??? w A 2 my1 , A 2 m x s w x 1 , x 2 x ??? w x my1 , x m x , x 2 m . Therefore, setting f Ž x 1 . s w x 1 , x 2 x ??? w x my1 , x m x, f Ž x 2 . s x 2 m , f Ž x 3 . s ??? s f Ž x 2 m . s 1, we see that w fails to be an M-test word. Clearly, a similar argument can be used to consider the case of odd m ) 1.

396

S. V. IVANOV

It is not clear whether or not there are M-test words in Fm , m G 3, among Zieschang’s words x 1k 1 ??? x mk m and this might be a rather interesting and difficult problem. If m s 2, then any word x 1k 1 x 2k 2 , where k 1 , k 2 are multiples of k ) 1, is an M-test word. This follows from a result of Lyndon wLS1x that elements of a solution to the equation and Schutzenberger ¨ x l 1 y l 2 z l 3 s 1 with l i ) 1 in Fm lie in a cyclic subgroup. On the other hand, it is easy to see from consideration of values of a C-test word w Ž x 1 , . . . , x n . on a cyclic subgroup of Fm that if n ) 1 then w g FnX , where FnX is the commutator subgroup of the free group Fn with basis  x 1 , . . . , x n4 . This implies that there are no C-test words among Zieschang’s words x 1k 1 ??? x mk m . Rips’s words w x 1 , . . . , x m x with m ) 2 are neither M- nor C-test ones. To see this put A1 s x 3 x 1 xy1 A 2 s x 3 x 2 xy1 A 3 s w x 1 , x 2 x. Then 3 , 3 , rank² A1 , A 2 , A 3 : s 3 and

w A1 , A 2 x , A 3 s x 3 w x 1 , x 3 x xy1 3 , w x1 , x 2 x s

x 3 , w x1 , x 2 x , w x1 , x 2 x .

Hence, upon setting f Ž x 1 . s x 3 , f Ž x 2 . s f Ž x 3 . s w x 1 , x 2 x we see that ww x 1 , x 2 x, x 3 x is not an M-test word and now our claim becomes obvious. Thus Definitions 2]3 look rather restrictive and the very existence of M-test words for Fm , m ) 2, and, especially, C-test words in n letters for Fm with m, n ) 1 is unclear. The main result of this note is the following. THEOREM. For arbitrary n G 2 there exists a non-tri¨ ial word wnŽ x 1 , . . . , x n . which is a C-test word in n letters for any free group Fm of rank m G 2. In addition, wnŽ x 1 , . . . , x n . is not a proper power. COROLLARY. There is an element u g Fm such that if f is an endomorphism, c is a monomorphism of Fm , and f Ž u. s c Ž u., then f is also a monomorphism and, more specifically, f s tukc , where tu is the inner automorphism of Fm defined by means of u and k is an integer. COROLLARY 2. There are two elements u1 , u 2 g Fm such that any monomorphism c of Fm is uniquely determined by c Ž u1 ., c Ž u 2 .. Corollaries 1]2 immediately follow from the Theorem: It suffices to put u s wmŽ x 1 , . . . , x m . in Corollary 1 and u1 s wm Ž x 1 , . . . , x m ., u 2 f ² wmŽ x 1 , . . . , x m .: in Corollary 2. The construction of the C-test word wnŽ x 1 , . . . , x n . goes as follows: If n s 2 then w 2 Ž x 1 , x 2 . s x 18 , x 28

100

x 1 x 18 , x 28

x 18 , x 28

500

x 2 x 18 , x 28

200

600

x 1 x 18 , x 28 x 2 x 18 , x 28

300 y1 x1 700 y1 x2

x 18 , x 28 x 18 , x 28

400 y1 x1 800 y1 x2 .

Ž 1.

ELEMENTS OF FREE GROUPS

397

Proceeding by induction on n, for n G 3 we put wn Ž x 1 , . . . , x n . s w 2 Ž wny1 Ž x 1 , . . . , x ny1 . , wny1 Ž x 2 , . . . , x n . . .

Ž 2.

