On Certain C-Test Words for Free Groups

Transcription

1 Journal of Algebra 247, doi: jabr , available online at http: on On Certain C-Test Words for Free Groups Donghi Lee Departent of Matheatics, Uni ersity of Illinois at Urbana Chapaign, 1409 West Green Street, Urbana, Illinois E-ail: Counicated by Efi Zelano Received October 15, 2000 Let F be a free group of a finite rank 2 and let X i, Yj be eleents in F. A non-epty word wž x,..., x. 1 n is called a C-test word in n letters for F if, whenever Ž X,..., X. wy,...,y 1, the two n-typles Ž X,..., X. 1 n 1 n 1 n and Ž Y,...,Y. 1 n are conjugate in F. In this paper we construct, for each n 2, a C-test word Ž x,..., x. with the additional property that Ž X,..., X. n 1 n n 1 n 1if and only if the subgroup of F generated by X 1,..., Xn is cyclic. Making use of such words Ž x,..., x. and Ž x,..., x , we provide a positive solution to the following proble raised by Shpilrain: There exist two eleents u 1, u2 F such that every endoorphis of F with non-cyclic iage is copletely deterined by Ž u., Ž u Elsevier Science INTRODUCTION Let F ² x,..., x : 1 be the free group of a finite rank 2 on the set x,..., x 4 1. The purpose of this paper is to present a positive solution to the following proble raised by Shpilrain 1 : Proble. Are there two eleents u 1, u2 in F such that any endoorphis of F with non-cyclic iage is uniquely deterined by Ž u., Ž u.? Ž 1 2 In other words, are there two eleents u 1, u2 in F such that whenever Ž u. Ž u. i i, i 1, 2, for endoorphiss, of F with non-cyclic iages, it follows that?. In 2, Ivanov solved in the affirative this proble in the case where is a onoorphis of F by constructing a so-called C-test word w Ž x,..., x. for each n 2. A C-test word is defined due to Ivanov n 1 n 2 as follows: DEFINITION. A non-epty word Ž x,..., x. 1 n is a C-test word in n letters for F if for any two n-tuples Ž X,..., X., Ž Y,...,Y. of eleents 1 n 1 n $ Elsevier Science All rights reserved.

2 510 DONGHI LEE of F the equality Ž X,..., X. Ž Y,...,Y. 1 n 1 n 1 iplies the existence of an eleent S F such that Yi SXiS 1 for all i 1, 2,..., n. According to the result of 2, Corollary 1, if is a C-test word in letters for F, is an endoorphis, is a onoorphis of F, and Ž. Ž., then we have S, where S F is such that ² S, Ž.: is cyclic, and S is the inner autoorphis of F defined by eans of S. Notice that this assertion is no longer true if is extended to an endoorphis of F with non-cyclic iage, since Ž. 1 is no longer guaranteed. The efforts to extend the result of 2, Corollary 1 to the case where is an endoorphis of F with non-cyclic iage have led us to proving the following THEOREM. For e ery n 2 there exists a C-test word Ž x,..., x. n 1 n in n letters for F with the additional property that Ž X,..., X. n 1 n 1 if and only if the subgroup ² X,..., X : of F generated by X,..., X is cyclic. 1 n 1 n We construct such a C-test word Ž x,..., x. n 1 n by cobining Ivanov s C-test word w Ž x, x. and an auxiliary word už x, x defined below. Here, let us recall Ivanov s C-test word w Ž x, x.: Ž w x, x x, x x x, x x x, x x x, x x x, x x x, x x x, x x x, x x. We define an auxiliary word už x, x. as follows: Ž 1.1. už x, x. x, x x x, x x x, x x x, x x x, x x x, x x x, x x x, x x. We then construct x,..., x as follows: If n 2 then n 1 n Ž Ž x 1, x2. w2ž x 1, x 2.. If n 3 then Ž Ž x 1, x 2, x3. už už 2Ž x 1, x 2., 2Ž x 2, x 3.., Ž.. u x, x, x, x

3 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 511 Inductively, for n 4, define Ž Ž 1.4 x,..., x u u x, x, x,..., x, For instance, n 1 n n n 1 n 1Ž x n 1, x 2, x 3,..., x n 2, x n.., už n 1 Ž x n 1, x 2, x 3,..., x n 2, x n., Ž x, x, x,..., x, x.... n 1 n 2 3 n 2 1 Ž Ž. 1.5 x, x, x, x u u x, x, x, x, x, x, Ž.. u x, x, x, x, x x In Section 2, we establish several technical leas concerning properties of Ivanov s word Ž x, x. and the auxiliary word už x, x which will be used throughout this paper. In Sections 3 5, we prove that, for each n 3, the word Ž x,..., x. n 1 n constructed above is indeed a C-test word with the property in the stateent of the Theore Žthe case n 2is already proved in. 2. We first treat the case n 3 in Section 3 and then proceed by siultaneous induction on n with base n 4 together with several necessary leas in Sections 4 and 5. Once the Theore is proved, our Corollary 1, which is an extended version of 2, Corollary 1 to the case where is an endoorphis of F with non-cyclic iage, follows iediately, as intended, by taking u Ž x,..., x.: 1 COROLLARY 1. There exists an eleent u F such that if is an endoorphis, is an endoorphis of F with non-cyclic iage, and Ž u. Ž u., then also has non-cyclic iage, ore precisely, S, where S F is such that ² S, Ž u.: is cyclic, and S is the inner autoor- phis of F defined by eans of S. In Corollary 2, we provide a positive solution to the above-entioned Shpilrain proble: COROLLARY 2. There exist two eleents u 1, u2 F such that any endoorphis of F with non-cyclic iage is uniquely deterined by Ž u., Ž u The proof of Corollary 2 akes use of the words Ž x,..., x. 1 and Ž x,..., x Its detailed proof is given in Section 6. The idea and the techniques used in 2 are developed further in the present paper.

4 512 DONGHI LEE 2. PRELIMINARY LEMMAS We begin this section by establishing soe notation and terinology. We let X, Y Ž with or without subscripts. be words in F throughout this paper. By X Y we denote the equality in F of words X and Y, and by X Y the graphical Ž letter-by-letter. equality of words X and Y. The Ž 1 length of a word X is denoted by X note x x We say that a word X is a proper power if X Y l for soe Y with l 1 and that a word A is siple if A is non-epty and cyclically reduced and is not a proper power. If A is siple, then an A-periodic word is a subword of A k with soe k 0. Now let us introduce several leas concerning properties of the word Ž x, x. defined in Ž For proofs of Leas 1 and 2, see LEMMA 1 2, Lea 3. If the subgroup ² X, X : 1 2 of F is non-cyclic, then Ž X, X is neither equal to the epty word nor a proper power. If ² X, X : is cyclic, then Ž X, X LEMMA 2 2, Lea 4. If the subgroup ² X, X : 1 2 of F is non-cyclic and Ž X, X. Ž Y, Y., then there exists a word Z F such that Y ZX Z 1 and Y ZX Z LEMMA 3. If the subgroup ² X, X : 1 2 of F is non-cyclic, then Ž X, X. 1 Ž Y, Y. for any words Y, Y Proof. By way of contradiction, suppose that Ž X, X. 1 Ž Y, Y for soe words Y, Y.If² Y, Y : is cyclic, then it follows fro Lea 1 that Ž Y, Y. 1, so that Ž X, X. 1, i.e., ² X, X : is cyclic. This contradiction to the hypothesis of the lea allows us to assue that ² Y, Y : is non-cyclic. As in 2, Leas , let W be a cyclically reduced word that is conjugate to Ž X, X , and let B be a siple word such that 8 X, X 8 is conjugate to B Žrecall fro 3, 4 that a coutator X, Y 1 2 of words X and Y is not a proper power.. Then according to 2, Lea 2, W has the for W RTRT RT, where R s are B-periodic words with Ž i B R i 100 B i i, 0 T 6 B, and 3488 B W 3648 B i. The sae holds for Ž Y, Y., and we attach the prie sign to the notations for Ž Y, Y Then the equality 2 X 1, X2 2 Y 1, Y2 yields that W is a cyclic 1 perutation of W, so that W W. At this point, apply the argu- 1 1 ents in 2, Lea 4 to W, W to get B B and that Ri B-over- laps only with R 1 for each i 1, 2,..., 8. But this is ipossible, for if R i i

