On Certain C-Test Words for Free Groups

Journal of Algebra 247, 509᎐540 Ž2002. doi:10.1006rjabr.2001.9001, available online at http:rrwww.idealibrary.com on

On Certain C-Test Words for Free Groups Donghi Lee Department of Mathematics, Uni¨ ersity of Illinois at Urbana᎐Champaign, 1409 West Green Street, Urbana, Illinois 61801 E-mail: [email protected] Communicated by Efim Zelmano¨ Received October 15, 2000

Let Fm be a free group of a finite rank m G 2 and let X i , Yj be elements in Fm . A non-empty word w Ž x 1 , . . . , x n . is called a C-test word in n letters for Fm if, whenever Ž X1 , . . . , X n . s w Ž Y1 , . . . , Yn . / 1, the two n-typles Ž X1 , . . . , X n . and Ž Y1 , . . . , Yn . are conjugate in Fm . In this paper we construct, for each n G 2, a C-test word ¨ nŽ x 1 , . . . , x n . with the additional property that ¨ nŽ X 1 , . . . , X n . s 1 if and only if the subgroup of Fm generated by X1 , . . . , X n is cyclic. Making use of such words ¨ m Ž x 1 , . . . , x m . and ¨ mq1Ž x 1 , . . . , x mq1 ., we provide a positive solution to the following problem raised by Shpilrain: There exist two elements u1 , u 2 g Fm such that every endomorphism ␺ of Fm with non-cyclic image is completely determined by ␺ Ž u1 ., ␺ Ž u 2 .. 䊚 2002 Elsevier Science

1. INTRODUCTION Let Fm s ² x 1 , . . . , x m : be the free group of a finite rank m G 2 on the set  x 1 , . . . , x m 4 . The purpose of this paper is to present a positive solution to the following problem raised by Shpilrain w1x: Problem. Are there two elements u1 , u 2 in Fm such that any endomorphism ␺ of Fm with non-cyclic image is uniquely determined by ␺ Ž u1 ., ␺ Ž u 2 .? ŽIn other words, are there two elements u1 , u 2 in Fm such that whenever ␾ Ž u i . s ␺ Ž u i ., i s 1, 2, for endomorphisms ␾ , ␺ of Fm with non-cyclic images, it follows that ␾ s ␺ ?. In w2x, Ivanov solved in the affirmative this problem in the case where ␺ is a monomorphism of Fm by constructing a so-called C-test word wnŽ x 1 , . . . , x n . for each n G 2. A C-test word is defined due to Ivanov w2x as follows: DEFINITION. A non-empty word ¨ Ž x 1 , . . . , x n . is a C-test word in n letters for Fm if for any two n-tuples Ž X 1 , . . . , X n ., Ž Y1 , . . . , Yn . of elements 509 0021-8693r02 $35.00 䊚 2002 Elsevier Science All rights reserved.

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DONGHI LEE

of Fm the equality ¨ Ž X 1 , . . . , X n . s ¨ Ž Y1 , . . . , Yn . / 1 implies the existence of an element S g Fm such that Yi s SX i Sy1 for all i s 1, 2, . . . , n. According to the result of w2, Corollary 1x, if ¨ is a C-test word in m letters for Fm , ␾ is an endomorphism, ␺ is a monomorphism of Fm , and ␾ Ž ¨ . s ␺ Ž ¨ ., then we have ␾ s ␶ S ( ␺ , where S g Fm is such that ² S, ␺ Ž ¨ .: is cyclic, and ␶ S is the inner automorphism of Fm defined by means of S. Notice that this assertion is no longer true if ␺ is extended to an endomorphism of Fm with non-cyclic image, since ␺ Ž ¨ . / 1 is no longer guaranteed. The efforts to extend the result of w2, Corollary 1x to the case where ␺ is an endomorphism of Fm with non-cyclic image have led us to proving the following THEOREM. For e¨ ery n G 2 there exists a C-test word ¨ nŽ x 1 , . . . , x n . in n letters for Fm with the additional property that ¨ nŽ X 1 , . . . , X n . s 1 if and only if the subgroup ² X 1 , . . . , X n : of Fm generated by X 1 , . . . , X n is cyclic. We construct such a C-test word ¨ nŽ x 1 , . . . , x n . by combining Ivanov’s C-test word w 2 Ž x 1 , x 2 . and an auxiliary word uŽ x 1 , x 2 . defined below. Here, let us recall Ivanov’s C-test word w 2 Ž x 1 , x 2 .: 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

300 y1 x1

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

400 y1 x1

x 18 , x 28

700 y1 x2

x 18 , x 28

800 y1 x2 .

We define an auxiliary word uŽ x 1 , x 2 . as follows:

Ž 1.1.

u Ž x 1 , x 2 . s x 148 , x 240

100

x 16 x 148 , x 240

200

x 16 x 148 , x 240

= x 148 , x 240

400

x 16 x 148 , x 240

500

x 25 x 148 , x 240

= x 148 , x 240

700

x 25 x 148 , x 240

800

x 25 .

We then construct ¨ nŽ x 1 , . . . , x n . as follows: If n s 2 then ¨ 2 Ž x1 , x 2 . s w2 Ž x1 , x 2 . .

Ž 1.2. If n s 3 then

Ž 1.3.

300

¨ 3 Ž x1 , x 2 , x 3 . s u Ž u Ž ¨ 2 Ž x1 , x 2 . , ¨ 2 Ž x 2 , x 3 . . ,

u Ž ¨ 2 Ž x 2 , x 3 . , ¨ 2 Ž x 3 , x1 . . . .

x 16 600

x 25

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511

Inductively, for n G 4, define

Ž 1.4.

¨ n Ž x 1 , . . . , x n . s u Ž u Ž ¨ ny1 Ž x 1 , x 2 , x 3 , . . . , x ny1 . , ¨ ny 1 Ž x ny1 , x 2 , x 3 , . . . , x ny2 , x n . . ,

u Ž ¨ ny 1 Ž x ny1 , x 2 , x 3 , . . . , x ny2 , x n . , ¨ ny 1 Ž x n , x 2 , x 3 , . . . , x ny2 , x 1 . . . .

For instance,

Ž 1.5.

¨ 4 Ž x1 , x 2 , x 3 , x 4 . s u Ž u Ž ¨ 3 Ž x1 , x 2 , x 3 . , ¨ 3 Ž x 3 , x 2 , x 4 . . ,

u Ž ¨ 3 Ž x 3 , x 2 , x 4 . , ¨ 3 Ž x 4 , x 2 x1 . . . . In Section 2, we establish several technical lemmas concerning properties of Ivanov’s word ¨ 2 Ž x 1 , x 2 . and the auxiliary word uŽ x 1 , x 2 . which will be used throughout this paper. In Sections 3᎐5, we prove that, for each n G 3, the word ¨ nŽ x 1 , . . . , x n . constructed above is indeed a C-test word with the property in the statement of the Theorem Žthe case n s 2 is already proved in w2x.. We first treat the case n s 3 in Section 3 and then proceed by simultaneous induction on n with base n s 4 together with several necessary lemmas in Sections 4 and 5. Once the Theorem is proved, our Corollary 1, which is an extended version of w2, Corollary 1x to the case where ␺ is an endomorphism of Fm with non-cyclic image, follows immediately, as intended, by taking u s ¨ m Ž x 1 , . . . , x m .: COROLLARY 1. There exists an element u g Fm such that if ␾ is an endomorphism, ␺ is an endomorphism of Fm with non-cyclic image, and ␾ Ž u. s ␺ Ž u., then ␾ also has non-cyclic image, more precisely, ␾ s ␶ S ( ␺ , where S g Fm is such that ² S, ␺ Ž u.: is cyclic, and ␶ S is the inner automorphism of Fm defined by means of S. In Corollary 2, we provide a positive solution to the above-mentioned Shpilrain problem: COROLLARY 2. There exist two elements u1 , u 2 g Fm such that any endomorphism ␺ of Fm with non-cyclic image is uniquely determined by ␺ Ž u1 ., ␺ Ž u 2 .. The proof of Corollary 2 makes use of the words ¨ mŽ x 1 , . . . , x m . and ¨ mq 1Ž x 1 , . . . , x mq1 .. Its detailed proof is given in Section 6. The idea and the techniques used in w2x are developed further in the present paper.

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2. PRELIMINARY LEMMAS We begin this section by establishing some notation and terminology. We let X, Y Žwith or without subscripts. be words in Fm throughout this paper. By X s Y we denote the equality in Fm of words X and Y, and by X ' Y the graphical Žletter-by-letter . equality of words X and Y. The < s 2.. We say that a length of a word X is denoted by < X < Žnote < x 1 xy1 1 word X is a proper power if X s Y l for some Y with l ) 1 and that a word A is simple if A is non-empty and cyclically reduced and is not a proper power. If A is simple, then an A-periodic word is a subword of Ak with some k ) 0. Now let us introduce several lemmas concerning properties of the word ¨ 2 Ž x 1 , x 2 . defined in Ž1.2.. For proofs of Lemmas 1 and 2, see w2x. LEMMA 1 w2, Lemma 3x. If the subgroup ² X 1 , X 2 : of Fm is non-cyclic, then ¨ 2 Ž X 1 , X 2 . is neither equal to the empty word nor a proper power. If ² X 1 , X 2 : is cyclic, then ¨ 2 Ž X 1 , X 2 . s 1. LEMMA 2 w2, Lemma 4x. If the subgroup ² X 1 , X 2 : of Fm is non-cyclic and ¨ 2 Ž X 1 , X 2 . s ¨ 2 Ž Y1 , Y2 ., then there exists a word Z g Fm such that Y1 s ZX 1 Zy1 LEMMA 3.

and Y2 s ZX 2 Zy1 .

If the subgroup ² X 1 , X 2 : of Fm is non-cyclic, then

¨ 2 Ž X 1 , X 2 .y1 / ¨ 2 Ž Y1 , Y2 . for any words Y1 , Y2 .