Comerford wCx gave an algorithm to recognize test words. It would be interesting to find an algorithm Žif any. to recognize C-test words and show Žif true. that the sets of C-test words in n letters are identical for all Fm , m G 2. These problems are of special interest because of their connections with Tarski’s problems on solvability of the elementary theory ThŽ Fm . of a non-abelian free group Fm and on whether or not ThŽ F2 . s ThŽ F3 .. The fact that wnŽ x 1 , . . . , x n . is a C-test word in Fm can be expressed by the following formula of ThŽ Fm . ' z ; y 1 . . . ; yn; x 1 ??? ; x n

Ž Ž wn Ž x 1 , . . . , x n . s wn Ž y1 , . . . , yn . / 1. « Ž x 1 s zy1 zy1 n ??? n x n s zyn zy1 . . . Hence if there were no algorithm to detect C-test words for Fm , then ThŽ Fm . would be unsolvable and if the sets of C-test words for F2 and F3 in n letters were different, then ThŽ F2 . / ThŽ F3 .. 1. SEVERAL LEMMAS Let Fm s ² x 1 , . . . , x m : be the free group of rank m over an alphabet X " 1 s  x 1" 1, . . . , x m" 14 . The writing X s Y means the equality of words X and Y over X " 1 in Fm . By writing X ' Y we mean the graphical Žletter-by-letter . equality of words X and Y. The length of a word X is < s 2.. A word over X " 1 is called simple if A denoted by < X < Žnote < x 1 xy1 1 is non-empty, cyclically reduced, and is not a proper power. If A is simple, then an A-periodic word V is a subword of A k with some k ) 0. The following two facts about periodic words are easy to prove and they will be repeatedly used in the sequel: ŽP1. < A < q < B <, with both ŽP2.

If U is both an A-periodic and B-periodic word with < U < G then A is a cyclic permutation of B. In addition, if U begins A and B, then A ' B. Suppose U, V are A-periodic words so that A k 1 ' S1UT1 ,

A k 2 ' S2 VT2

and T with < T < G < A < is a subword of both U and V such that U ' U1TU2 ,

V ' V1TV2 .

Then < S1U1 < y < S2 V1 < is a multiple of < A <.

398

S. V. IVANOV

LEMMA 1. Suppose X 1 , X 2 g Fm , the subgroup ² X 1 , X 2 : of Fm is not cyclic, and the commutator w X 18 , X 28 x is conjugate to B l , where B is simple and l ) 0. Then there is a word S such that SX 1 Sy1 s Y1 , SX 2 Sy1 s Y2 , maxŽ< Y1 <, < Y2 <. - 6 < B l <, and w Y18 , Y28 x s B l. In addition, neither of Y1 , Y2 belongs to the subgroup ² B :. Proof. It is well known that a commutator w A, B x / 1 of two words A, B is never a proper power, whence l s 1 Žsee wLS2, Kx.. However, we will neither need nor use this fact. Conjugating if necessary, we can assume that X 1 ' A1k 1 and X 2 ' UAk2 2 Uy1 , where A1 , A 2 are simple and the word U is chosen so that U has the minimal length over all words in the double coset ² A1 :U² A 2 :. Then k k 2 y1 X 18 , X 28 ' A81 k 1 UA82 k 2 Uy1Ay8 UAy8 U . 1 2

Ž 3.

First suppose that < U < ) 12 Ž< A1 < q < A 2 <.. Denote the eight distinguished subwords in the right part of Ž3. by P1 , Q1 , . . . , P4 , Q4 , respectively. Consider a subword of the form Pi Q i Piq1 Žindexes modulo 4. of the cyclic word P1 Q1 ??? P4 Q4 . It follows from the choice of U Ž' Q i" 1 . and inequalities 8 k 1 , 8 k 2 ) 2 that there can be at most 12 Ž< A1 < q < A 2 <. cancellations in Pi Q i Piq1 and so some letters of each Q i , Pi remain uncancelled in the cyclically reduced word B l. Therefore, < B l < G 2 Ž < A81 k 1 < q < A82 k 2 < q 2 < U < . y 4 Ž < A1 < q < A 2 < . ) < A81 k 1 < q < A82 k 2 < q 2 < U < .

Ž 4.