5 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 513 B -overlapped only with R 1, then R would have to B i i 1 -overlap only with Ri 1 indices odulo 8 by the order of indices in W T8 R T R T R and W RTRT RT We also establish several leas concerning properties of the auxiliary word už x, x. defined in Ž LEMMA 4. If the subgroup ² X, X : of F is non-cyclic, then už X, X is neither equal to the epty word nor a proper power. If ² X, X : 1 2 is cyclic, 6 5 then either u X 1, X2 is equal to the epty word pro ided X1 X2 or otherwise už X, X. is a proper power. 1 2 Proof. The proof of the first part is siilar to that of 2, Lea 3, and the second part is iediate fro definition Ž 1.1. of už x, x LEMMA 5. If the subgroup Ž X, X. of F is non-cyclic and už X, X uy, Ž Y., then there exists a word Z F such that 1 2 Y1 ZX1Z 1 and Y2 ZX2Z 1. Proof. Applying the sae arguent as in 2, Lea 4 to už X, X and uy, 1 Y 2, we deduce that Y1 ZX1Z and Y2 ZX2Z for soe word Z F. Since extraction of roots is unique in a free group, it follows 1 1 that Y ZX Z and Y ZX Z, as required LEMMA 6. If the subgroup ² X, X : 1 2 of F is non-cyclic, then už X, X. 1 uy, Ž Y. for any words Y, Y Proof. Suppose to the contrary that už X, X. 1 uy, Ž Y for soe words Y, Y. Then the subgroup ² Y, Y : of F is non-cyclic, for otherwise the equality už X, X. 1 uy, Ž Y. would yield that už X, X Ž 1 1 uy, Y uy, Y , contrary to Lea 4. Fro here on, follow the proof of Lea 3 to arrive at a contradiction. LEMMA 7. If the subgroup ² X, X : 1 2 of F is non-cyclic, then, for each i 1, 2, the subgroup ² X, už X, X.. of F is also non-cyclic. i 1 2 Proof. Suppose on the contrary that ² X, už X, X.: i 1 2 is cyclic for soe i 1, 2. Then by Lea 4 we have Ž 2.1. Xi už X 1, X2. for soe non-zero integer l. Let U be a cyclically reduced word that is conjugate to u X, X, and let C be a siple word such that X, X is conjugate to C. By X, we denote a cyclically reduced word that is conjugate to a word X in F. Then by the sae arguent as in 2, Le- a 1, ax X, X 6 C, 1 2 l

6 514 DONGHI LEE and also by the sae arguent as in 2, Lea 2, U has the for U QSQS QS, where Q are C-periodic words with Ž i C Q i 100 C i i, 0 S 6 C, and 3488 C U 3648 C. This yields that i l X 6 C 3488 C U i už X 1, X2. už X 1, X 2., which contradicts equality 2.1. In the following lea, which will be useful in Sections 4 6, the word Ž x,..., x. with n 3 is defined in Ž 1.3. Ž n 1 n LEMMA 8. If both Ž X, X. and Ž Y,...,Y n 1 n with n 3 are neither equal to the epty word nor proper powers, then the subgroup ² Ž X, X , Ž Y,...,Y.: of F is non-cyclic. n 1 n Proof. Suppose on the contrary that ² Ž X, X., Ž Y,...,Y.: n 1 n is cyclic. It then follows fro the hypothesis of the lea that either 2Ž X 1, X2. nž Y 1,...,Yn. or 1 2Ž 1 2. nž 1 n. X, X Y,...,Y. This iplies, by definitions Ž 1.3. Ž 1.4. of Ž x,..., x. n 1 n, the existence of words Z 1, Z2 in F such that 1 2Ž 1 2. Ž Ž 1 2. Ž 1 2. either X, X u Z, Z or X, X u Z, Z. But since an arguent siilar to that in Lea 3 shows that the latter equality cannot hold, the forer ust hold. Fro here on, follow the arguent in 2, Lea 4 to obtain that B X B Z 6 and B X 1 B Z , where X, Z i i are conjugates of X i, Z i, respectively, and B is a siple word such that B X 8, X 8. This yields X 2 1, i.e., X , contrary to the hypothesis Ž X, X THE CASE n 3 In this section, we prove that Ž x, x, x is a C-test word with the additional property that Ž X, X, X if and only if the subgroup ² X, X, X : of F is cyclic. We begin with leas that play crucial roles in proving this assertion.

7 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 515 LEMMA 9. If už Ž X, X., Ž Y, Y.. 1, then Ž X, X and Ž Y, Y Proof. The hypothesis of the lea iplies by Lea 4 that Ž X, X. 6 Ž Y, Y. 5 ; hence if one of Ž X, X. and Ž Y, Y is equal to the epty word, then so is the other. So assue that Ž X, X and Ž Y, Y. 1. Notice that the equality Ž X, X. 6 Ž Y 1, Y. 5 iplies that ² Ž X, X., Ž Y, Y.: is cyclic. Hence, in view of Leas 1 and 3, we have Ž X, X. Ž Y, Y This together with Ž X, X. 6 Ž Y, Y. 5 yields that Ž X, X. Ž Y, Y , contrary to our assuption. LEMMA 10. Suppose that the subgroup ² X, X, X : of F is non-cyclic. Then Ž X, X, X. 1. Futherore, either Ž X, X, X is not a proper power or it has one of the following three fors: A1 X, X, X X, X and X 1; A2 X, X, X X, X and X 1; A3 X, X, X X, X and X 1. Reark. In view of Lea 1, Ž X, X., Ž X, X., Ž X, X in Ž A1., Ž A2., Ž A3., respectively, are neither equal to the epty word nor proper powers. Proof. Recall fro Ž 1.3. that Ž Ž X, X, X. u u Ž X, X., Ž X, X., Ž.. u X, X, X, X In the case where the subgroup ² už Ž X, X., Ž X, X.., už Ž X 2, X., Ž X, X..: of F is non-cyclic, the assertion that Ž X, X, X is neither equal to the epty word nor a proper power, as desired, follows iediately fro Lea 4. So we only need to consider the case where Ž 3.1. the subgroup² už 2Ž X 1, X 2., 2Ž X 2, X 3.., Ž.: u X, X, X, X is cyclic Here, if už Ž X, X., Ž X, X.. už Ž X, X., Ž X, X , then Lea 9 iplies that Ž X, X. Ž X, X. Ž X, X ; hence, by Lea 1, ² X, X :, ² X, X :, and ² X, X : are all cyclic. This yields that ² X, X, X : is cyclic, contrary to the hypothesis of the lea. Thus, at least one of the words už Ž X, X., Ž X, X.. and už Ž X, X.,