Proof. By way of contradiction, suppose that ¨ 2 Ž X 1 , X 2 .y1 s ¨ 2 Ž Y1 , Y2 . for some words Y1 , Y2 . If ² Y1 , Y2 : is cyclic, then it follows from Lemma 1 that ¨ 2 Ž Y1 , Y2 . s 1, so that ¨ 2 Ž X 1 , X 2 . s 1, i.e., ² X 1 , X 2 : is cyclic. This contradiction to the hypothesis of the lemma allows us to assume that ² Y1 , Y2 : is non-cyclic. As in w2, Lemmas 1᎐4x, let W be a cyclically reduced word that is conjugate to ¨ 2 Ž X 1 , X 2 ., and let B be a simple word such that w X 18 , X 28 x is conjugate to B Žrecall from w3, 4x that a commutator w X, Y x of words X and Y is not a proper power.. Then according to w2, Lemma 2x, W has the form W ' R1T1 R 2 T2 ⭈⭈⭈ R 8 T8 , where R i ’s are B-periodic words with Ž i ⭈ 100 y 14.< B < - < R i < F i ⭈ 100 < B <, 0 F < Ti < - 6 < B <, and 3488 < B < - < W < - 3648 < B <. The same holds for ¨ 2 Ž Y1 , Y2 ., and we attach the prime sign ⬘ to the notations for ¨ 2 Ž Y1 , Y2 .. Then the equality ¨ 2 Ž X 1 , X 2 .y1 s ¨ 2 Ž Y1 , Y2 . yields that Wy1 is a cyclic permutation of W X , so that < Wy1 < s < W X <. At this point, apply the arguments in w2, Lemma 4x to Wy1 , W X to get By1 ' BX and that RXi BX-overlaps only with Ry1 for each i s 1, 2, . . . , 8. But this is impossible, for if RXi i

ON CERTAIN C-TEST WORDS FOR FREE GROUPS

513

X X BX-overlapped only with Ry1 i , then R iq1 would have to B -overlap only y1 y1 y1 Ž with R iy1 indices modulo 8. by the order of indices in W ' Ty1 8 R 8 ⭈⭈⭈ X X X X X X X y1 y1 y1 y1 T2 R 2 T1 R1 and W ' R1T1 R 2 T2 ⭈⭈⭈ R 8 T8 .

We also establish several lemmas concerning properties of the auxiliary word uŽ x 1 , x 2 . defined in Ž1.1.. LEMMA 4. If the subgroup ² X 1 , X 2 : of Fm is non-cyclic, then uŽ X 1 , X 2 . is neither equal to the empty word nor a proper power. If ² X 1 , X 2 : is cyclic, then either uŽ X 1 , X 2 . is equal to the empty word pro¨ ided X 16 s Xy5 or 2 otherwise uŽ X 1 , X 2 . is a proper power. Proof. The proof of the first part is similar to that of w2, Lemma 3x, and the second part is immediate from definition Ž1.1. of uŽ x 1 , x 2 .. LEMMA 5. If the subgroup Ž X 1 , X 2 . of Fm is non-cyclic and uŽ X 1 , X 2 . s Ž u Y1 , Y2 ., then there exists a word Z g Fm such that Y1 s ZX 1 Zy1

and Y2 s ZX 2 Zy1 .

Proof. Applying the same argument as in w2, Lemma 4x to uŽ X 1 , X 2 . and uŽ Y1 , Y2 ., we deduce that Y16 s ZX 16 Zy1 and Y25 s ZX 25 Zy1 for some word Z g Fm . Since extraction of roots is unique in a free group, it follows that Y1 s ZX 1 Zy1 and Y2 s ZX 2 Zy1, as required. LEMMA 6. If the subgroup ² X 1 , X 2 : of Fm is non-cyclic, then uŽ X 1 , X 2 .y1 / uŽ Y1 , Y2 . for any words Y1 , Y2 . Proof. Suppose to the contrary that uŽ X 1 , X 2 .y1 s uŽ Y1 , Y2 . for some words Y1 , Y2 . Then the subgroup ² Y1 , Y2 : of Fm is non-cyclic, for otherwise the equality uŽ X 1 , X 2 .y1 s uŽ Y1 , Y2 . would yield that uŽ X 1 , X 2 . s ., contrary to Lemma 4. From here on, follow the uŽ Y1 , Y2 .y1 s uŽ Y1y1 , Yy1 2 proof of Lemma 3 to arrive at a contradiction. LEMMA 7. If the subgroup ² X 1 , X 2 : of Fm is non-cyclic, then, for each i s 1, 2, the subgroup ² X i , uŽ X 1 , X 2 .. of Fm is also non-cyclic. Proof. Suppose on the contrary that ² X i , uŽ X 1 , X 2 .: is cyclic for some i s 1, 2. Then by Lemma 4 we have

Ž 2.1.

X i s u Ž X1 , X 2 .

l

for some non-zero integer l . Let U be a cyclically reduced word that is conjugate to uŽ X 1 , X 2 ., and let C be a simple word such that w X 148 , X 240 x is conjugate to C. By X, we denote a cyclically reduced word that is conjugate to a word X in Fm . Then by the same argument as in w2, Lemma 1x, max Ž < X 1< , < X 2< . - 6 < C < ,

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DONGHI LEE

and also by the same argument as in w2, Lemma 2x, U has the form U ' Q1 S1 Q 2 S2 ⭈⭈⭈ Q 8 S8 , where Q i are C-periodic words with Ž i ⭈ 100 y 14.< C < - < Q i < F i ⭈ 100 < C <, 0 F < Si < - 6 < C <, and 3488 < C < - < U < - 3648 < C <. This yields that < X i< - 6 < C < - 3488 < C < - < U < s u Ž X 1 , X 2 . F u Ž X 1 , X 2 . l , which contradicts equality Ž2.1.. In the following lemma, which will be useful in Sections 4᎐6, the word ¨ nŽ x 1 , . . . , x n . with n G 3 is defined in Ž1.3. ᎐ Ž1.4..

LEMMA 8. If both ¨ 2 Ž X 1 , X 2 . and ¨ nŽ Y1 , . . . , Yn . with n G 3 are neither equal to the empty word nor proper powers, then the subgroup ² ¨ 2 Ž X 1 , X 2 ., ¨ nŽ Y1 , . . . , Yn .: of Fm is non-cyclic. Proof. Suppose on the contrary that ² ¨ 2 Ž X 1 , X 2 ., ¨ nŽ Y1 , . . . , Yn .: is cyclic. It then follows from the hypothesis of the lemma that either

¨ 2 Ž X 1 , X 2 . s ¨ n Ž Y1 , . . . , Yn .

¨ 2 Ž X1 , X 2 .

y1

or

s ¨ n Ž Y1 , . . . , Yn . .

This implies, by definitions Ž1.3. ᎐ Ž1.4. of ¨ nŽ x 1 , . . . , x n ., the existence of words Z1 , Z2 in Fm such that either

¨ 2 Ž X 1 , X 2 . s u Ž Z1 , Z 2 .

or

¨ 2 Ž X1 , X 2 .

y1

s u Ž Z1 , Z 2 . .

But since an argument similar to that in Lemma 3 shows that the latter equality cannot hold, the former must hold. From here on, follow the argument in w2, Lemma 4x to obtain that By␣ X˜1 B ␣ s Z˜16

and

␣ ˜6 By ␣ X˜y1 1 B s Z1 ,

where X˜i , Z˜i are conjugates of X i , Zi , respectively, and B is a simple word such that B s w X˜18 , X˜28 x. This yields X˜12 s 1, i.e., X˜1 s 1, contrary to the hypothesis ¨ 2 Ž X 1 , X 2 . / 1. 3. THE CASE n s 3 In this section, we prove that ¨ 3 Ž x 1 , x 2 , x 3 . is a C-test word with the additional property that ¨ 3 Ž X 1 , X 2 , X 3 . s 1 if and only if the subgroup ² X 1 , X 2 , X 3 : of Fm is cyclic. We begin with lemmas that play crucial roles in proving this assertion.

ON CERTAIN C-TEST WORDS FOR FREE GROUPS

LEMMA 9. ¨ 2 Ž Y1 , Y2 . s 1.

515

If uŽ ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž Y1 , Y2 .. s 1, then ¨ 2 Ž X 1 , X 2 . s 1 and

Proof. The hypothesis of the lemma implies by Lemma 4 that

¨ 2 Ž X 1 , X 2 . 6 s ¨ 2 Ž Y1 , Y2 .y5 ; hence if one of ¨ 2 Ž X 1 , X 2 . and ¨ 2 Ž Y1 , Y2 . is equal to the empty word, then so is the other. So assume that ¨ 2 Ž X 1 , X 2 . / 1 and ¨ 2 Ž Y1 , Y2 . / 1. Notice that the equality ¨ 2 Ž X 1 , X 2 . 6 s ¨ 2 Ž Y1 , Y2 .y5 implies that ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž Y1 , Y2 .: is cyclic. Hence, in view of Lemmas 1 and 3, we have ¨ 2 Ž X 1 , X 2 . s ¨ 2 Ž Y1 , Y2 .. This together with ¨ 2 Ž X 1 , X 2 . 6 s ¨ 2 Ž Y1 , Y2 .y5 yields that ¨ 2 Ž X 1 , X 2 . s ¨ 2 Ž Y1 , Y2 . s 1, con-

trary to our assumption. LEMMA 10. Suppose that the subgroup ² X 1 , X 2 , X 3 : of Fm is non-cyclic. Then ¨ 3 Ž X 1 , X 2 , X 3 . / 1. Futhermore, either ¨ 3 Ž X 1 , X 2 , X 3 . is not a proper power or it has one of the following three forms:

Ž A1.

¨ 3 Ž X1 , X 2 , X 3 . s ¨ 2 Ž X 2 , X 3 .

960

and X 1 s 1;

Ž A2.

¨ 3 Ž X1 , X 2 , X 3 . s ¨ 2 Ž X 3 , X1 .

400

and X 2 s 1;

Ž A3.

¨ 3 Ž X1 , X 2 , X 3 . s ¨ 2 Ž X1 , X 2 .

576

and X 3 s 1.

Remark. In view of Lemma 1, ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 ., ¨ 2 Ž X 1 , X 2 . in ŽA1., ŽA2., ŽA3., respectively, are neither equal to the empty word nor proper powers. Proof. Recall from Ž1.3. that ¨ 3 Ž X1 , X 2 , X 3 . s u Ž u Ž ¨ 2 Ž X1 , X 2 . , ¨ 2 Ž X 2 , X 3 . . ,

u Ž ¨ 2 Ž X 2 , X 3 . , ¨ 2 Ž X 3 , X1 . . . . In the case where the subgroup ² uŽ ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .., uŽ ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 ..: of Fm is non-cyclic, the assertion that ¨ 3 Ž X 1 , X 2 , X 3 . is neither equal to the empty word nor a proper power, as desired, follows immediately from Lemma 4. So we only need to consider the case where

Ž 3.1.

the subgroup ² u Ž ¨ 2 Ž X 1 , X 2 . , ¨ 2 Ž X 2 , X 3 . . , u Ž ¨ 2 Ž X 2 , X 3 . , ¨ 2 Ž X 3 , X 1 . .: is cyclic.

Here, if uŽ ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .. s uŽ ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .. s 1, then Lemma 9 implies that ¨ 2 Ž X 1 , X 2 . s ¨ 2 Ž X 2 , X 3 . s ¨ 2 Ž X 3 , X 1 . s 1; hence, by Lemma 1, ² X 1 , X 2 :, ² X 2 , X 3 :, and ² X 3 , X 1 : are all cyclic. This yields that ² X 1 , X 2 , X 3 : is cyclic, contrary to the hypothesis of the lemma. Thus, at least one of the words uŽ ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .. and uŽ ¨ 2 Ž X 2 , X 3 .,

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DONGHI LEE

¨ 2 Ž X 3 , X 1 .. has to be not equal to the empty word. We divide this situation

into three cases. Case I. s 1.

uŽ ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .. / 1 and uŽ ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 ..