Clearly, B l is conjugate to w A81 k 1 , UA82 k 2 Uy1 x by a word S with
1

Ž < B< q 2

A81 k 1 , UA82 k 2 Uy1

3

. - 2 < Bl <

following from Ž4.. Referring again to Ž4., we have < SA81 k 1 Sy1 < , < SUA82 k 2 Uy1 Sy1 < F 2 < S < q < B l < - 4 < B l < . Now suppose < U < F 12 Ž< A1 < q < A 2 <.. Notice that < B l < ) 2 max Ž < A81 k 1 < y < A1 < , < A82 k 2 < y < A 2 < . . To easily see this it suffices to interpret the commutator w A81 k 1 , UA82 k 2 Uy1 x by a commutator diagram which looks like a van Kampen diagram consistk1 Ž k2 . ing of two 2-cells with boundary labels A81 k 1 , Ay8 or A82 k 2 , Ay8 which 1 2

ELEMENTS OF FREE GROUPS

399

are joined by a simple path with label UA82 k 2 Uy1 Žor Uy1A81 k 1 U, respectively. and then make use of ŽP2. and B l / 1 Žfor more details see wLS2, O1x.. Therefore, < B l < ) < A81 k 1 < q < A82 k 2 < y Ž < A1 < q < A 2 < . G < A81 k 1 < q < A82 k 2 < q 2 < U < y 2 Ž < A1 < q < A 2 < . G

3

Ž < A81 k < q < A82k 1

4

2

< q 2
Arguing as above, we get the existence of a desired word S with
1 2

< B < q < A81 k 1 < q < A82 k 2 < q 2 < U < - 2 < B l < .

Hence < SA81 k 1 Sy1 < , < SUA82 k 2 Uy1 Sy1 < F 2 < S < q

4 3

< B l < - 6 < B l <,

as required. To prove the additional claim of Lemma 1, note that if, say, Y1 s B k , then w B 8 k , Y2 x s B l , whence Y2 g ² B : as well as so B l s 1, contrary to l ) 0. LEMMA 2. Suppose the subgroup ² X 1 , X 2 : of Fm is not cyclic, the words S, Y1 , Y2 , B l are defined for X 1 , X 2 as in Lemma 1, and w 2 Ž x 1 , x 2 . is defined by Ž1.. Then the cyclically reduced word W that is conjugate to w 2 Ž X 1 , X 2 . has the form W ' R1T1 R 2 T2 ??? R 8 T8 ,

Ž 5.

where R i are B-periodic words Ž R i is what is left from B i?100 l after all cancellations. with

Ž i ? 100 y 14 . < B l < - < R i < F i ? 100 < B l < , and 0 F < Ti < - 6 < B l < . In particular, 3488 < B l < - < W < - 3648 < B l <. Proof. By Lemma 1, the word w 2 Ž X 1 , X 2 . is conjugate to a Žnot cyclically reduced. word w 2 Ž Y1 , Y2 . s W0 such that 800 l y1 W0 s B 100 l Y1 B 200 l Y1 B 300 l Y1y1 B 400 l Y1y1 B 500 l Y2 B 600 l Y2 B 700 l Yy1 Y2 , 2 B

Ž 6.