8 516 DONGHI LEE Ž X, X has to be not equal to the epty word. We divide this situation into three cases. Case I. už Ž X, X., Ž X, X.. 1 and už Ž X, X., Ž X, X It follows fro už Ž X, X., Ž X, X and Lea 9 that Ž X, X. Ž X, X. 1; so, by Lea 1, ² X, X : and ² X, X : are cyclic. Since ² X, X, X : is non-cyclic, X3 ust be equal to the epty word; hence we have 24 3Ž Ž Ž 2Ž Ž 2Ž X, X, X u u X, X,1,1 u X, X, X, X X, X. Therefore X, X, X has for A3 in this case Case II. už Ž X, X., Ž X, X.. 1 and už Ž X, X., Ž X, X Since už Ž X, X., Ž X, X , we have, by Leas 1 and 9, that ² X, X : and ² X, X : are cyclic, so that X 1; thus Ž Ž Ž 2Ž Ž 2Ž X, X, X u 1, u 1, X, X u 1, X, X X, X X, X. Therefore X, X, X has for A2 in this case Case III. už Ž X, X., Ž X, X.. 1 and už Ž X, X., Ž X, X In this case, we want to prove CLAIM. This case is reduced to the following two cases: Ž i. ² Ž X, X., Ž X, X., Ž X, X.: is cyclic; Ž ii. both ² Ž X, X., Ž X, X.: and ² Ž X, X., Ž X, X.: are non-cyclic. Proof of the Clai. Assuing at least one of ² Ž X, X., Ž X, X.: and ² Ž X, X., Ž X, X.: is cyclic, we want to show that Case Ž i occurs. Let us say that ² Ž X, X., Ž X, X.: is cyclic Ž the case where ² Ž X, X., Ž X, X.: is cyclic is analogous.. If Ž X, X , then už Ž X, X., Ž X, X.. Ž X, X. 24 and už Ž X, X., Ž X, X Ž X, X. 20. It then follows fro Ž 3.1. that ² Ž X, X., Ž X, X.: is cyclic, which eans that Case Ž. i occurs. Now let Ž X, X Then since ² Ž X, X., Ž X, X.: is cyclic, Ž 3.1. yields by Lea 7 that

9 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 517 ² Ž X, X., Ž X, X.: is also cyclic; hence ² Ž X, X., Ž X, X , Ž X, X.: is cyclic; that is, Case Ž i. occurs as well We divide Case Ž. i further into subcases according to the nuber of non-epty words aong Ž X, X., Ž X, X., and Ž X, X Here, we note that if there exists only one non-epty word, then it has to be Ž X, X , for otherwise we would have a contradiction of the hypothesis of Case III. Therefore, Case III is decoposed into the following six subcases. Case III.1. Ž X, X. Ž X, X. 1 and Ž X, X In this case, it follows fro Lea 1 that ² X, X : and ² X, X : are cyclic, so that X1 1; hence we have Ž. Ž X, X, X. u u 1, Ž X, X., u Ž X, X., už 2Ž X 2, X 3., 2Ž X 2, X X, X X, X. Thus, Ž X, X, X. has for Ž A Case III.2. Ž X, X. 1 and ² 1 Ž X, X., 1 Ž X, X.: is cyclic. Since Ž X, X. 1, we have by Lea 1 that ² X, X : is cyclic. Also since ² 1 Ž X, X.,1 Ž X, X.: is cyclic, we have 1 Ž X, X Ž X, X by Leas 1 and 3. Apply Lea 2 to this equality: there exists a word S F such that X SX S 1 and X SX S 1, which yields that S 1 X2S SX1S 1, so that X2 S 2 XS 1 2. It then follows fro ² X, X : being cyclic that ² S, X, X : is cyclic. This together 1 with the equality X SX S iplies that ² X, X, X : is cyclic, contrary to the hypothesis of the lea. Therefore this case cannot occur. Case III.3. Ž X, X. 1 and ² 1 Ž X, X., 1 Ž X, X.: is cyclic. Repeat an arguent siilar to that in Case III.2 to conclude that this case cannot occur. Case III.4. Ž X, X. 1 and ² 1 Ž X, X., 1 Ž X, X.: is cylcic. Also repeat an arguent siilar to that in Case III.2 to conclude that this case cannot occur.

10 518 DONGHI LEE ² : Case III.5. 1 X, X,1 X, X,1 X, X is cyclic In this case, it follows fro Leas 1 and 3 that Ž Ž X 1, X2. 2Ž X 2, X3. 2Ž X 3, X 1.. Applying Lea 2 to these equalities, we have the existence of words T 1 and T in F such that 2 Ž 3.3. X T X T 1, X T X T 1 ; X T X T 1, X T X T Cobining Ž 3.2. and Ž 3.3. yields that ² T 1 T, X :, ² T, Ž X, X.: , and ² T, Ž X, X.: are all cyclic. At this point, we apply Ivanov s arguent Žsee 2, pp to obtain the following CLAIM Ivanov. T T. 1 2 Proof of the Clai. Suppose on the contrary that T1 T 2. It then follows fro ² T 1 T, X : being cyclic that l 1 1 Ž 3.4. X3 Ž T1 T2. with nonzero integers l and l. It also follows fro ² T, Ž X, X.: and ² T, Ž X, X.: being cyclic that l 3 l4 T1 2Ž X 2, X3. and T2 2Ž X 3, X 1., with integers l and l at least one of which is non-zero, so that 3 4 l 2 Hence, by 3.4, 1 l 3 l4 T1 T2 2Ž X 2, X3. 2Ž X 3, X 1.. l 1 l 3 l X3 2 X 2, X3 2 X 3, X 1. Note, by definition Ž 1.2. of Ž x, x , that the right-hand side of equality Ž 3.5. belongs to the subgroup F, N 3, where N3 is the noral closure in F of the word X. So inside the relation odule Nˆ 3 3 N N, N of the one-relator group equality 3.5 can be expressed as ² : G x,..., x X, 1 3 Ž l P. Xˆ 3 0, 1 l 2

11 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 519 where P is an eleent of the augentation ideal of the group ring Ž G. of G over the integers and Xˆ 3 is the canonical generator of the relation odule Nˆ 3 of G. By Lyndon s result on the relation odule R of a one-relator group ² x,..., x R:Žsee 4, 5., which says that if Q Rˆ 1 0 in R then Q is an eleent of the augentation ideal of Ž² x 1,..., x R :., we ust have l 1 0. This contradiction of l 1 0 copletes the proof of the clai. If T T 1, then equalities Ž yield that X1 X2 X 3, contrary to the hypothesis of the lea. If T1 T2 1, then we derive fro equalities Ž 3.3. that X T 3 XT 3, X T 3 XT 3, and X T 3 XT 3, so that ² T, X :, ² T, X :, and ² T, X : are all cyclic; therefore, ² X 1, X, X : is cyclic. A contradiction iplies that this case cannot occur. 2 3 Case III.6. Both ² Ž X, X., Ž X, X.: and ² Ž X, X., Ž X 3, X.: are non-cyclic. 1 In this case, in view of 3.1 and Leas 4 and 6, we have that Ž. Ž. u X, X, X, X u X, X, X, X, and so by Lea 5 there exists a word T F Ž 3.6. such that 1 2Ž X 1, X2. T 2Ž X 2, X3. T 1 and 1 2Ž X 2, X3. T 2Ž X 3, X1. T 1. Apply Lea 2 to these equalities: there exist words U and U in F 1 2 such that Ž 3.7. X UXU 1, X UXU 1 ; X U X U 1, X U X U Cobining the equalities in Ž 3.6. and Ž 3.7., we deduce that ² U 1 U, X : 1 2 3, ² T 1 U, Ž X, X.:, and ² T 1 U, Ž X, X.: are cyclic. Here, apply Ivanov s arguent used in Case III.5 to get U1 U 2. Then reasoning as in Case III.5, we conclude that this case cannot occur. The proof of Lea 10 is coplete. Now we are ready to prove the Theore for the case n 3. Proof of the Theore Ž n 3.. The additional property that Ž 3 X 1, X, X. 1 if and only if the subgroup ² X, X, X : of F is cyclic follows iediately fro definition Ž 1.3. of Ž x, x, x. and Lea