It follows from uŽ ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .. s 1 and Lemma 9 that ¨ 2 Ž X 2 , X 3 . s ¨ 2 Ž X 3 , X 1 . s 1; so, by Lemma 1, ² X 2 , X 3 : and ² X 3 , X 1 : are cyclic. Since ² X 1 , X 2 , X 3 : is non-cyclic, X 3 must be equal to the empty word; hence we have 24

¨ 3 Ž X1 , X 2 , X 3 . s u Ž u Ž ¨ 2 Ž X1 , X 2 . , 1. , 1. s u ¨ 2 Ž X1 , X 2 . , 1

Ž

s ¨ 2 Ž X1 , X 2 .

24⭈24

s ¨ 2 Ž X1 , X 2 .

576

.

.

Therefore ¨ 3 Ž X 1 , X 2 , X 3 . has form ŽA3. in this case. Case II. / 1.

uŽ ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .. s 1 and uŽ ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 ..

Since uŽ ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .. s 1, we have, by Lemmas 1 and 9, that ² X 1 , X 2 : and ² X 2 , X 3 : are cyclic, so that X 2 s 1; thus ¨ 3 Ž X 1 , X 2 , X 3 . s u Ž 1, u Ž 1, ¨ 2 Ž X 3 , X 1 . . . s u 1, ¨ 2 Ž X 3 , X 1 .

Ž

s ¨ 2 Ž X 3 , X1 .

20⭈20

s ¨ 2 Ž X 3 , X1 .

400

20

.

.

Therefore ¨ 3 Ž X 1 , X 2 , X 3 . has form ŽA2. in this case. Case III. / 1.

uŽ ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .. / 1 and uŽ ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 ..

In this case, we want to prove CLAIM. This case is reduced to the following two cases: Ži. ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: is cyclic; Žii. both ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .: and ² ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: are non-cyclic. Proof of the Claim. Assuming at least one of ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .: and ² ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: is cyclic, we want to show that Case Ži. occurs. Let us say that ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .: is cyclic Žthe case where ² ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: is cyclic is analogous.. If ¨ 2 Ž X 2 , X 3 . s 1, then uŽ ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .. s ¨ 2 Ž X 1 , X 2 . 24 and uŽ ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .. s ¨ 2 Ž X 3 , X 1 . 20 . It then follows from Ž3.1. that ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 3 , X 1 .: is cyclic, which means that Case Ži. occurs. Now let ¨ 2 Ž X 2 , X 3 . / 1. Then since ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .: is cyclic, Ž3.1. yields by Lemma 7 that

ON CERTAIN C-TEST WORDS FOR FREE GROUPS

517

² ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: is also cyclic; hence ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: is cyclic; that is, Case Ži. occurs as well. We divide Case Ži. further into subcases according to the number of non-empty words among ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 ., and ¨ 2 Ž X 3 , X 1 .. Here, we note that if there exists only one non-empty word, then it has to be ¨ 2 Ž X 2 , X 3 ., for otherwise we would have a contradiction of the hypothesis of Case III. Therefore, Case III is decomposed into the following six subcases. Case III.1. ¨ 2 Ž X 1 , X 2 . s ¨ 2 Ž X 3 , X 1 . s 1 and ¨ 2 Ž X 2 , X 3 . / 1. In this case, it follows from Lemma 1 that ² X 1 , X 2 : and ² X 3 , X 1 : are cyclic, so that X 1 s 1; hence we have ¨ 3 Ž X 1 , X 2 , X 3 . s u Ž u Ž 1, ¨ 2 Ž X 2 , X 3 . . , u Ž ¨ 2 Ž X 2 , X 3 . , 1 . . 20

s u Ž ¨ 2 Ž X2 , X3 . , ¨ 2 Ž X2 , X3 . s ¨ 2 Ž X2 , X3 .

20⭈24q24⭈20

24

.

s ¨ 2 Ž X2 , X3 .

960

.

Thus, ¨ 3 Ž X 1 , X 2 , X 3 . has form ŽA1.. Case III.2. ¨ 2 Ž X 1 , X 2 . s 1 and ²1 / ¨ 2 Ž X 2 , X 3 ., 1 / ¨ 2 Ž X 3 , X 1 .: is cyclic. Since ¨ 2 Ž X 1 , X 2 . s 1, we have by Lemma 1 that ² X 1 , X 2 : is cyclic. Also since ²1 / ¨ 2 Ž X 2 , X 3 ., 1 / ¨ 2 Ž X 3 , X 1 .: is cyclic, we have 1 / ¨ 2 Ž X 2 , X 3 . s ¨ 2 Ž X 3 , X 1 . by Lemmas 1 and 3. Apply Lemma 2 to this equality: there exists a word S g Fm such that X 2 s SX 3 Sy1

and

X 3 s SX 1 Sy1 ,

which yields that Sy1 X 2 S s SX 1 Sy1 , so that X 2 s S 2 X 1 Sy2 . It then follows from ² X 1 , X 2 : being cyclic that ² S, X 1 , X 2 : is cyclic. This together with the equality X 3 s SX 1 Sy1 implies that ² X 1 , X 2 , X 3 : is cyclic, contrary to the hypothesis of the lemma. Therefore this case cannot occur. Case III.3. ¨ 2 Ž X 2 , X 3 . s 1 and ²1 / ¨ 2 Ž X 1 , X 2 ., 1 / ¨ 2 Ž X 3 , X 1 .: is cyclic. Repeat an argument similar to that in Case III.2 to conclude that this case cannot occur. Case III.4. ¨ 2 Ž X 3 , X 1 . s 1 and ²1 / ¨ 2 Ž X 1 , X 2 ., 1 / ¨ 2 Ž X 2 , X 3 .: is cylcic. Also repeat an argument similar to that in Case III.2 to conclude that this case cannot occur.

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Case III.5. ²1 / ¨ 2 Ž X 1 , X 2 ., 1 / ¨ 2 Ž X 2 , X 3 ., 1 / ¨ 2 Ž X 3 , X 1 .: is cyclic. In this case, it follows from Lemmas 1 and 3 that

Ž 3.2.

1 / ¨ 2 Ž X1 , X 2 . s ¨ 2 Ž X 2 , X 3 . s ¨ 2 Ž X 3 , X1 . .

Applying Lemma 2 to these equalities, we have the existence of words T1 and T2 in Fm such that

Ž 3.3.

X 1 s T1 X 2 T1y1 ,

X 2 s T1 X 3T1y1 ;

X 2 s T2 X 3 Ty1 2 ,

X 3 s T2 X 1Ty1 2 .

Combining Ž3.2. and Ž3.3. yields that ²T1y1 T2 , X 3 :, ²T1 , ¨ 2 Ž X 2 , X 3 .:, and ²T2 , ¨ 2 Ž X 3 , X 1 .: are all cyclic. At this point, we apply Ivanov’s argument Žsee w2, pp. 403᎐404x. to obtain the following CLAIM ŽIvanov.. T1 s T2 . Proof of the Claim. Suppose on the contrary that T1 / T2 . It then follows from ²T1y1 T2 , X 3 : being cyclic that X 3l 1 s Ž T1y1 T2 .

Ž 3.4.

l2

with nonzero integers l 1 and l 2 . It also follows from ²T1 , ¨ 2 Ž X 2 , X 3 .: and ²T2 , ¨ 2 Ž X 3 , X 1 .: being cyclic that T1 s ¨ 2 Ž X 2 , X 3 .

l3

T2 s ¨ 2 Ž X 3 , X 1 .

and

l4

,

with integers l 3 and l 4 at least one of which is non-zero, so that T1y1 T2 s ¨ 2 Ž X 2 , X 3 .

yl 3

¨ 2 Ž X 3 , X1 .

l4

.

Hence, by Ž3.4.,

Ž 3.5.

X 3l 1 s ¨ 2 Ž X 2 , X 3 .

yl 3

¨ 2 Ž X 3 , X1 .

l4 l 2

.

Note, by definition Ž1.2. of ¨ 2 Ž x 1 , x 2 ., that the right-hand side of equality Ž3.5. belongs to the subgroup w Fm , N3 x, where N3 is the normal closure in Fm of the word X 3 . So inside the relation module Nˆ3 s N3rw N3 , N3 x of the one-relator group G s ² x1 , . . . , x m 5 X3 : , equality Ž3.5. can be expressed as

Ž l 1 y P . ⭈ Xˆ3 s 0,

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where P is an element of the augmentation ideal of the group ring ⺪Ž G . of G over the integers and Xˆ3 is the canonical generator of the relation module Nˆ3 of G. By Lyndon’s result on the relation module R of a one-relator group ² x 1 , . . . , x m 5 R : Žsee w4, 5x., which says that if Q ⭈ Rˆ s 0 in R then Q is an element of the augmentation ideal of ⺪Ž² x 1 , . . . , x m 5 R :., we must have l 1 s 0. This contradiction of l 1 s 0 completes the proof of the claim. If T1 s T2 s 1, then equalities Ž3.3. yield that X 1 s X 2 s X 3 , contrary to the hypothesis of the lemma. If T1 s T2 / 1, then we derive from equalities Ž3.3. that X 1 s T13 X 1T1y3 ,

X 2 s T13 X 2 T1y3 , and

X 3 s T13 X 3T1y3 ,

so that ²T1 , X 1 :, ²T1 , X 2 :, and ²T1 , X 3 : are all cyclic; therefore, ² X 1 , X 2 , X 3 : is cyclic. A contradiction implies that this case cannot occur. Case III.6. Both ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .: and ² ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: are non-cyclic. In this case, in view of Ž3.1. and Lemmas 4 and 6, we have that u Ž ¨ 2 Ž X1 , X 2 . , ¨ 2 Ž X 2 , X 3 . . s u Ž ¨ 2 Ž X 2 , X 3 . , ¨ 2 Ž X 3 , X1 . . , and so by Lemma 5 there exists a word T g Fm such that

Ž 3.6.

1 / ¨ 2 Ž X 1 , X 2 . s T¨ 2 Ž X 2 , X 3 . Ty1

and

1 / ¨ 2 Ž X 2 , X 3 . s T¨ 2 Ž X 3 , X 1 . Ty1 .

Apply Lemma 2 to these equalities: there exist words U1 and U2 in Fm such that

Ž 3.7.

X 1 s U1 X 2 U1y1 ,

X 2 s U1 X 3 U1y1 ;

X 2 s U2 X 3 U2y1 ,

X 3 s U2 X 1U2y1 .

Combining the equalities in Ž3.6. and Ž3.7., we deduce that ²U1y1 U2 , X 3 :, ²Ty1 U1 , ¨ 2 Ž X 2 , X 3 .:, and ²Ty1 U2 , ¨ 2 Ž X 3 , X 1 .: are cyclic. Here, apply Ivanov’s argument used in Case III.5 to get U1 s U2 . Then reasoning as in Case III.5, we conclude that this case cannot occur. The proof of Lemma 10 is complete. Now we are ready to prove the Theorem for the case n s 3. Proof of the Theorem Ž n s 3.. The additional property that ¨ 3 Ž X 1 , X 2 , X 3 . s 1 if and only if the subgroup ² X 1 , X 2 , X 3 : of Fm is cyclic follows immediately from definition Ž1.3. of ¨ 3 Ž x 1 , x 2 , x 3 . and Lemma 10.