400

S. V. IVANOV

where Y1 , Y2 are not in ² B : and are shorter than 6 < B l <. Therefore, no more than Ž6 q 1.< B l < letters can cancel from both sides of each subword B i?100 l and Lemma is proven. LEMMA 3. The word W gi¨ en by Ž5. is not a proper power. Proof. Assuming the contrary, put that W ' C k , where C is simple, k ) 1, and W is a cyclic permutation of W that begins with R 82 , where R 8 ' R 81 R 82 , R 82 begins with B and < R 81 < - < B <. By Lemma 2, < R 82 < ) Ž 800 y 15 . < B l < s 785 < B l < . Hence, if < C < F 784 < B l <, then R 82 is both a B- and C-periodic word of length < R 82 < ) < C < q < B <, whence C ' B following from ŽP1.. But then we have Y1 , Y2 g ² B :, contrary to Lemma 1. Thus it is proven that < C < ) 784 < B l <. Consider the second occurrence of C in W ' C k and denote the subword of W corresponding to this occurrence by Q. Let Q1 be the beginning of Q with < Q1 < s 784 < B l <. Then Q1 ' B 784 l for C begins with B 784 l and Q ' C. Since < R i < F 700 < B l < for i / 8 and < Tj < - 6 < B l < for all j s 1, . . . , 8, we see that Q1 has common subwords with R i 0 and R i 0q1 for some i 0 of lengths ) < B <, because 700 q 2Ž6 q 0.5. - 784. This, in view of ŽP2., implies that R i 0 Ti 0 R i 0q1 is a B-periodic word, whence one of Y1 , Y2 must be in ² B :. A contradiction to Lemma 1 completes the proof. LEMMA 4. Suppose the subgroup ² X 1 , X 2 : of Fm is not cyclic and w 2 Ž X 1 , X 2 . is conjugate to w 2 Ž X 1X , X 2X .. Then there is a word Z such that X 1X s ZX 1 Zy1 , X 2X s ZX 2 Zy1 . Proof. Let us apply Lemmas 1, 2 to w 2 Ž X 1 , X 2 ., w 2 Ž X 1X , X 2X . and keep their notation Žthe notation related to the pair X 1X , X 2X is supplied with the prime sign 9.. In addition, due to Lemma 1, we can assume that if B is conjugate to B9, then B ' B9. Also suppose that < B l < G <Ž B9. l9 <. Since W is a cyclic permutation W9 of W9 and so < W < s < W9 <, we have from Lemma 2 that < Bl < -

3648 3488

Ž B9 .

l9

- 1.05 Ž B9 .

l9

.

Ž 7.

Consider the subword R i of W. By Lemma 2, < R i < ) Ž i ? 100 y 14 . < B l < G Ž i ? 100 y 14 . Ž B9 . l9 . Since R i is also a subword of the cyclic word W9 Žbeing a cyclic permutation of W . and < R i < ) 86 <Ž B9. l9 <, we see that R i and one of RXj have a

ELEMENTS OF FREE GROUPS

401

common subword Q i j of length ) ŽŽ86 y 6.r2.<Ž B9. l9 < s 40 <Ž B9. l9 <. Then, by Ž7., < Q i j < ) 40 Ž B9 . l9 )

40 1.05

< B l < ) 38 < B l < .

Consequently, Q i j is both a B- and B9-periodic word with < Q i j < ) < B < q < B9 <. Hence, B9 is a cyclic permutation of B, whence B9 ' B by the assumption. Now it follows from Ž7. that l s l9 as well. We will say that the subwords R i and RXj of the cyclic word W B-o¨ erlap if R i and RXj have a common subword of length G < B <. The argument above shows that every R i B-overlaps with at least one RXj . It is easy to see that R i cannot B-overlap with two distinct RXj1 , RXj2 Žotherwise, one of Y1X , Y2X would be in ² B :, contrary to Lemma 1. and, analogously, that every RXj B-overlaps with a unique R i . Suppose R 8 B-overlaps with RXj and j / 8. By Lemma 2, < R 8 < ) 786 < B l < ,

< RXj < F 700 < B l < .

Consequently, it follows from the inequality Ž86 y 6.r2 ) 1 that R 8 also B-overlaps with one of RXjy1 , RXjq1 Žindexes modulo 8.. A contradiction shows that R 8 B-overlaps with RX8 . Note R 7 cannot B-overlap with RX8 for the same reason. Hence, a similar argument yields that R 7 B-overlaps with RX7 . Repeating the reasoning five more times, we conclude that every R i B-overlaps with RXi , i s 1, . . . , 8. Since it is proven that every pair R i , RXi , i s 1, . . . , 8, has a common subword of length G < B <, we can introduce integers a 1 , . . . , a 8 such that a i measures the ‘‘shift’’ of the subword B i?100 l of W0 Žsee Ž6.. relative to the subword B i?100 l of W0X in the cyclic word W. More specifically, let B i?100 l ' Pi1 R i Pi2 ' Pi1X RXi Pi2X be the factorizations according to the cancellations that lead from Ž6. to Ž5. and R i ' Si1 Q ii Si2 ,

RXi ' SXi1 QXii SXi2 ,

where Q ii , QXii are the subwords of R i , RXi , respectively, of length ) < B < that make R i , RXi B-overlap Žnote Q ii s QXi i in cyclic W .. Clearly, the words Ž Pi1 Si1 .Ž Pi1X SXi1 .y1 and Ž SXi2 Pi2X .Ž Si2 Pi2 .y1 are mutually inverse powers of B. Then the number a i , i s 1, . . . , 8, is defined to be k provided Ž Pi1 Si1 .Ž Pi1X SXi1 .y1 s B k .