12 520 DONGHI LEE So we only need to prove that Ž x, x, x is a C-test word; that is, supposing 1 Ž X, X, X. Ž Y, Y, Y , we want to prove the existence of a word Z F such that Yi ZXiZ 1 for all i 1,2,3. We begin by distinguishing two cases according to whether Ž X, X, X is a proper power or not. Case I. X, X, X is a proper power Applying Lea 10 to Ž X, X, X. and Ž Y, Y, Y , we have that Ž X, X, X. has one of three types Ž A1., Ž A2., and Ž A ; besides, by the equality Ž X, X, X. Ž Y, Y, Y., Ž Y, Y, Y has the sae type as Ž X, X, X. does, because the exponents in Ž A1., Ž A2., and Ž A are all distinct. This gives us only three possibilities, Ž A1.& Ž A1., Ž A2.& Ž A2., and Ž A3.& Ž A3., for the types of Ž X, X, X.& Ž Y, Y, Y If Ž X, X, X.& Ž Y, Y, Y. is of type Ž A1.& Ž A1. ŽŽ A2.& Ž A or Ž A3.& Ž A3. is siilar., then X Y 1 and 1 X, X Y, Y. Applying Lea 2 to the equality 1 Ž X, X. Ž Y, Y , we have that two 2-tuples Ž X, X. and Ž Y, Y are conjugate in F, which together with X Y 1 yields that two 3-tuples Ž X, X, X. and Ž Y, Y, Y are conjugate in F, as desired. Case II. X, X, X is not a proper power In this case, it follows fro Lea 4 that Ž 3.8. the subgroup² už 2Ž X 1, X 2., 2Ž X 2, X 3.., Ž.: u X, X, X, X of F is non-cyclic This enables us to apply Lea 5 to the equality Ž X, X, X Ž Y, Y, Y.: for soe word S F, we have Ž 3.9. Here, we consider two subcases. Ž. 1 u X, X, X, X Su Ž Y, Y., Ž Y, Y. S 1, Ž. 1 u X, X, X, X Su Ž Y, Y., Ž Y, Y. S

13 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 521 Case II.1. One of ² Ž X, X., Ž X, X.: and ² Ž X, X., Ž X 3, X.: is non-cyclic. 1 Let us say that ² Ž X, X., Ž X, X.: is non-cyclic Ž the case where ² Ž X, X., Ž X, X.: is non-cyclic is analogous Then, by Lea 5, the first equality of Ž 3.9. iplies the existence of a word W F such that Ž Ž X 1, X2. W 2Ž Y 1, Y2. W 1 and 1 2Ž X 2, X3. W 2Ž Y 2, Y3. W 1. Apply Lea 2 to these equalities: there exist words T and T in F 1 2 such that Ž X TYT 1, X TYT 1 ; X TYT 1, X TYT Now applying Ivanov s arguent introduced in Case III.5 of Lea 10 to equalities Ž Ž 3.11., we get T T, by which Ž yields the desired result. Case II.2. Both ² Ž X, X., Ž X, X.: and ² Ž X, X., Ž X, X.: are cyclic. In this case, if Ž X, X. 1, then the subgroup ² Ž X, X., Ž X 2, X., Ž X, X.: would be cyclic, contrary to Ž So Ž X, X ust be equal to the epty word. On the other hand, equalities Ž 3.9. iply by Lea 4 that both ² Ž Y, Y., Ž Y, Y.: and ² Ž Y, Y., Ž Y, Y.: are also cyclic. Then, for the sae reason as with Ž X, X., Ž Y, Y also has to be equal to the epty word. Thus, it follows fro Ž 3.9. that that is, X, X S Y, Y S and X, X S Y, Y S, 1 2Ž X 1, X2. S 2Ž Y 1, Y2. S 1 and 1 2Ž X 3, X1. S 2Ž Y 3, Y1. S 1, which is a situation siilar to So fro here on, we can follow the proof of Case II.1 to obtain the desired result. The proof of the Theore for the case n 3 is coplete.

14 522 DONGHI LEE 4. THE BASE n 4 OF SIMULTANEOUS INDUCTION In this section, we prove the base step n 4 of siultaneous induction which we use in Leas and the Theore. LEMMA 11 Ž n 4.. If both Ž X, X. and Ž Y, Y, Y are neither equal to the epty word nor proper powers, then ² Ž X, X., Ž Y, Y, Y.: is non-cyclic. Proof. This is a special case of Lea 8. LEMMA 12 Ž n 4.. If už Ž X, X, X., Ž Y, Y, Y.. 1, Ž X 1, X, X. Ž Y, Y, Y Proof. By Lea 4, the hypothesis of the lea iplies that X, X, X Y, Y, Y, so that if one of Ž X, X, X. or Ž Y, Y, Y is equal to the epty word, then so is the other. Hence assue that Ž X, X, X and Ž Y, Y, Y If one of Ž X, X, X. or Ž Y, Y, Y is a proper power, then so is the other by Ž 4.1., because 6 and 5 are relatively prie. Then by Lea 10 Ž X, X, X.& Ž Y, Y, Y has nine possible types, and we can easily check that 6 ties any of 960, 400, or 576 never equals 5 ties any of these, which eans that equality Ž 4.1. cannot hold in any case, a contradiction. If neither Ž X, X, X. nor Ž Y, Y, Y is a proper power, then by Lea 6 we have Ž X, X, X. Ž Y, Y, Y., because Ž iplies that ² Ž X, X, X., Ž Y, Y, Y.: is cyclic. This equality together with Ž 4.1. yields that Ž X, X, X. Ž Y, Y, Y , contrary to our assuption. This copletes the proof. LEMMA 13 Ž n 4.. Suppose that ² X, X, X, X : is non-cyclic. Then Ž X, X, X, X. 1. Furtherore, either Ž X, X, X, X is not a proper power or it has one of the following four fors: B1 X, X, X, X X, X and X X 1; B2 X, X, X, X X, X and X X 1; B3 X, X, X, X X, X and X X 1; Ž B Ž X 1, X 2, X 3, X4. 2Ž X 1, X 2, X3. and X1 X3 X4 1.

15 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 523 Reark. In view of Leas 1 and 10, Ž X, X., Ž X, X , Ž X, X., and Ž X, X, X. in Ž B1., Ž B2., Ž B3., and Ž B , respectively, are neither equal to the epty word nor proper powers. Proof. Recall fro Ž 1.5. that Ž Ž. u X, X, X, X u u X, X, X, X, X, X, Ž.. u X, X, X, X, X, X If ² už Ž X, X, X., Ž X, X, X.., už Ž X, X, X., Ž X, X, X..: is non-cyclic, then the assertion that Ž X, X, X, X is neither equal to the epty word nor a proper power, as desired, follows directly fro Lea 4. So we only need to consider the case where Ž 4.2. ² už 3Ž X 1, X 2, X 3., 3Ž X 3, X 2, X 4.., Ž.: u X, X, X, X, X, X is cyclic Here, to avoid a contradiction of the hypothesis that ² X, X, X, X : is non-cyclic, at least one of už Ž X, X, X., Ž X, X, X.. or už Ž X 3, X, X., Ž X, X, X has to be not equal to the epty word. So we have three cases to consider. Case I. už Ž X, X, X., Ž X, X, X.. 1 and už Ž X, X, X , Ž X, X, X In this case, we have, by Lea 12 Ž n 4. and the Theore Ž n 3,. that both ² X, X, X : and ² X, X, X : are cyclic. Since ² X 1, X 2, X 3, X : 4 is non-cyclic, X2 and X4 ust be equal to the epty word; hence we have Ž. Ž X, X, X, X. u u Ž X, X, X.,1, u X, X, X,1 X, X, X Ž 3 1. X, X by for A2. Therefore, Ž X, X, X, X. has for Ž B in this case. Case II. už Ž X, X, X., Ž X, X, X.. 1 and už Ž X, X, X , Ž X, X, X In this case, by Lea 12 Ž n 4. and the Theore Ž n 3., we have that both ² X, X, X : and ² X, X, X : are cyclic, so that X X 1;