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So we only need to prove that ¨ 3 Ž x 1 , x 2 , x 3 . is a C-test word; that is, supposing 1 / ¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž Y1 , Y2 , Y3 ., we want to prove the existence of a word Z g Fm such that Yi s ZX i Zy1

for all i s 1, 2, 3.

We begin by distinguishing two cases according to whether ¨ 3 Ž X 1 , X 2 , X 3 . is a proper power or not. Case I. ¨ 3 Ž X 1 , X 2 , X 3 . is a proper power. Applying Lemma 10 to ¨ 3 Ž X 1 , X 2 , X 3 . and ¨ 3 Ž Y1 , Y2 , Y3 ., we have that ¨ 3 Ž X 1 , X 2 , X 3 . has one of three types ŽA1., ŽA2., and ŽA3.; besides, by the equality ¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž Y1 , Y2 , Y3 ., ¨ 3 Ž Y1 , Y2 , Y3 . has the same type as ¨ 3 Ž X 1 , X 2 , X 3 . does, because the exponents in ŽA1., ŽA2., and ŽA3. are all distinct. This gives us only three possibilities, ŽA1. & ŽA1., ŽA2. & ŽA2., and ŽA3. & ŽA3., for the types of ¨ 3 Ž X 1 , X 2 , X 3 . & ¨ 3 Ž Y1 , Y2 , Y3 .. If ¨ 3 Ž X 1 , X 2 , X 3 . & ¨ 3 Ž Y1 , Y2 , Y3 . is of type ŽA1. & ŽA1. ŽŽA2. & ŽA2. or ŽA3. & ŽA3. is similar., then X 1 s Y1 s 1 and 1 / ¨ 2 Ž X 2 , X 3 .

960

s ¨ 2 Ž Y2 , Y3 .

960

.

Applying Lemma 2 to the equality 1 / ¨ 2 Ž X 2 , X 3 . s ¨ 2 Ž Y2 , Y3 ., we have that two 2-tuples Ž X 2 , X 3 . and Ž Y2 , Y3 . are conjugate in Fm , which together with X 1 s Y1 s 1 yields that two 3-tuples Ž X 1 , X 2 , X 3 . and Ž Y1 , Y2 , Y3 . are conjugate in Fm , as desired. Case II. ¨ 3 Ž X 1 , X 2 , X 3 . is not a proper power. In this case, it follows from Lemma 4 that

Ž 3.8. the subgroup ² u Ž ¨ 2 Ž X1 , X 2 . , ¨ 2 Ž X 2 , X 3 . . , u Ž ¨ 2 Ž X 2 , X 3 . , ¨ 2 Ž X 3 , X 1 . .: of Fm is non-cyclic. This enables us to apply Lemma 5 to the equality ¨ 3 Ž X 1 , X 2 , X 3 . s

¨ 3 Ž Y1 , Y2 , Y3 .: for some word S g Fm , we have

1 / u Ž ¨ 2 Ž X1 , X 2 . , ¨ 2 Ž X 2 , X 3 . .

Ž 3.9.

s Su Ž ¨ 2 Ž Y1 , Y2 . , ¨ 2 Ž Y2 , Y3 . . Sy1 , 1 / u Ž ¨ 2 Ž X 2 , X 3 . , ¨ 2 Ž X 3 , X1 . . s Su Ž ¨ 2 Ž Y2 , Y3 . , ¨ 2 Ž Y3 , Y1 . . Sy1 .

Here, we consider two subcases.

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Case II.1. One of ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .: and ² ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: is non-cyclic. Let us say that ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .: is non-cyclic Žthe case where ² ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: is non-cyclic is analogous.. Then, by Lemma 5, the first equality of Ž3.9. implies the existence of a word W g Fm such that

Ž 3.10.

1 / ¨ 2 Ž X 1 , X 2 . s W¨ 2 Ž Y1 , Y2 . Wy1

and

1 / ¨ 2 Ž X 2 , X 3 . s W¨ 2 Ž Y2 , Y3 . Wy1 .

Apply Lemma 2 to these equalities: there exist words T1 and T2 in Fm such that

Ž 3.11.

X 1 s T1Y1T1y1 ,

X 2 s T1Y2 T1y1 ;

X 2 s T2 Y2 Ty1 2 ,

X 3 s T2 Y3 Ty1 2 .

Now applying Ivanov’s argument introduced in Case III.5 of Lemma 10 to equalities Ž3.10. ᎐ Ž3.11., we get T1 s T2 , by which Ž3.11. yields the desired result. Case II.2. Both ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 .: and ² ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: are cyclic. In this case, if ¨ 2 Ž X 2 , X 3 . / 1, then the subgroup ² ¨ 2 Ž X 1 , X 2 ., ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž X 3 , X 1 .: would be cyclic, contrary to Ž3.8.. So ¨ 2 Ž X 2 , X 3 . must be equal to the empty word. On the other hand, equalities Ž3.9. imply by Lemma 4 that both ² ¨ 2 Ž Y1 , Y2 ., ¨ 2 Ž Y2 , Y3 .: and ² ¨ 2 Ž Y2 , Y3 ., ¨ 2 Ž Y3 , Y1 .: are also cyclic. Then, for the same reason as with ¨ 2 Ž X 2 , X 3 ., ¨ 2 Ž Y2 , Y3 . also has to be equal to the empty word. Thus, it follows from Ž3.9. that 1 / ¨ 2 Ž X1 , X 2 .

24

s S¨ 2 Ž Y1 , Y2 . Sy1

1 / ¨ 2 Ž X 3 , X1 .

24

20

and

s S¨ 2 Ž Y3 , Y1 . Sy1 , 20

that is, 1 / ¨ 2 Ž X 1 , X 2 . s S¨ 2 Ž Y1 , Y2 . Sy1

and

1 / ¨ 2 Ž X 3 , X 1 . s S¨ 2 Ž Y3 , Y1 . Sy1 , which is a situation similar to Ž3.10.. So from here on, we can follow the proof of Case II.1 to obtain the desired result. The proof of the Theorem for the case n s 3 is complete.

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4. THE BASE n s 4 OF SIMULTANEOUS INDUCTION In this section, we prove the base step n s 4 of simultaneous induction which we use in Lemmas 11᎐13 and the Theorem. LEMMA 11 Ž n s 4.. If both ¨ 2 Ž X 1 , X 2 . and ¨ 3 Ž Y1 , Y2 , Y3 . are neither equal to the empty word nor proper powers, then ² ¨ 2 Ž X 1 , X 2 ., ¨ 3 Ž Y1 , Y2 , Y3 .: is non-cyclic. Proof. This is a special case of Lemma 8. LEMMA 12 Ž n s 4.. If uŽ ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž Y1 , Y2 , Y3 .. s 1, ¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž Y1 , Y2 , Y3 . s 1. Proof. By Lemma 4, the hypothesis of the lemma implies that

Ž 4.1.

6

¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž Y1 , Y2 , Y3 .

y5

,

so that if one of ¨ 3 Ž X 1 , X 2 , X 3 . or ¨ 3 Ž Y1 , Y2 , Y3 . is equal to the empty word, then so is the other. Hence assume that ¨ 3 Ž X 1 , X 2 , X 3 . / 1 and ¨ 3 Ž Y1 , Y2 , Y3 . / 1. If one of ¨ 3 Ž X 1 , X 2 , X 3 . or ¨ 3 Ž Y1 , Y2 , Y3 . is a proper power, then so is the other by Ž4.1., because 6 and 5 are relatively prime. Then by Lemma 10 ¨ 3 Ž X 1 , X 2 , X 3 . & ¨ 3 Ž Y1 , Y2 , Y3 . has nine possible types, and we can easily check that 6 times any of 960, 400, or 576 never equals 5 times any of these, which means that equality Ž4.1. cannot hold in any case, a contradiction. If neither ¨ 3 Ž X 1 , X 2 , X 3 . nor ¨ 3 Ž Y1 , Y2 , Y3 . is a proper power, then by Lemma 6 we have ¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž Y1 , Y2 , Y3 ., because Ž4.1. implies that ² ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž Y1 , Y2 , Y3 .: is cyclic. This equality together with Ž4.1. yields that ¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž Y1 , Y2 , Y3 . s 1, contrary to our assumption. This completes the proof. LEMMA 13 Ž n s 4.. Suppose that ² X 1 , X 2 , X 3 , X 4 : is non-cyclic. Then

¨ 4 Ž X 1 , X 2 , X 3 , X 4 . / 1. Furthermore, either ¨ 4 Ž X 1 , X 2 , X 3 , X 4 . is not a

proper power or it has one of the following four forms: 960⭈400 and X 1 s X 2 s 1; Ž B1 . ¨ 4 Ž X1 , X 2 , X 3 , X4 . s ¨ 2 Ž X4 , X 3 .

Ž B2 .

¨ 4 Ž X1 , X 2 , X 3 , X4 . s ¨ 2 Ž X1 , X4 .

400 2

and X 2 s X 3 s 1;

576⭈400 and X 2 s X 4 s 1; Ž B3 . ¨ 4 Ž X1 , X 2 , X 3 , X4 . s ¨ 2 Ž X 3 , X1 .

Ž B4 .

¨ 4 Ž X1 , X 2 , X 3 , X4 . s ¨ 2 Ž X1 , X 2 , X 3 .

X 1 s X 3 s X 4 / 1.

1936

and

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523

Remark. In view of Lemmas 1 and 10, ¨ 2 Ž X 4 , X 3 ., ¨ 2 Ž X 1 , X 4 ., ¨ 2 Ž X 3 , X 1 ., and ¨ 3 Ž X 1 , X 2 , X 3 . in ŽB1., ŽB2., ŽB3., and ŽB4., respectively, are neither equal to the empty word nor proper powers. Proof. Recall from Ž1.5. that u 4 Ž X1 , X 2 , X 3 , X4 . s u Ž u Ž ¨ 3 Ž X1 , X 2 , X 3 . , ¨ 3 Ž X 3 , X 2 , X4 . . , u Ž ¨ 3 Ž X 3 , X 2 , X4 . , ¨ 3 Ž X4 , X 2 , X1 . . . . If ² uŽ ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .., uŽ ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 ..: is non-cyclic, then the assertion that ¨ 4 Ž X 1 , X 2 , X 3 , X 4 . is neither equal to the empty word nor a proper power, as desired, follows directly from Lemma 4. So we only need to consider the case where

Ž 4.2.

² u Ž ¨ 3 Ž X1 , X 2 , X 3 . , ¨ 3 Ž X 3 , X 2 , X4 . . , u Ž ¨ 3 Ž X 3 , X 2 , X 4 . , ¨ 3 Ž X 4 , X 2 , X 1 . .:

is cyclic.