402

S. V. IVANOV

Referring to Ž5. ] Ž6. and using what was proven and defined above, we have the eight equations Bya 1 Y1 B a 2 s Y1X ,

Bya 5 Y2 B a 6 s Y2X ,

Bya 2 Y1 B a 3 s Y1X ,

Bya 6 Y2 B a 7 s Y2X ,

By a 3 Y1y1 B a 4 s Ž Y1X .

y1

By a 4 Y1y1 B a 5 s Ž Y1X .

y1

,

a8 By a 7 Yy1 s Ž Y2X . 2 B

y1

,

a1 By a 8 Yy1 s Ž Y2X . 2 B

y1

, .

Since Y1 , Y2 f ² B : by Lemma 1, we have from the first two equations that Bya 1 Y1 B a 2 s Bya 2 Y1 B a 3 , whence a 1 s a 2 , a 2 s a 3 . Analogously, we obtain

a1 s a2 , a2 s a3 , a 3 s a4 , a 4 s a5 ,

a5 s a6 , a6 s a 7 , a7 s a8 , a8 s a1 .

Now we see that all the a i ’s are equal, therefore, we can drop their indexes and write that Y1X s By a Y1 B a , Y2X s By a Y2 B a , as required. 2. PROOF OF THE THEOREM First let us show that wnŽ x 1 , . . . , x n . is not equal to 1 in the free group Fn s ² x 1 , . . . , x n : of rank n. Proceeding by induction on n G 2 Žwith a reference to formula Ž1. for n s 2., assume that wnq 1 Ž x 1 , . . . , x nq1 . s w 2 Ž wn Ž x 1 , . . . , x n . , wn Ž x 2 , . . . , x nq1 . . s 1. By induction hypothesis, this implies that ² wn Ž x 1 , . . . , x n ., wnŽ x 2 , . . . , x nq1 .: is cyclic and that both wnŽ x 1 , . . . , x n . and wnŽ x 2 , . . . , x nq1 . are non-trivial. Therefore, it follows from Lemma 3 that wn Ž x 1 , . . . , x ny1 . s wn Ž x 2 , . . . , x nq1 .

"1

.

Ž 8.

Consider an endomorphism p 1: Fn ª Fn given by p 1Ž x 1 . s 1 and p 1Ž x i . s x i for i ) 1. Applying p 1 to Eq. Ž8., we have wn Ž x 2 , . . . , x nq1 . s wn Ž 1, x 2 , . . . , x n .

"1

.

403

ELEMENTS OF FREE GROUPS

By another induction on n, it is easy to see that wnŽ1, x 2 , . . . , x n . s 1. Hence it follows from Ž8. that wn Ž x 2 , . . . , x nq1 . s 1, contrary to the induction hypothesis. Suppose now that X 1 , . . . , X n and Y1 , . . . , Yn are some words in Fm such that the words wnŽ X 1 , . . . , X n . / 1 and wnŽ Y1 , . . . , Yn . are conjugate. By induction on n, let us prove the existence of a word Z such that Yi s ZX i Uy1 for all i s 1, . . . , n. If n s 2 it suffices to refer to Lemma 4. Suppose n s 3. Then, by definition Ž2., we have that w 3 Ž X1 , X 2 , X 3 . s w 2 Ž w 2 Ž X1 , X 2 . , w 2 Ž X 2 , X 3 . . / 1 is conjugate to w 2 Ž w 2 Ž Y1 , Y2 ., w 2 Ž Y2 , Y3 ... By Lemma 4, there is a word S1 such that w 2 Ž Y1 , Y2 . s S1w 2 Ž X 1 , X 2 . S1y1 / 1,

Ž 9.

w 2 Ž Y2 , Y3 . s S1w 2 Ž X 2 , X 3 . S1y1 / 1. Applying Lemma 4 to these two new conjugacies, we further have Y1 s T1 X 1T1y1 ,

Y2 s T1 X 2 T1y1 ,

Y2 s T2 X 2 Ty1 2 ,

Y3 s T2 X 3Ty1 2

Ž 10 . with some T1 , T2 . It follows from Ž9. ] Ž10. that T1w 2 Ž X 1 , X 2 . T1y1 s S1w 2 Ž X 1 , X 2 . S1y1 , y1 T2 w 2 Ž X 2 , X 3 . Ty1 2 s S1 w 2 Ž X 2 , X 3 . S1 .