16 524 DONGHI LEE thus Ž. 4Ž X 1, X 2, X 3, X4. u 1, u 1, 3Ž X 4, X 2, X1. u 1, Ž X, X, X. Ž X, X, X Ž 1 4. X, X by for A2. Therefore, Ž X, X, X, X. has for Ž B in this case. Case III. už Ž X, X, X., Ž X, X, X.. 1 and už Ž X 3, X 2, X., Ž X, X, X By reasoning as in Case III of Lea 10, we break this case into the following six subcases. Case III.1. Ž X, X, X. Ž X, X, X. 1 and Ž X, X, X It follows fro the Theore Ž n 3. that ² X, X, X : and ² X, X, X : are cyclic, so that X1 X2 1; hence we have 4Ž X 1, X 2, X 3, X4. už už 1, 3Ž X 3, X2 X 4.., už 3Ž X 3, X 2, X 4., u Ž X, X, X., Ž X, X, X. Ž Ž X 3, X 2, X Ž X, X. by for Ž A Thus, Ž X, X, X, X. has for Ž B Case III.2. Ž X, X, X. 1 and ² 1 Ž X, X, X., 1 Ž X 4, X, X.: 2 1 is cyclic. Since Ž X, X, X. 1, we have, by the Theore Ž n , that Ž 4.3. ² X, X, X : is cyclic. Also since ² 1 Ž X, X, X.,1 Ž X, X, X.: is cyclic, in view of Leas 10 and 11 Ž n 4., this case is reduced to the following two cases: Ž. i both Ž X, X, X. and Ž X, X, X are proper powers; Ž ii. neither Ž X, X, X. nor Ž X, X, X is a proper power. Case Ž. i is divided further into subcases according to the types of Ž X, X, X.& Ž X, X, X by Lea 10. Of nine possible types of Ž X, X, X.& Ž X, X, X., Ž A3.& Ž A1., Ž A3.& Ž A2., and Ž A & Ž A3. cannot occur, for if Ž X, X, X. were of type Ž A3., then X 1,

17 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 525 which together with Ž 4.3. yields a contradiction of the hypothesis of the lea. Also, Ž A1.& Ž A1. and Ž A2.& Ž A1. cannot occur, for if Ž 3 X 4, X, X. were of type Ž A1. 2 1, then X4 1, again a contradiction. Moreover, Ž A1.& Ž A2. cannot occur, for this iplies that X3 X2 1, so that Ž X, X, X. 1, a contradiction. Also, Ž A2.& Ž A cannot occur, for this iplies that X X 1, so that Ž X, X, X , a contradiction as well. For this reason, in Case Ž. i, we only need to consider Ž A1.& Ž A3. and Ž A2.& Ž A2. for the types of Ž X, X, X.& Ž X, X, X Therefore, Case III.2 is decoposed into the following three subcases. Case III.2.1. X, X, X & X, X X is of type A1 & A In this case, it follows fro Lea 10 that X X 1, Ž X 3, X 2, X. Ž X, X. 960, and Ž X, X, X. Ž X, X Since ² Ž X 3, X, X., Ž X, X, X.: is cyclic by the hypothesis of Case III.2, we have that ² Ž X, X., Ž X, X.: is cyclic, so that 1 Ž X, X. Ž X 4, X. 2 by Leas 1 and 3. Apply Lea 2 to this equality: there is a word S F such that 4.4 X SX S 1 and X SX S If S 1, then fro Ž 4.4. we have X X, which together with Ž yields a contradiction of the hypothesis of the lea. Now let S 1. We derive fro Ž 4.4. that X S 2 X S 2 and X S 2 XS 2, so that ² S, X : and ² S, X : are cyclic; thus ² X, X : is cyclic. This, together with Ž 4.3., also yields a contradiction Ž because X Therefore, we conclude that this case cannot occur. Case III.2.2. X, X, X & X, X, X is of type A2 & A It follows fro Lea 10 that X 2 1, 3 X 3, X 2, X4 2 X 4, X 3 400, and Ž X, X, X. Ž X, X Since ² Ž X, X, X., Ž X 4, X 2, X.: is cyclic, ² Ž X, X., Ž X, X.: is cyclic; hence, by Leas 1 and 3, 1 Ž X, X. Ž X, X Then by Lea 2 there exists a word U F such that 4.5 X UX U 1 and X UX U If U 1, then it follows fro Ž 4.5. that X1 X4 X 3, which together with Ž 4.3. yields a contradiction of the hypothesis of the lea. Now let 1 1 U 1. We have fro 4.5 that UX1U U X3U. This equality iplies by Ž 4.3. that ² U, X, X, X : is cyclic; thus, by the first equality of Ž 4.5., we 1 2 3

18 526 DONGHI LEE have that ² X, X, X, X : is cyclic. A contradiction iplies that this case cannot occur. Case III.2.3. Neither Ž X, X, X. nor Ž X, X, X is a proper power. Since ² Ž X, X, X., Ž X, X, X.: is cyclic, we have, by Lea 6, that 1 3Ž X 3, X 2, X4. 3Ž X 4, X 2, X 1.. Apply the Theore Ž n 3. to this equality: there is a word T F such that 4.6 X TX T 1, X TX T 1, and X TX T The second equality of Ž 4.6. iplies that ² T, X : is cyclic; hence, by Ž , ² T, X, X, X : is cyclic Ž because X Then by the third equality of Ž 4.6., we have that ² X, X, X, X : is cyclic. A contradiction iplies that this case cannot occur. Case III.3. Ž X, X, X. 1 and ² 1 Ž X, X, X., 1 Ž X 4, X, X.: is cyclic. 2 1 Repeat arguents siilar to those in Case III.2 to conclude that this case cannot occur. Case III.4. Ž X, X, X. 1 and ² 1 Ž X, X, X., 1 Ž X 3, X, X.: is cyclic. 2 4 Also, repeat arguents siilar to those in Case III.2 to conclude that this case cannot occur. Case III.5. ² 1 Ž X, X, X., 1 Ž X, X, X., 1 Ž X 4, X 2, X.: is cyclic. 1 In this case, we want to prove CLAIM. 1 X, X, X X, X, X X, X, X Proof of the Clai. If none of these is a proper power, then the assertion follows iediately fro Lea 6. So assue that one of these is a proper power. Then, in view of Leas 10 and 11 Ž n 4., the other two also have to be proper powers; thus two of X 1, X 3, and X4 ust be equal to the epty word, unless X2 1. However, if two of X 1, X 3, and X4 were equal to the epty word, then we would have a contradiction of the non-triviality of Ž X, X, X., Ž X, X, X., or Ž X, X, X