Here, to avoid a contradiction of the hypothesis that ² X 1 , X 2 , X 3 , X 4 : is non-cyclic, at least one of uŽ ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .. or uŽ ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .. has to be not equal to the empty word. So we have three cases to consider. Case I. uŽ ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .. / 1 and uŽ ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .. s 1. In this case, we have, by Lemma 12 Ž n s 4. and the Theorem Ž n s 3., that both ² X 3 , X 2 , X 4 : and ² X 4 , X 2 , X 1 : are cyclic. Since ² X 1 , X 2 , X 3 , X 4 : is non-cyclic, X 2 and X 4 must be equal to the empty word; hence we have ¨ 4 Ž X1 , X 2 , X 3 , X4 . s u Ž u Ž ¨ 3 Ž X1 , X 2 , X 3 . , 1. , 1 . 24

s u Ž ¨ 3 Ž X1 , X 2 , X 3 . , 1 . s ¨ 3 Ž X1 , X 2 , X 3 . s ¨ 2 Ž X 3 , X1 .

576.400

24⭈24

by form Ž A2. .

Therefore, ¨ 4Ž X 1 , X 2 , X 3 , X 4 . has form ŽB3. in this case. Case II. uŽ ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .. s 1 and uŽ ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .. / 1. In this case, by Lemma 12 Ž n s 4. and the Theorem Ž n s 3., we have that both ² X 1 , X 2 , X 3 : and ² X 3 , X 2 , X 4 : are cyclic, so that X 2 s X 3 s 1;

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thus ¨ 4 Ž X 1 , X 2 , X 3 , X 4 . s u Ž 1, u Ž 1, ¨ 3 Ž X 4 , X 2 , X 1 . . .

s u Ž 1, ¨ 3 Ž X 4 , X 2 , X 1 . s ¨ 2 Ž X1 , X4 .

400⭈400

20

. s ¨ 3 Ž X4 , X2 , X1 . 20⭈20

by form Ž A2. .

Therefore, ¨ 4Ž X 1 , X 2 , X 3 , X 4 . has form ŽB2. in this case. Case III. uŽ ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .. / 1 and uŽ ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .. / 1. By reasoning as in Case III of Lemma 10, we break this case into the following six subcases. Case III.1. ¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž X 4 , X 2 , X 1 . s 1 and ¨ 3 Ž X 3 , X 2 , X 4 . / 1. It follows from the Theorem Ž n s 3. that ² X 1 , X 2 , X 3 : and ² X 4 , X 2 , X 1 : are cyclic, so that X 1 s X 2 s 1; hence we have ¨ 4 Ž X 1 , X 2 , X 3 , X 4 . s u Ž u Ž 1, ¨ 3 Ž X 3 , X 2 X 4 . . , u Ž ¨ 3 Ž X 3 , X 2 , X 4 . , 1 . . 20

s u Ž ¨ 3 Ž X 3 , X 2 , X4 . , ¨ 3 Ž X 3 , X 2 , X4 . s ¨ 3 Ž X 3 , X 2 , X4 . s ¨ 2 Ž X4 , X 3 .

24

.

20⭈24q24⭈20

960⭈400

by form Ž A2. .

Thus, ¨ 4Ž X 1 , X 2 , X 3 , X 4 . has form ŽB1.. Case III.2. ¨ 3 Ž X 1 , X 2 , X 3 . s 1 and ²1 / ¨ 3 Ž X 3 , X 2 , X 4 ., 1 / ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic. Since ¨ 3 Ž X 1 , X 2 , X 3 . s 1, we have, by the Theorem Ž n s 3., that

Ž 4.3.

² X 1 , X 2 , X 3 : is cyclic.

Also since ²1 / ¨ 3 Ž X 3 , X 2 , X 4 ., 1 / ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic, in view of Lemmas 10 and 11 Ž n s 4., this case is reduced to the following two cases: Ži. both ¨ 3 Ž X 3 , X 2 , X 4 . and ¨ 3 Ž X 4 , X 2 , X 1 . are proper powers; Žii. neither ¨ 3 Ž X 3 , X 2 , X 4 . nor ¨ 3 Ž X 4 , X 2 , X 1 . is a proper power. Case Ži. is divided further into subcases according to the types of ¨ 3 Ž X 3 , X 2 , X 4 . & ¨ 3 Ž X 4 , X 2 , X 1 . by Lemma 10. Of nine possible types of ¨ 3 Ž X 3 , X 2 , X 4 . & ¨ 3 Ž X 4 , X 2 , X 1 ., ŽA3. & ŽA1., ŽA3. & ŽA2., and ŽA3. & ŽA3. cannot occur, for if ¨ 3 Ž X 3 , X 2 , X 4 . were of type ŽA3., then X 4 s 1,

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525

which together with Ž4.3. yields a contradiction of the hypothesis of the lemma. Also, ŽA1. & ŽA1. and ŽA2. & ŽA1. cannot occur, for if ¨ 3 Ž X 4 , X 2 , X 1 . were of type ŽA1., then X 4 s 1, again a contradiction. Moreover, ŽA1. & ŽA2. cannot occur, for this implies that X 3 s X 2 s 1, so that ¨ 3 Ž X 3 , X 2 , X 4 . s 1, a contradiction. Also, ŽA2. & ŽA3. cannot occur, for this implies that X 2 s X 1 s 1, so that ¨ 3 Ž X 4 , X 2 , X 1 . s 1, a contradiction as well. For this reason, in Case Ži., we only need to consider ŽA1. & ŽA3. and ŽA2. & ŽA2. for the types of ¨ 3 Ž X 3 , X 2 , X 4 . & ¨ 3 Ž X 4 , X 2 , X 1 .. Therefore, Case III.2 is decomposed into the following three subcases. Case III.2.1. ¨ 3 Ž X 3 , X 2 , X 4 . & ¨ 3 Ž X 4 , X 2 X 1 . is of type ŽA1. & ŽA3.. In this case, it follows from Lemma 10 that X 3 s X 1 s 1, ¨ 3 Ž X 3 , X 2 , X 4 . s ¨ 2 Ž X 2 , X 4 . 960 , and ¨ 3 Ž X 4 , X 2 , X 1 . s ¨ 2 Ž X 4 , X 2 . 576 . Since ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic by the hypothesis of Case III.2, we have that ² ¨ 2 Ž X 2 , X 4 ., ¨ 2 Ž X 4 , X 2 .: is cyclic, so that 1 / ¨ 2 Ž X 2 , X 4 . s ¨ 2 Ž X 4 , X 2 . by Lemmas 1 and 3. Apply Lemma 2 to this equality: there is a word S g Fm such that

Ž 4.4.

X 2 s SX 4 Sy1

and

X 4 s SX 2 Sy1 .

If S s 1, then from Ž4.4. we have X 2 s X 4 , which together with Ž4.3. yields a contradiction of the hypothesis of the lemma. Now let S / 1. We derive from Ž4.4. that X 2 s S 2 X 2 Sy2

and

X 4 s S 2 X 4 Sy2 ,

so that ² S, X 2 : and ² S, X 4 : are cyclic; thus ² X 2 , X 4 : is cyclic. This, together with Ž4.3., also yields a contradiction Žbecause X 2 / 1.. Therefore, we conclude that this case cannot occur. Case III.2.2. ¨ 3 Ž X 3 , X 2 , X 4 . & ¨ 3 Ž X 4 , X 2 , X 1 . is of type ŽA2. & ŽA2.. It follows from Lemma 10 that X 2 s 1, ¨ 3 Ž X 3 , X 2 , X 4 . s ¨ 2 Ž X 4 , X 3 . 400 , and ¨ 3 Ž X 4 , X 2 , X 1 . s ¨ 2 Ž X 1 , X 4 . 400 . Since ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic, ² ¨ 2 Ž X 4 , X 3 ., ¨ 2 Ž X 1 , X 4 .: is cyclic; hence, by Lemmas 1 and 3, 1 / ¨ 2 Ž X 4 , X 3 . s ¨ 2 Ž X 1 , X 4 .. Then by Lemma 2 there exists a word U g Fm such that

Ž 4.5.

X 4 s UX1Uy1

and

X 3 s UX4 Uy1 .

If U s 1, then it follows from Ž4.5. that X 1 s X 4 s X 3 , which together with Ž4.3. yields a contradiction of the hypothesis of the lemma. Now let U / 1. We have from Ž4.5. that UX1Uy1 s Uy1 X 3U. This equality implies by Ž4.3. that ²U, X 1 , X 2 , X 3 : is cyclic; thus, by the first equality of Ž4.5., we

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have that ² X 1 , X 2 , X 3 , X 4 : is cyclic. A contradiction implies that this case cannot occur. Case III.2.3. Neither ¨ 3 Ž X 3 , X 2 , X 4 . nor ¨ 3 Ž X 4 , X 2 , X 1 . is a proper power. Since ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic, we have, by Lemma 6, that 1 / ¨ 3 Ž X 3 , X 2 , X4 . s ¨ 3 Ž X4 , X 2 , X1 . . Apply the Theorem Ž n s 3. to this equality: there is a word T g Fm such that

Ž 4.6.

X 3 s TX 4 Ty1 ,

X 2 s TX 2 Ty1 , and

X 4 s TX 1Ty1 .

The second equality of Ž4.6. implies that ²T, X 2 : is cyclic; hence, by Ž4.3., ²T, X 1 , X 2 , X 3 : is cyclic Žbecause X 2 / 1.. Then by the third equality of Ž4.6., we have that ² X 1 , X 2 , X 3 , X 4 : is cyclic. A contradiction implies that this case cannot occur. Case III.3. ¨ 3 Ž X 3 , X 2 , X 4 . s 1 and ²1 / ¨ 3 Ž X 1 , X 2 , X 3 ., 1 / ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic. Repeat arguments similar to those in Case III.2 to conclude that this case cannot occur. Case III.4. ¨ 3 Ž X 4 , X 2 , X 1 . s 1 and ²1 / ¨ 3 Ž X 1 , X 2 , X 3 ., 1 / ¨ 3 Ž X 3 , X 2 , X 4 .: is cyclic. Also, repeat arguments similar to those in Case III.2 to conclude that this case cannot occur. Case III.5. ²1 / ¨ 3 Ž X 1 , X 2 , X 3 ., 1 / ¨ 3 Ž X 3 , X 2 , X 4 ., 1 / ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic. In this case, we want to prove CLAIM. 1 / ¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž X 3 , X 2 , X 4 . s ¨ 3 Ž X 4 , X 2 , X 1 .. Proof of the Claim. If none of these is a proper power, then the assertion follows immediately from Lemma 6. So assume that one of these is a proper power. Then, in view of Lemmas 10 and 11 Ž n s 4., the other two also have to be proper powers; thus two of X 1 , X 3 , and X 4 must be equal to the empty word, unless X 2 s 1. However, if two of X 1 , X 3 , and X 4 were equal to the empty word, then we would have a contradiction of the non-triviality of ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 ., or ¨ 3 Ž X 4 , X 2 , X 1 ..