In view of Lemma 3, it follows from these equations that l

T1y1S1 s w 2 Ž X 1 , X 2 . 1 ,

Ty1 2 S1 s w 2 Ž X 2 , X 3 .

l2

Ž 11 .

with some integers l 1 , l 2 . ² y1 : In view of Ž10., we also have T1 X 2 T1y1 s T2 X 2 Ty1 2 , whence T1 T2 , X 2 is cyclic. Since X 2 / 1, there are integers l 3 and l 4 / 0 such that l

X 2l 3 s Ž T1y1 T2 . 4 .

Ž 12 .

If l 3 s 0, then T1 s T2 and equalities Ž10. imply the existence of a desired word Z Žs T1 ..

404

S. V. IVANOV

Assume l 3 / 0. It follows from equalities Ž11. that l

T1y1 T2 s w 2 Ž X 1 , X 2 . 1 w 2 Ž X 2 , X 3 .

yl 2

and, therefore, by Ž12., l

X 2l 3 s Ž w 2 Ž X 1 , X 2 . 1 w 2 Ž X 2 , X 3 .

yl 2 l 4

.

Ž 13 .

with l 3 / 0. It follows from definition Ž1. of w 2 Ž x 1 , x 2 . that the right part of equality Ž13. is an element of the subgroup w Fm , N2 x, where N2 is the normal closure in Fm of the word X 2 . Consequently, considering the relation corresponding to Ž13. in the relation module of the one-relator group G s ² x1 , . . . , x n 5 X2 : , we have an equality of the form

Ž l 3 y P . ? Xˆ2 s 0, where P is an element of the augmentation ideal of the group ring ZŽ G . of G over the integers and Xˆ2 is the canonical generator of the relation module of G. However, it follows from the well-known Lyndon’s description of the relation module R of a one-relator group ² x 1 , . . . , x n 5 R : Žsee wLS2, Lx. that if Q ? Rˆ s 0 in R then Q is an element of the augmentation ideal of ZŽ² x 1 , . . . , x n 5 R :.. A contradiction to the assumption l 3 / 0 completes the proof for n s 3. Suppose n G 4 and wnŽ X 1 , . . . , X n . / 1 is conjugate to wnŽ Y1 , . . . , Yn .. By the inductive definition of wnŽ x 1 , . . . , x n . and Lemma 4, we have with some words S1 , S2 that wny 1 Ž Y1 , . . . , Yny1 . s S1wny1 Ž X 1 , . . . , X ny1 . S1y1 / 1, wny 1 Ž Y2 , . . . , Yn . s S2 wny1 Ž X 2 , . . . , X n . Sy1 2 / 1. By induction hypothesis, we have some words T1 , T2 such that Yi s T1 X i T1y1 , i s 1, . . . , n y 1,

Yj s T2 X j Ty1 2 , j s 2, . . . , n. Ž 14 .

Since n G 4, we can consider the equations Y2 s T1 X 2 T1y1 s T2 X 2 Ty1 2 ,

Y3 s T1 X 3 T1y1 s T2 X 3 Ty1 2

which imply that T1y1 T2 belongs to the intersection M2 l M3 , where M2 , M3 are the maximal cyclic subgroups of Fm that contain X 2 , X 3 ,

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respectively. If M2 l M3 is trivial then T1 s T2 and the Theorem is proven. Otherwise, we have that M2 s M3 and so ² X 2 , X 3 : is cyclic. Then, obviously, 1 s w 2 Ž X 2 , X 3 . s w 3 Ž X 1 , X 2 , X 3 . s ??? s wn Ž X 1 , . . . , X n . . A contradiction and a reference to Lemma 3 complete the proof of the Theorem.

REFERENCES wCx wKx wLx wLS1x wLS2x wMx wNx wO1x wRx wSx wTx wZ1x wZ2x

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