19 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 527 Hence we ust have X 1. Then 2 Ž Ž X, X, X. Ž X, X., Ž X, X, X. Ž X, X., Ž X, X, X. Ž X, X. ; hence the hypothesis of Case III.5 iplies that ² Ž X, X., Ž X, X , Ž X, X.: is cyclic. It then follows fro Leas 1 and 3 that Ž X 3, X. Ž X, X. Ž X, X., which together with Ž 4.7. proves the clai Now apply the Theore n 3 to the equalities in the Clai: there exist words Z and Z in F such that 1 2 Ž 4.8. X Z X Z 1, X Z X Z 1, X Z X Z 1 ; X Z X Z 1, X Z X Z 1, X Z X Z ² : ² : ² 2 2 We deduce fro these equalities that Z 1, X 2, Z 2, X 2, ZZ, 1 2 Z1Z 2, : ² 2 2 : ² 2 2 X, ZZ, Z Z, X, and ZZ, Z Z, X : are all cyclic. Here, if 2 2 either ZZ 1orZ Z 1, then we would have that ² X, X, X, X : is cyclic, a contradiction. Hence we ust have that ZZ Z1 2 Z2 1, that is, Z Z 1; thus, by Ž 4.8., 1 2 X X X In addition, it follows fro the Clai that Ž Ž X, X, X, X. u u Ž X, X, X., Ž X, X, X., Ž.. u X, X, X, X, X, X Ž 3Ž Ž u X, X, X, X, X, X Ž X, X, X X, X, X Therefore, in this case, X, X, X, X has for B Case III.6. Both ² Ž X, X, X., Ž X, X, X.: and ² Ž X 3, X 2, X., Ž X, X, X.: are non-cyclic. In view of Ž 4.2. and Leas 4 and 6, we have that Ž. u X, X, X, X, X, X Ž. u X, X, X, X, X, X

20 528 DONGHI LEE Then by Lea 5 applied to this equality, there is a word W F such that 1 3Ž X 1, X 2, X3. W 3Ž X 3, X 2, X4. W 1 and 1 3Ž X 3, X 2, X4. W 3Ž X 4, X 2, X1. W 1. Applying the Theore Ž n 3. to these equalities yields the existence of words V1 and V2 in F such that X1 VXV , X2 VXV , X3 VXV ; X3 V2X4V 1 2, X2 V2X2V 1 2, X4 V2X1V 1 2. This is the sae situation as Ž 4.8.; hence, reasoning as in Case III.5, we have X X X. But then Ž X, X, X. Ž X, X, X. Ž X 4, X, X. 2 1, which yields a contradiction to the hypothesis of Case III.6. Therefore, we conclude that this case cannot occur. The proof of Lea 13 Ž n 4. is now coplete. Proof of the Theore Ž n 4.. The additional property that Ž 4 X 1, X 2, X, X. 1 if and only if the subgroup ² X, X, X, X : of F is cyclic is iediate fro definition Ž 1.5. of Ž x, x, x, x. and Lea 13 Ž n Now we want to prove that Ž x, x, x, x is a C-test word, that is, supposing 1 Ž X, X, X, X. Ž Y, Y, Y, Y , we want to prove the existence of a word Z in F such that Yi ZX Z 1 for all i 1,2,3,4. i We consider two cases corresponding to whether Ž X, X, X, X is a proper power or not. Case I. Ž X, X, X, X is a proper power. Apply Lea 13 Ž n 4. to Ž X,..., X. and Ž Y,...,Y.: Ž X 1,..., X. has one of the four types Ž B1. Ž B4. 4 ; furtherore, by the equality Ž X,..., X. Ž Y,...,Y., Ž Y,...,Y. has the sae type as Ž X 1,..., X. does, since the exponents in Ž B1. Ž B4. 4 are all distinct. This gives us only four possibilities, Ž B1.& Ž B1.,..., Ž B4.& Ž B4., for the types of Ž X,..., X.& Ž Y,...,Y If Ž X,..., X. & Ž Y,...,Y. is of type Ž B1. & Ž B1. ŽŽ B2. & Ž B or Ž B3.& Ž B3. is analogous., then Ž Ž 4 3. X X Y Y 1 and 1 X, X Y, Y. Applying Lea 2 to the equality 1 Ž X, X. Ž Y, Y , we have that two 2-tuples Ž X, X. and Ž Y, Y are conjugate in F, which together with X1 X2 Y1 Y2 1 yields the desired result.

21 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 529 If X,..., X & Y,...,Y is of type B4 & B4, then X X X, Y Y Y, and X, X, X Y, Y, Y. The equality 1 Ž X, X, X. Ž Y, Y, Y yields, by the Theore Ž n 3., that two 3-tuples Ž X, X, X. and Ž Y, Y, Y are conjugate in F. Then the result follows fro X1 X3 X4 and Y1 Y3 Y 4. Case II. X, X, X, X is not a proper power In view of Lea 4, it follows that Ž 4.9. ² už 3Ž X 1, X 2, X 3., 3Ž X 3, X 2, X 4.., Ž.: u X, X, X, X, X, X is non-cyclic This enables us to apply Lea 5 to the equality Ž X,..., X Ž Y,...,Y.: there exists a word S F such that Ž Ž 3Ž Ž Ž. 1 u X, X, X, X, X, X 1 Su 3 Y 1, Y 2, Y 3, 3 Y 3, Y 2, Y4 S, Ž 3Ž Ž Ž. 1 u X, X, X, X, X, X 1 Su 3 Y 3, Y 2, Y 4, 3 Y 4, Y 2, Y1 S. Here, we have four subcases to consider. Case II.1. Both ² Ž X, X, X., Ž X, X, X.: and ² Ž X 3, X 2, X., Ž X, X, X.: are non-cyclic The hypothesis of this case enables us to apply Lea 5 to equalities Ž 4.10.: there exist words T and T in F such that Ž Ž X 1, X 2, X3. T1 3Ž Y 1, Y 2, Y3. T1 1, 1 3Ž X 3, X 2, X4. T1 3Ž Y 3, Y 2, Y4. T1 1 ; 1 3Ž X 3, X 2, X4. T2 3Ž Y 3, Y 2, Y4. T 1 2, 1 3Ž X 4, X 2, X1. T2 3Ž Y 4, Y 2, Y1. T 1 2. Then by the Theore Ž n 3. applied to Ž 4.11., there exist words U 1, U 2, and U3 such that Ž X UYU 1, X UYU 1, X UYU 1 ; X UYU 1, X UYU 1, X UYU 1 ; X UYU 1, X UYU 1, X UYU

22 530 DONGHI LEE Here, if one of U1 U 2, U2 U 3, and U3 U1 is true, then the required result follows directly fro Ž So assue that U 1, U 2, and U3 are pairwise distinct. Cobining the equalities in Ž 4.12., we deduce that ² U 1 U, X :, ² U 1 U, X :, ² U 1 U, X :, ² U 1 U, X :, ² U 1 U, X : , and ² U 1 U, X : are all cyclic, so that ² X, X :, ² X, X :, and ² X, X : are cyclic. Since ² X,..., X : is non-cyclic Žthis follows fro Ž X,..., X , we ust have X 1; so, by Ž , Y2 1. Then by Lea 10, the equalities on the first line of Ž yield that naely, X, X T Y, Y T, X, X T Y, Y T, Ž Ž X 3, X1. T1 2Ž Y 3, Y1. T1 1, 1 2Ž X 4, X3. T1 2Ž Y 4, Y3. T1 1. This is a situation siilar to Ž 3.10., so fro here on, we can follow the proof of Case II.1 of the Theore Ž n 3. to obtain that two 3-tuples Ž X, X, X. and Ž Y, Y, Y are conjugate in F. Since X2 Y2 1, the desired result follows. Case II.2. ² Ž X, X, X., Ž X, X, X.: is non-cyclic, and ² Ž X 3, X, X., Ž X, X, X.: is cyclic In this case, we can apply Lea 5 to the first equality of Ž 4.10.: there exists a word V F such that Ž Ž X 1, X 2, X3. V 3Ž Y 1, Y 2, Y3. V 1, 1 3Ž X 3, X 2, X4. V 3Ž Y 3, Y 2, Y4. V 1. Then the Theore Ž n 3. applied to Ž yields the existence of words W1 and W2 in F such that Ž X WYW 1, X WYW 1, X WYW 1 ; X WYW 1, X WYW 1, X WYW If W W, then the required result follows fro Ž Now assue that W W. We deduce fro Ž that ² W 1 W, X : and ² W 1 W, X : are cyclic, so that ² : 4.16 X, X is cyclic. 2 3