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Hence we must have X 2 s 1. Then

Ž 4.7.

¨ 3 Ž X1 , X 2 , X 3 . s ¨ 2 Ž X 3 , X1 .

400

,

¨ 3 Ž X 3 , X 2 , X4 . s ¨ 2 Ž X4 , X 3 .

400

,

¨ 3 Ž X4 , X 2 , X1 . s ¨ 2 Ž X1 , X4 .

400

;

hence the hypothesis of Case III.5 implies that ² ¨ 2 Ž X 3 , X 1 ., ¨ 2 Ž X 4 , X 3 ., ¨ 2 Ž X 1 , X 4 .: is cyclic. It then follows from Lemmas 1 and 3 that ¨ 2 Ž X 3 , X 1 . s ¨ 2 Ž X 4 , X 3 . s ¨ 2 Ž X 1 , X 4 ., which together with Ž4.7. proves the claim. Now apply the Theorem Ž n s 3. to the equalities in the Claim: there exist words Z1 and Z2 in Fm such that

Ž 4.8.

X 1 s Z1 X 3 Zy1 1 ,

X 2 s Z1 X 2 Zy1 1 ,

X 3 s Z1 X 4 Zy1 1 ;

X 3 s Z2 X 4 Zy1 2 ,

X 2 s Z2 X 2 Zy1 2 ,

X 4 s Z2 X 1 Zy1 2 .

We deduce from these equalities that ² Z1 , X 2 :, ² Z2 , X 2 :, ² Z1 Z22 , Z12 Z2 , X 1 :, ² Z1 Z22 , Z12 Z2 , X 3 :, and ² Z1 Z22 , Z12 Z2 , X 4 : are all cyclic. Here, if either Z1 Z22 / 1 or Z12 Z2 / 1, then we would have that ² X 1 , X 2 , X 3 , X 4 : is cyclic, a contradiction. Hence we must have that Z1 Z22 s Z12 Z2 s 1, that is, Z1 s Z2 s 1; thus, by Ž4.8., X1 s X 3 s X4 . In addition, it follows from the Claim that ¨ 4 Ž X1 , X 2 , X 3 , X4 . s u Ž u Ž ¨ 3 Ž X1 , X 2 , X 3 . , ¨ 3 Ž X1 , X 2 , X 3 . . ,

u Ž ¨ 3 Ž X1 , X 2 , X 3 . , ¨ 3 Ž X1 , X 2 , X 3 . . . 44

s u Ž ¨ 3 Ž X1 , X 2 , X 3 . , ¨ 3 Ž X1 , X 2 , X 3 . s ¨ 3 Ž X1 , X 2 , X 3 .

44⭈24q44⭈20

s ¨ 3 Ž X1 , X 2 , X 3 .

1936

44

.

.

Therefore, in this case, ¨ 4 Ž X 1 , X 2 , X 3 , X 4 . has form ŽB4.. Case III.6. Both ² ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .: and ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: are non-cyclic. In view of Ž4.2. and Lemmas 4 and 6, we have that u Ž ¨ 3 Ž X1 , X 2 , X 3 . , ¨ 3 Ž X 3 , X 2 , X4 . . s u Ž ¨ 3 Ž X 3 , X 2 , X4 . , ¨ 3 Ž X4 , X 2 , X1 . . .

528

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Then by Lemma 5 applied to this equality, there is a word W g Fm such that 1 / ¨ 3 Ž X 1 , X 2 , X 3 . s W¨ 3 Ž X 3 , X 2 , X 4 . Wy1

and

1 / ¨ 3 Ž X 3 , X 2 , X 4 . s W¨ 3 Ž X 4 , X 2 , X 1 . Wy1 . Applying the Theorem Ž n s 3. to these equalities yields the existence of words V1 and V2 in Fm such that X 1 s V1 X 3V1y1 ,

X 2 s V1 X 2 V1y1 ,

X 3 s V1 X 4 V1y1 ;

X 3 s V2 X 4 Vy1 2 ,

X 2 s V2 X 2 Vy1 2 ,

X 4 s V2 X 1Vy1 2 .

This is the same situation as Ž4.8.; hence, reasoning as in Case III.5, we have X 1 s X 3 s X 4 . But then ¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž X 3 , X 2 , X 4 . s ¨ 3 Ž X 4 , X 2 , X 1 ., which yields a contradiction to the hypothesis of Case III.6. Therefore, we conclude that this case cannot occur. The proof of Lemma 13 Ž n s 4. is now complete. Proof of the Theorem Ž n s 4.. The additional property that ¨ 4 Ž X 1 , X 2 , X 3 , X 4 . s 1 if and only if the subgroup ² X 1 , X 2 , X 3 , X 4 : of Fm is cyclic is immediate from definition Ž1.5. of ¨ 4Ž x 1 , x 2 , x 3 , x 4 . and Lemma 13 Ž n s 4.. Now we want to prove that ¨ 4 Ž x 1 , x 2 , x 3 , x 4 . is a C-test word, that is, supposing 1 / ¨ 4 Ž X 1 , X 2 , X 3 , X 4 . s ¨ 4Ž Y1 , Y2 , Y3 , Y4 ., we want to prove the existence of a word Z in Fm such that Yi s ZX i Zy1

for all i s 1, 2, 3, 4.

We consider two cases corresponding to whether ¨ 4Ž X 1 , X 2 , X 3 , X 4 . is a proper power or not. Case I. ¨ 4Ž X 1 , X 2 , X 3 , X 4 . is a proper power. Apply Lemma 13 Ž n s 4. to ¨ 4 Ž X 1 , . . . , X 4 . and ¨ 4Ž Y1 , . . . , Y4 .: ¨ 4 Ž X 1 , . . . , X 4 . has one of the four types ŽB1. ᎐ ŽB4.; furthermore, by the equality ¨ 4 Ž X 1 , . . . , X 4 . s ¨ 4 Ž Y1 , . . . , Y4 ., ¨ 4 Ž Y1 , . . . , Y4 . has the same type as ¨ 4 Ž X 1 , . . . , X 4 . does, since the exponents in ŽB1. ᎐ ŽB4. are all distinct. This gives us only four possibilities, ŽB1. & ŽB1., . . . , ŽB4. & ŽB4., for the types of ¨ 4 Ž X 1 , . . . , X 4 . & ¨ 4 Ž Y1 , . . . , Y4 .. If ¨ 4Ž X 1 , . . . , X 4 . & ¨ 4 Ž Y1 , . . . , Y4 . is of type ŽB1. & ŽB1. ŽŽB2. & ŽB2. or ŽB3. & ŽB3. is analogous., then X 1 s X 2 s Y1 s Y2 s 1 and 1 / ¨ 2 Ž X 4 , X 3 .

960⭈400

s ¨ 2 Ž Y4 , Y3 .

960⭈400

.

Applying Lemma 2 to the equality 1 / ¨ 2 Ž X 4 , X 3 . s ¨ 2 Ž Y4 , Y3 ., we have that two 2-tuples Ž X 4 , X 3 . and Ž Y4 , Y3 . are conjugate in Fm , which together with X 1 s X 2 s Y1 s Y2 s 1 yields the desired result.

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If ¨ 4 Ž X 1 , . . . , X 4 . & ¨ 4 Ž Y1 , . . . , Y4 . is of type ŽB4. & ŽB4., then X 1 s X 3 s X 4 , Y1 s Y3 s Y4 , and 1 / ¨ 3 Ž X1 , X 2 , X 3 .

1936

s ¨ 3 Ž Y1 , Y2 , Y3 .

1936

.

The equality 1 / ¨ 3 Ž X 1 , X 2 , X 3 . s ¨ 3 Ž Y1 , Y2 , Y3 . yields, by the Theorem Ž n s 3., that two 3-tuples Ž X 1 , X 2 , X 3 . and Ž Y1 , Y2 , Y3 . are conjugate in Fm . Then the result follows from X 1 s X 3 s X 4 and Y1 s Y3 s Y4 . Case II. ¨ 4Ž X 1 , X 2 , X 3 , X 4 . is not a proper power. In view of Lemma 4, it follows that

Ž 4.9.

² u Ž ¨ 3 Ž X1 , X 2 , X 3 . , ¨ 3 Ž X 3 , X 2 , X4 . . , u Ž ¨ 3 Ž X 3 , X 2 , X 4 . , ¨ 3 Ž X 4 , X 2 , X 1 . .:

is non-cyclic.

This enables us to apply Lemma 5 to the equality ¨ 4 Ž X 1 , . . . , X 4 . s

¨ 4 Ž Y1 , . . . , Y4 .: there exists a word S g Fm such that

1 / u Ž ¨ 3 Ž X1 , X 2 , X 3 . , ¨ 3 Ž X 3 , X 2 , X4 . .

Ž 4.10.

s Su Ž ¨ 3 Ž Y1 , Y2 , Y3 . , ¨ 3 Ž Y3 , Y2 , Y4 . . Sy1 , 1 / u Ž ¨ 3 Ž X 3 , X 2 , X4 . , ¨ 3 Ž X4 , X 2 , X1 . . s Su Ž ¨ 3 Ž Y3 , Y2 , Y4 . , ¨ 3 Ž Y4 , Y2 , Y1 . . Sy1 .

Here, we have four subcases to consider. Case II.1. Both ² ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .: and ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: are non-cyclic. The hypothesis of this case enables us to apply Lemma 5 to equalities Ž4.10.: there exist words T1 and T2 in Fm such that 1 / ¨ 3 Ž X 1 , X 2 , X 3 . s T1¨ 3 Ž Y1 , Y2 , Y3 . T1y1 ,

Ž 4.11.

1 / ¨ 3 Ž X 3 , X 2 , X 4 . s T1¨ 3 Ž Y3 , Y2 , Y4 . T1y1 ; 1 / ¨ 3 Ž X 3 , X 2 , X 4 . s T2¨ 3 Ž Y3 , Y2 , Y4 . Ty1 2 , 1 / ¨ 3 Ž X 4 , X 2 , X 1 . s T2¨ 3 Ž Y4 , Y2 , Y1 . Ty1 2 .

Then by the Theorem Ž n s 3. applied to Ž4.11., there exist words U1 , U2 , and U3 such that

Ž 4.12.

X 1 s U1Y1U1y1 ,

X 2 s U1Y2 U1y1 ,

X 3 s U1Y3U1y1 ;

X 3 s U2 Y3 U2y1 ,

X 2 s U2 Y2 U2y1 ,

X 4 s U2 Y4 U2y1 ;

X 4 s U3 Y4 U3y1 ,

X 2 s U3 Y2 U3y1 ,

X 1 s U3 Y1U3y1 .