23 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 531 On the other hand, since ² Ž X, X, X., Ž X, X, X.: is cyclic by the hypothesis of Case II.2, in view of Leas 10 and 11 Ž n 4., we have that Ž X, X, X. is a proper power if and only if Ž X, X, X is a proper power. For this reason, this case is reduced to the following three subcases. Case II.2.1. Ž X, X, X. 1 Ž Ž X, X, X by the hypothesis of Case II.2.. It follows fro the Theore Ž n 3. that ² X, X, X : is cyclic. Since ² X, X : is cyclic by Ž , to avoid a contradiction of the fact that ² X,..., X : is non-cyclic, we ust have X 1; 2 thus Y 1 by Then by Lea 10, equalities 4.14 yield that 2 1 2Ž X 3, X1. V 2Ž Y 3, Y1. V 1, 1 2Ž X 4, X3. V 2Ž Y 4, Y3. V 1. This is the sae situation as in 4.13 ; hence fro here on, repeating the proof of Case II.1, we obtain the desired result. Case II.2.2. Both Ž X, X, X. and Ž X, X, X are proper powers. In view of Lea 10, we have nine possibilities for the types of Ž X, X, X.& Ž X, X, X.. Of these nine possible types, Ž Al.& Ž A , Ž A1.& Ž A2., Ž A1.& Ž A3., Ž A2.& Ž A1., Ž A2.& Ž A3., Ž A3.& Ž A2., and Ž A3. & Ž A3. cannot occur, for if one of these occurred, then two of X 1, X 2, X 3, and X4 should be equal to the epty word, which yields a contradiction of the non-triviality of Ž X, X, X., Ž X, X, X.,or Ž X, X, X So only Ž A2.& Ž A2. and Ž A3.& Ž A1. can actually occur. If Ž X, X, X.& Ž X, X, X. is of type Ž A2.& Ž A , then X2 1, which is the sae situation as in Ž Hence fro here on, following the proof of Case II.2.1, we arrive at the desired result. If Ž X, X, X & Ž X, X, X. is of type Ž A3.& Ž A , then X4 1. The result then follows fro Ž Case II.2.3. Neither Ž X, X, X. 1 nor Ž X, X, X is a proper power. Since ² Ž X, X, X., Ž X, X, X.: is cyclic, it follows fro Lea 6 that Ž X, X, X. Ž X, X, X , and so fro the second equality of Ž that Ž X, X, X Su Y, Y, Y, Y, Y, Y S.

24 532 DONGHI LEE This equality iplies by Lea 4 that ² Ž Y, Y, Y., Ž Y, Y, Y.: is also cyclic. We then observe that equality Ž can hold only when neither Ž Y, Y, Y. nor Ž Y, Y, Y. is a proper power and Ž Y, Y, Y Ž Y, Y, Y., by which Ž yields that naely, X, X, X S Y, Y, Y S, 1 3Ž X 4, X 2, X1. S 3Ž Y 4, Y 2, Y1. S 1. Now apply the Theore Ž n 3. to this equality: there exists a word W F such that 3 X WYW 1, X WYW 1, X WYW Putting this together with Ž 4.15., we have the sae situation as in Ž 4.12., except that we already assued W1 W2 in Case II.2. Therefore, fro here on, we can follow the proof of Case II.1 to derive the result. Case II.3. ² Ž X, X, X., Ž X, X, X.: is cyclic, and ² Ž X 3, X 2, X., Ž X, X, X.: is non-cyclic It is sufficient to repeat arguents siilar to those in Case II.2 to arrive at the desired result. Case II.4. Both ² Ž X, X, X., Ž X, X, X.: and ² Ž X 3, X 2, X., Ž X, X, X.: are cyclic Arguing as in the proof of Case II.2 of the Theore Ž n 3., replacing Ž 3.8. and Ž 3.9. by Ž 4.9. and Ž 4.10., respectively, we deduce that Ž 3 X 3, X 2, X. Ž Y, Y, Y. 1. So ² : 4.19 X, X, X is cyclic; oreover, it follows fro 4.10 that naely, X, X, X S Y, Y, Y S, X, X, X S Y, Y, Y S, 1 3Ž X 1, X 2, X3. S 3Ž Yi, Y 2, Y3. S 1, 1 3Ž X 4, X 2, X1. S 3Ž Y 4, Y 2, Y1. S 1.

25 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 533 Then by the Theore n 3 applied to these equalities, we have the existence of words Z and Z in F such that 1 2 Ž X ZYZ 1, X ZY Z 1, X ZYZ 1 ; X ZYZ 1, X ZYZ 1, X ZYZ If Z Z, then the result follows fro Ž Now assue that Z1 Z 2. Then equalities Ž yield that ² Z 1 Z, X : and ² Z 1 Z, X : are cyclic, so that ² X, X : is cyclic. Since ² X, X, X : is cyclic by Ž , we ust have X 1, which is the sae situation as in Ž Thus, fro here on, we can follow the proof of Case II.2.1 to get the required result. The Theore Ž n 4. is now copletely proved. 5. THE INDUCTIVE STEP In this section, we prove the inductive step of siultaneous induction which we use in Leas and the Theore. Let n 5 throughout this section. LEMMA 11. If both Ž X,..., X. and Ž Y,...,Y. n 2 1 n 2 n 1 1 n 1 are nei- ther equal to the epty word nor proper powers, then ² Ž X,..., X. n 2 1 n 2, Ž Y,...,Y.: is non-cyclic. n 1 1 n 1 Proof. By way of contradiction, suppose that ² Ž X,..., X. n 2 1 n 2, Ž Y,...,Y.: is cyclic. Since both Ž X,..., X. and Ž n 1 1 n 1 n 2 1 n 2 n 1 Y 1,..., Y. are non-proper powers, it follows fro Lea 6 that n 1 n 2Ž X 1,..., Xn 2. n 1Ž Y 1,...,Y n 1., and so fro Ž 1.3. Ž 1.4. and Leas 4 and 5 that there exists a word S F such that Ž 5.1. Ž n 3Ž 1 n 3. n 3Ž n 3 n 2.. Ž n 2Ž 1 n 2. n 2Ž n 2 n 1.. Ž n 3Ž n 3 n 2. n 3Ž n Ž. u X,..., X, X,..., X 1 Su Y,...,Y, Y,...,Y S, u X,..., X, X,..., X 1 Su n 2 Y n 2,...,Y n 1, n 2 Y n 1,...,Y1 S. We first assue that ² Ž X,..., X., Ž X,..., X.: n 3 1 n 3 n 3 n 3 n 2 is non-cyclic. This enables us to apply Lea 5 to the first equality of Ž 5.1. :

26 534 there exists a word T F DONGHI LEE such that Ž 5.2. n 3Ž X 1,..., Xn 3. T n 2Ž Y 1,...,Yn 2. T 1 n 3Ž X n 3,..., Xn 2. T n 2Ž Y n 2,...,Yn 1. T 1. If both sides of the first equality of Ž 5.2. are non-proper powers, then this equality yields a contradiction the induction hypothesis Lea 11; if both sides of the first equality of Ž 5.2. are proper powers, then we see fro Lea 10 and the induction hypothesis of Lea 13 that they cannot be the sae proper powers Žbecause the exponents on the two sides cannot be identical., a contradiction as well. We next assue that ² Ž X,..., X., Ž X,..., X.: n 3 1 n 3 n 3 n 3 n 2 is cyclic. Then the first equality of Ž 5.1. yields by Lea 4 that ² 5.3 X,..., X, X,..., X, n 3 1 n 3 n 3 n 3 n 2 S Ž Y,...,Y. S 1, S Ž Y,...,Y. S 1 is cyclic. n 2 1 n 2 n 2 n 2 n 1 Here, in view of the induction hypothesis of Lea 13, we see that there are only two ways to avoid a contradiction of Lea 8 and the induction hypothesis of Lea 11: Ž. i any non-trivial word of S n 2 Y 1,...,Yn 2 S 1 and S Ž Y,...,Y. S 1 is of type Ž B4.; Ž ii. n 2 n 2 n 1 any non-trivial word in Ž 5.3. is of one of types Ž B1., Ž B2., and Ž B3.. ŽFor n 5, there is only one way: any non-trivial word of S Ž Y,...,Y. S 1 and S Ž n 2 1 n 2 n 2 Y n 2,..., Y. S 1 is of one of types Ž A1., Ž A2., and Ž A3.. n 1. However, in either case, we can observe that equalities Ž 5.2. cannot hold. This contradiction copletes the proof. LEMMA 12. If už Ž X,..., X., Ž Y,...,Y.. n 1 1 n 1 n 1 1 n 1 1, then Ž X,..., X. 1 and Ž Y,...,Y. 1. n 1 1 n 1 n 1 1 n 1 Proof. By Lea 4, the hypothesis of the lea yields that : 6 5 n 1 1 n 1 n 1 1 n X,..., X Y,...,Y. If one of Ž X,..., X. and Ž Y,...,Y. n 1 1 n 1 n 1 1 n 1 is equal to the epty word, then by Ž 5.4. there is nothing to prove. So assue that Ž n 1 X 1,..., X. 1 and Ž Y,...,Y. n 1 n 1 1 n 1 1. Since 6 and 5 are relative- ly prie, equality Ž 5.4. iplies that both Ž X,..., X. n 1 1 n 1 and Ž Y,...,Y. n 1 1 n 1 either proper powers or non-proper powers. If both are proper powers, then a contradiction of equality Ž 5.4. follows fro the induction hypothesis of Lea 13, for 6 ties any of Ž n 4., 400 Ž n 3., Ž n 4., and 1936 cannot be equal to 5 ties any of these.

27 ON CERTAIN C-TEST WORDS FOR FREE GROUPS 535 If both are non-proper powers, then, since ² Ž X,..., X., Ž n 1 1 n 1 n 1 Y 1,...,Y.: is cyclic, we have by Lea 6 that Ž X,..., X. n 1 n 1 1 n 1 Ž Y,...,Y.. This together with Ž 5.4. yields that Ž X,..., X. n 1 1 n 1 n 1 1 n 1 Ž Y,...,Y. n 1 1 n 1 1. This contradiction to our assuption copletes the proof. LEMMA 13. Suppose that ² X,..., X : is non-cyclic. Then Ž 1 n n X 1,..., X. 1. Furtherore, either Ž X,..., X. n n 1 n is not a proper power or it has one of the following four fors: X, X Ž n 3. and 2 Ž n 1. X1 X2 Xn 2 1, for n e en B1 n X 1,..., Xn Ž n Ž X n 1, Xn. and X1 X2 Xn 2 1, for n odd; 400 X, X Ž n 2. and 2 Ž 1 n. X2 X3 Xn 1 1, for n e en B2 n X 1,..., Xn Ž n X, X and 2Ž n 1. X2 X3 Xn 1 1, for n odd; X, X Ž n 3. and 2 Ž n 1 1. X2 Xn 2 Xn 1, ne en B3 n X 1,..., Xn Ž n Ž X 1, Xn 1. and X2 Xn 2 Xn 1, n odd; Ž B Ž X,..., X. Ž X, X,..., X. and n 1 n n n 1 X X X 1. 1 n 1 n Reark. In view of Lea 1 and the induction hypothesis of Lea 13, Ž X, X. and Ž X, X. in Ž B1., Ž X, X. and Ž X, X. 2 n n 1 2 n 1 n 2 1 n 2 n 1 in Ž B2., Ž X, X. and Ž X, X. in Ž B3., and Ž X, X,..., X. 2 n n 1 n n 1 in Ž B4. are neither equal to the epty word nor proper powers.

28 536 DONGHI LEE Proof. Closely follow the proof of Lea 13 Ž n 4. replacing references to Leas 10, 11 Ž n 4., and 12 Ž n 4., and the Theore Ž n 3. by references to the induction hypothesis of Lea 13, Leas 8 and 11, Lea 12, and the induction hypothesis of the Theore, respectively. Different situations fro Lea 13 Ž n 4. can possibly occur only in Case III.2 and Case III.5, which we reconsider below. Case III.2. ² 1 Ž X, X, X,..., X, X., 1 Ž n 1 n n 2 n n 1 X n, X 2, X,..., X, X.: is cyclic, and Ž X, X,..., X n 2 1 n n 1 Since Ž X, X,..., X. n n 1 1, we have, by the induction hypothesis of the Theore, that ² : 5.5 X, X,..., X is cyclic. 1 2 n 1 Also since ² 1 Ž X, X,..., X, X., 1 Ž n 1 n 1 2 n 2 n n 1 X n, X 2,..., X, X.: n 2 1 is cyclic, in view of Leas 8 and 11 and the induction hypothesis of Lea 13, this case is reduced to the following two cases: Ž. i both Ž X, X,..., X, X. and Ž n 1 n 1 2 n 2 n n 1 X n, X 2,..., X, X. n 2 1 are proper powers; Ž ii. neither Ž X, X,..., X, X. nor Ž n 1 n 1 2 n 2 n n 1 X n, X 2,..., X, X. is a proper power. n 2 1 Case Ž. i is divided further into subcases according to the types of Ž X, X,..., X, X.& Ž X, X,..., X, X. n 1 n 1 2 n 2 n n 1 n 2 n 2 1 by the induc- tion hypothesis of Lea 13. However, the forer word cannot be of type Ž B3. or Ž B4., for this type together with Ž 5.5. yields a contradiction of the hypothesis of Lea 13. For the sae reason, the latter cannot be of type Ž B1. or Ž B4.. Also Ž B1.& Ž B2. and Ž B2.& Ž B3. cannot occur, for these types yield a contradiction of the non-triviality of Ž n 1 X n 1, X 2,..., X, X. and Ž X, X,..., X, X., respectively. Thus, in Case Ž i. n 2 n n 1 n 2 n 2 1, only Ž B1.& Ž B3. and Ž B2.& Ž B2. need to be considered. This allows us to follow the proof of Lea 13 Ž n 4. fro here on. Case III.5. ² 1 Ž X, X,..., X., 1 Ž n n 1 n 1 X n 1, X 2, X 3,..., X, X.,1 Ž X, X, X,..., X, X.: is cyclic. n 2 n n 1 n 2 3 n 2 1 In this case, to be able to keep following the proof of Lea 13 n 4, it is sufficient prove the following CLAIM. 1 Ž X, X,..., X. Ž X,..., X, X. n n 1 n 1 n 1 n 2 n Ž X,..., X, X.. n 1 n n 2 1 Proof of the Clai. If none of these is a proper power, then the assertion follows iediately fro Lea 6. So assue that one of these is a proper power. Then, in view of the induction hypothesis of Lea 13