530

DONGHI LEE

Here, if one of U1 s U2 , U2 s U3 , and U3 s U1 is true, then the required result follows directly from Ž4.12.. So assume that U1 , U2 , and U3 are pairwise distinct. Combining the equalities in Ž4.12., we deduce that ²U3y1 U1 , X 1 :, ²U1y1 U2 , X 2 :, ²U2y1 U3 , X 2 :, ²U3y1 U1 , X 2 :, ²U1y1 U2 , X 3 :, and ²U2y1 U3 , X 4 : are all cyclic, so that ² X 1 , X 2 :, ² X 3 , X 2 :, and ² X 4 , X 2 : are cyclic. Since ² X 1 , . . . , X 4 : is non-cyclic Žthis follows from ¨ 4Ž X 1 , . . . , X 4 . / 1., we must have X 2 s 1; so, by Ž4.12., Y2 s 1. Then by Lemma 10, the equalities on the first line of Ž4.11. yield that 1 / ¨ 2 Ž X 3 , X1 .

400

s T1¨ 2 Ž Y3 , Y1 .

400

T1y1 ,

1 / ¨ 2 Ž X4 , X 3 .

400

s T1¨ 2 Ž Y4 , Y3 .

400

T1y1 ,

namely,

Ž 4.13.

1 / ¨ 2 Ž X 3 , X 1 . s T1¨ 2 Ž Y3 , Y1 . T1y1 , 1 / ¨ 2 Ž X 4 , X 3 . s T1¨ 2 Ž Y4 , Y3 . T1y1 .

This is a situation similar to Ž3.10., so from here on, we can follow the proof of Case II.1 of the Theorem Ž n s 3. to obtain that two 3-tuples Ž X 1 , X 3 , X 4 . and Ž Y1 , Y3 , Y4 . are conjugate in Fm . Since X 2 s Y2 s 1, the desired result follows. Case II.2. ² ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .: is non-cyclic, and ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic. In this case, we can apply Lemma 5 to the first equality of Ž4.10.: there exists a word V g Fm such that

Ž 4.14.

1 / ¨ 3 Ž X 1 , X 2 , X 3 . s V¨ 3 Ž Y1 , Y2 , Y3 . Vy1 , 1 / ¨ 3 Ž X 3 , X 2 , X 4 . s V¨ 3 Ž Y3 , Y2 , Y4 . Vy1 .

Then the Theorem Ž n s 3. applied to Ž4.14. yields the existence of words W1 and W2 in Fm such that

Ž 4.15.

X 1 s W1Y1W1y1 ,

X 2 s W1Y2W1y1 ,

X 3 s W1Y3W1y1 ;

X 3 s W2 Y3Wy1 2 ,

X 2 s W2 Y2Wy1 2 ,

X 4 s W2 Y4Wy1 2 .

If W1 s W2 , then the required result follows from Ž4.15.. Now assume that W1 / W2 . We deduce from Ž4.15. that ²W1y1 W2 , X 2 : and ²W1y1 W2 , X 3 : are cyclic, so that

Ž 4.16.

² X 2 , X 3 : is cyclic.

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On the other hand, since ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic by the hypothesis of Case II.2, in view of Lemmas 10 and 11 Ž n s 4., we have that ¨ 3 Ž X 3 , X 2 , X 4 . is a proper power if and only if ¨ 3 Ž X 4 , X 2 , X 1 . is a proper power. For this reason, this case is reduced to the following three subcases. Case II.2.1. ¨ 3 Ž X 4 , X 2 , X 1 . s 1 Ž ¨ 3 Ž X 3 , X 2 , X 4 . / 1 by the hypothesis of Case II.2.. It follows from the Theorem Ž n s 3. that ² X 4 , X 2 , X 1 : is cyclic. Since ² X 2 , X 3 : is cyclic by Ž4.16., to avoid a contradiction of the fact that ² X 1 , . . . , X 4 : is non-cyclic, we must have

Ž 4.17.

X 2 s 1;

thus Y2 s 1 by Ž4.15.. Then by Lemma 10, equalities Ž4.14. yield that 1 / ¨ 2 Ž X 3 , X 1 . s V¨ 2 Ž Y3 , Y1 . Vy1 , 1 / ¨ 2 Ž X 4 , X 3 . s V¨ 2 Ž Y4 , Y3 . Vy1 . This is the same situation as in Ž4.13.; hence from here on, repeating the proof of Case II.1, we obtain the desired result. Case II.2.2. Both ¨ 3 Ž X 3 , X 2 , X 4 . and ¨ 3 Ž X 4 , X 2 , X 1 . are proper powers. In view of Lemma 10, we have nine possibilities for the types of ¨ 3 Ž X 3 , X 2 , X 4 . & ¨ 3 Ž X 4 , X 2 , X 1 .. Of these nine possible types, ŽAl. & ŽA1.,

ŽA1. & ŽA2., ŽA1. & ŽA3., ŽA2. & ŽA1., ŽA2. & ŽA3., ŽA3. & ŽA2., and ŽA3. & ŽA3. cannot occur, for if one of these occurred, then two of X 1 , X 2 , X 3 , and X 4 should be equal to the empty word, which yields a contradiction of the non-triviality of ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 ., or ¨ 3 Ž X 4 , X 2 , X 1 .. So only ŽA2. & ŽA2. and ŽA3. & ŽA1. can actually occur. If ¨ 3 Ž X 3 , X 2 , X 4 . & ¨ 3 Ž X 4 , X 2 , X 1 . is of type ŽA2. & ŽA2., then X 2 s 1, which is the same situation as in Ž4.17.. Hence from here on, following the proof of Case II.2.1, we arrive at the desired result. If ¨ 3 Ž X 3 , X 2 , X 4 . & ¨ 3 Ž X 4 , X 2 , X 1 . is of type ŽA3. & ŽA1., then X 4 s 1. The result then follows from Ž4.15.. Case II.2.3. Neither ¨ 3 Ž X 3 , X 2 , X 4 . / 1 nor ¨ 3 Ž X 4 , X 2 , X 1 . / 1 is a proper power. Since ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: is cyclic, it follows from Lemma 6 that ¨ 3 Ž X 3 , X 2 , X 4 . s ¨ 3 Ž X 4 , X 2 , X 1 ., and so from the second equality of Ž4.10. that 44 Ž 4.18. 1 / ¨ 3 Ž X4 , X 2 , X1 . s Su Ž ¨ 3 Ž Y3 , Y2 , Y4 . , ¨ 3 Ž Y4 , Y2 , Y1 . . Sy1 .

532

DONGHI LEE

This equality implies by Lemma 4 that ² ¨ 3 Ž Y3 , Y2 , Y4 ., ¨ 3 Ž Y4 , Y2 , Y1 .: is also cyclic. We then observe that equality Ž4.18. can hold only when neither ¨ 3 Ž Y3 , Y2 , Y4 . nor ¨ 3 Ž Y4 , Y2 , Y1 . is a proper power and ¨ 3 Ž Y3 , Y2 , Y4 . s ¨ 3 Ž Y4 , Y2 , Y1 ., by which Ž4.18. yields that 1 / ¨ 3 Ž X4 , X 2 , X1 .

44

s S¨ 3 Ž Y4 , Y2 , Y1 . Sy1 , 44

namely, 1 / ¨ 3 Ž X 4 , X 2 , X 1 . s S¨ 3 Ž Y4 , Y2 , Y1 . Sy1 . Now apply the Theorem Ž n s 3. to this equality: there exists a word W3 g Fm such that X 4 s W3 Y4Wy1 3 ,

X 2 s W3 Y2Wy1 3 ,

X 1 s W3 Y1Wy1 3 .

Putting this together with Ž4.15., we have the same situation as in Ž4.12., except that we already assumed W1 / W2 in Case II.2. Therefore, from here on, we can follow the proof of Case II.1 to derive the result. Case II.3. ² ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .: is cyclic, and ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: is non-cyclic. It is sufficient to repeat arguments similar to those in Case II.2 to arrive at the desired result. Case II.4. Both ² ¨ 3 Ž X 1 , X 2 , X 3 ., ¨ 3 Ž X 3 , X 2 , X 4 .: and ² ¨ 3 Ž X 3 , X 2 , X 4 ., ¨ 3 Ž X 4 , X 2 , X 1 .: are cyclic. Arguing as in the proof of Case II.2 of the Theorem Ž n s 3., replacing Ž3.8. and Ž3.9. by Ž4.9. and Ž4.10., respectively, we deduce that ¨ 3 Ž X 3 , X 2 , X 4 . s ¨ 3 Ž Y3 , Y2 , Y4 . s 1. So

Ž 4.19.

² X 3 , X 2 , X 4 : is cyclic;

moreover, it follows from Ž4.10. that 1 / ¨ 3 Ž X1 , X 2 , X 3 .

24

s S¨ 3 Ž Y1 , Y2 , Y3 . Sy1 ,

1 / ¨ 3 Ž X4 , X 2 , X1 .

20

s S¨ 3 Ž Y4 , Y2 , Y1 . Sy1 ,

24

20

namely, 1 / ¨ 3 Ž X 1 , X 2 , X 3 . s S¨ 3 Ž Yi, Y2 , Y3 . Sy1 , 1 / ¨ 3 Ž X 4 , X 2 , X 1 . s S¨ 3 Ž Y4 , Y2 , Y1 . Sy1 .

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533

Then by the Theorem Ž n s 3. applied to these equalities, we have the existence of words Z1 and Z2 in Fm such that

Ž 4.20.

X 1 s Z1Y1 Zy1 1 ,

X 2 s Z1Y2 Zy1 1 ,

X 3 s Z1Y3 Zy1 1 ;

X 4 s Z2 Y4 Zy1 2 ,

X 2 s Z2 Y2 Zy1 2 ,

X 1 s Z2 Y1 Zy1 2 .

If Z1 s Z2 , then the result follows from Ž4.20.. Now assume that Z1 / Z2 . : ² y1 : Then equalities Ž4.20. yield that ² Zy1 1 Z 2 , X 1 and Z1 Z 2 , X 2 are cyclic, so that ² X 1 , X 2 : is cyclic. Since ² X 3 , X 2 , X 4 : is cyclic by Ž4.19., we must have X 2 s 1, which is the same situation as in Ž4.17.. Thus, from here on, we can follow the proof of Case II.2.1 to get the required result. The Theorem Ž n s 4. is now completely proved.

5. THE INDUCTIVE STEP In this section, we prove the inductive step of simultaneous induction which we use in Lemmas 11᎐13 and the Theorem. Let n G 5 throughout this section. LEMMA 11. If both ¨ ny 2 Ž X 1 , . . . , X ny2 . and ¨ ny1Ž Y1 , . . . , Yny1 . are neither equal to the empty word nor proper powers, then ² ¨ ny 2 Ž X 1 , . . . , X ny2 ., ¨ ny 1Ž Y1 , . . . , Yny1 .: is non-cyclic. Proof. By way of contradiction, suppose that ² ¨ ny 2 Ž X 1 , . . . , X ny2 ., ¨ ny 1Ž Y1 , . . . , Yny1 .: is cyclic. Since both ¨ ny2 Ž X 1 , . . . , X ny2 . and ¨ ny1Ž Y1 , . . . , Yny 1 . are non-proper powers, it follows from Lemma 6 that ¨ ny 2 Ž X 1 , . . . , X ny2 . s ¨ ny1 Ž Y1 , . . . , Yny1 . ,

and so from Ž1.3. ᎐ Ž1.4. and Lemmas 4 and 5 that there exists a word S g Fm such that u Ž ¨ ny 3 Ž X 1 , . . . , X ny3 . , ¨ ny3 Ž X ny3 , . . . , X ny2 . .

Ž 5.1.

s Su Ž ¨ ny 2 Ž Y1 , . . . , Yny2 . , ¨ ny2 Ž Yny2 , . . . , Yny1 . . Sy1 , u Ž ¨ ny 3 Ž X ny3 , . . . , X ny2 . , ¨ ny3 Ž X ny2 , . . . , X 1 . . s Su Ž ¨ ny 2 Ž Yny2 , . . . , Yny1 . , ¨ ny2 Ž Yny1 , . . . , Y1 . . Sy1 .

We first assume that ² ¨ ny 3 Ž X 1 , . . . , X ny3 ., ¨ ny3 Ž X ny3 , . . . , X ny2 .: is non-cyclic. This enables us to apply Lemma 5 to the first equality of Ž5.1.:

534

DONGHI LEE

there exists a word T g Fm such that

Ž 5.2.

¨ ny 3 Ž X 1 , . . . , X ny3 . s T¨ ny2 Ž Y1 , . . . , Yny2 . Ty1 ¨ ny 3 Ž X ny3 , . . . , X ny2 . s T¨ ny2 Ž Yny2 , . . . , Yny1 . Ty1 .

If both sides of the first equality of Ž5.2. are non-proper powers, then this equality yields a contradiction the induction hypothesis Lemma 11; if both sides of the first equality of Ž5.2. are proper powers, then we see from Lemma 10 and the induction hypothesis of Lemma 13 that they cannot be the same proper powers Žbecause the exponents on the two sides cannot be identical., a contradiction as well. We next assume that ² ¨ ny 3 Ž X 1 , . . . , X ny3 ., ¨ ny3 Ž X ny3 , . . . , X ny2 .: is cyclic. Then the first equality of Ž5.1. yields by Lemma 4 that

Ž 5.3.

² ¨ ny 3 Ž X1 , . . . , X ny3 . , ¨ ny3 Ž X ny3 , . . . , X ny2 . , S¨ ny 2 Ž Y1 , . . . , Yny2 . Sy1 , S¨ ny2 Ž Yny2 , . . . , Yny1 . Sy1: is cyclic.

Here, in view of the induction hypothesis of Lemma 13, we see that there are only two ways to avoid a contradiction of Lemma 8 and the induction hypothesis of Lemma 11: Ži. any non-trivial word of S¨ ny 2 Ž Y1 , . . . , Yny2 . Sy1 and S¨ ny 2 Ž Yny2 , . . . , Yny1 . Sy1 is of type ŽB4.; Žii. any non-trivial word in Ž5.3. is of one of types ŽB1., ŽB2., and ŽB3.. ŽFor n s 5, there is only one way: any non-trivial word of S¨ ny 2 Ž Y1 , . . . , Yny2 . Sy1 and S¨ ny2 Ž Yny2 , . . . , Yny 1 . Sy1 is of one of types ŽA1., ŽA2., and ŽA3... However, in either case, we can observe that equalities Ž5.2. cannot hold. This contradiction completes the proof. LEMMA 12. If uŽ ¨ ny 1Ž X 1 , . . . , X ny1 ., ¨ ny1Ž Y1 , . . . , Yny1 .. s 1, then ¨ ny 1Ž X 1 , . . . , X ny1 . s 1 and ¨ ny1Ž Y1 , . . . , Yny1 . s 1. Proof. By Lemma 4, the hypothesis of the lemma yields that

Ž 5.4.

6

¨ ny 1 Ž X 1 , . . . , X ny1 . s ¨ ny1 Ž Y1 , . . . , Yny1 .

y5

.

If one of ¨ ny 1Ž X 1 , . . . , X ny1 . and ¨ ny1Ž Y1 , . . . , Yny1 . is equal to the empty word, then by Ž5.4. there is nothing to prove. So assume that ¨ ny 1Ž X 1 , . . . , X ny 1 . / 1 and ¨ ny1Ž Y1 , . . . , Yny1 . / 1. Since 6 and 5 are relatively prime, equality Ž5.4. implies that both ¨ ny 1Ž X 1 , . . . , X ny1 . and ¨ ny 1Ž Y1 , . . . , Yny1 . either proper powers or non-proper powers. If both are proper powers, then a contradiction of equality Ž5.4. follows from the induction hypothesis of Lemma 13, for 6 times any of 960 ⭈ 400 Ž ny4., 400 Ž ny3., 576 ⭈ 400 Ž ny4., and 1936 cannot be equal to 5 times any of these.

ON CERTAIN C-TEST WORDS FOR FREE GROUPS

535

If both are non-proper powers, then, since ² ¨ ny 1Ž X 1 , . . . , X ny1 ., ¨ ny1Ž Y1 , . . . , Yny 1 .: is cyclic, we have by Lemma 6 that ¨ ny1Ž X 1 , . . . , X ny1 . s ¨ ny 1Ž Y1 , . . . , Yny1 .. This together with Ž5.4. yields that ¨ ny1Ž X 1 , . . . , X ny1 . s ¨ ny 1Ž Y1 , . . . , Yny1 . s 1. This contradiction to our assumption completes the proof. LEMMA 13. Suppose that ² X 1 , . . . , X n : is non-cyclic. Then ¨ nŽ X 1 , . . . , X n . / 1. Furthermore, either ¨ nŽ X 1 , . . . , X n . is not a proper power or it has one of the following four forms:

¡¨ Ž X , X . 2

Ž B1 .

¨ n Ž X1 , . . . , X n . s

~

1

¨ 2 Ž X ny1 , X n .

2

¨ n Ž X1 , . . . , X n . s

~

Ž B4 .

n

¨ 2 Ž X n , X1 .

400 Ž ny 2.

and X 2 s X 3 s ⭈⭈⭈ s X ny1 s 1, for n odd; Ž ny 3.

~

X 1 . 576⭈400 and X 2 s ⭈⭈⭈ s X ny2 s X n s 1, n e¨ en

¢

and X 2 s ⭈⭈⭈ s X ny2 s X n s 1, n odd;

2

¨ n Ž X1 , . . . , X n . s

400 Ž ny 2.

and X 2 s X 3 s ⭈⭈⭈ s X ny1 s 1, for n e¨ en 1

¢ ¡¨ Ž X Ž B3 .

960⭈400 Ž ny 3.

and X 1 s X 2 s ⭈⭈⭈ s X ny2 s 1, for n odd;

¢ ¡¨ Ž X , X . Ž B2 .

960⭈400 Ž ny 3.

and X 1 s X 2 s ⭈⭈⭈ s X ny2 s 1, for n e¨ en n

ny1 ,

¨ 2 Ž X 1 , X ny1 .

576⭈400 Ž ny 3.

¨ n Ž X 1 , . . . , X n . s ¨ ny1 Ž X 1 , X 2 , . . . , X ny1 .

1936

and

X 1 s X ny1 s X n / 1.

Remark. In view of Lemma 1 and the induction hypothesis of Lemma 13, ¨ 2 Ž X n , X ny1 . and ¨ 2 Ž X ny1 , X n . in ŽB1., ¨ 2 Ž X 1 , X n . and ¨ 2 Ž X n , X 1 . in ŽB2., ¨ 2 Ž X ny1 , X 1 . and ¨ 2 Ž X 1 , X ny1 . in ŽB3., and ¨ ny1Ž X 1 , X 2 , . . . , X ny1 . in ŽB4. are neither equal to the empty word nor proper powers.

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DONGHI LEE

Proof. Closely follow the proof of Lemma 13 Ž n s 4. replacing references to Lemmas 10, 11 Ž n s 4., and 12 Ž n s 4., and the Theorem Ž n s 3. by references to the induction hypothesis of Lemma 13, Lemmas 8 and 11, Lemma 12, and the induction hypothesis of the Theorem, respectively. Different situations from Lemma 13 Ž n s 4. can possibly occur only in Case III.2 and Case III.5, which we reconsider below. Case III.2. ²1 / ¨ ny 1Ž X ny1 , X 2 , X 3 , . . . , X ny2 , X n ., 1 / ¨ ny1Ž X n , X 2 , X 3 , . . . , X ny2 , X 1 .: is cyclic, and ¨ ny1Ž X 1 , X 2 , . . . , X ny1 . s 1. Since ¨ ny 1Ž X 1 , X 2 , . . . , X ny1 . s 1, we have, by the induction hypothesis of the Theorem, that

Ž 5.5.

² X 1 , X 2 , . . . , X ny1 : is cyclic.

Also since ²1 / ¨ ny 1Ž X ny1 , X 2 , . . . , X ny2 , X n ., 1 / ¨ ny1Ž X n , X 2 , . . . , X ny 2 , X 1 .: is cyclic, in view of Lemmas 8 and 11 and the induction hypothesis of Lemma 13, this case is reduced to the following two cases: Ži. both ¨ ny 1Ž X ny1 , X 2 , . . . , X ny2 , X n . and ¨ ny1Ž X n , X 2 , . . . , X ny 2 , X 1 . are proper powers; Žii. neither ¨ ny 1Ž X ny1 , X 2 , . . . , X ny2 , X n . nor ¨ ny1Ž X n , X 2 , . . . , X ny 2 , X 1 . is a proper power. Case Ži. is divided further into subcases according to the types of ¨ ny 1Ž X ny1 , X 2 , . . . , X ny2 , X n . & ¨ ny1Ž X n , X 2 , . . . , X ny2 , X 1 . by the induc-

tion hypothesis of Lemma 13. However, the former word cannot be of type ŽB3. or ŽB4., for this type together with Ž5.5. yields a contradiction of the hypothesis of Lemma 13. For the same reason, the latter cannot be of type ŽB1. or ŽB4.. Also ŽB1. & ŽB2. and ŽB2. & ŽB3. cannot occur, for these types yield a contradiction of the non-triviality of ¨ ny 1Ž X ny1 , X 2 , . . . , X ny 2 , X n . and ¨ ny1Ž X n , X 2 , . . . , X ny2 , X 1 ., respectively. Thus, in Case Ži., only ŽB1. & ŽB3. and ŽB2. & ŽB2. need to be considered. This allows us to follow the proof of Lemma 13 Ž n s 4. from here on. Case III.5. ²1 / ¨ ny 1Ž X 1 , X 2 , . . . , X ny1 ., 1 / ¨ ny1Ž X ny1 , X 2 , X 3 , . . . , X ny 2 , X n ., 1 / ¨ ny1Ž X n , X 2 , X 3 , . . . , X ny2 , X 1 .: is cyclic. In this case, to be able to keep following the proof of Lemma 13 Ž n s 4., it is sufficient prove the following CLAIM. 1 / ¨ ny 1Ž X 1 , X 2 , . . . , X ny1 . s ¨ ny1Ž X ny1 , . . . , X ny2 , X n . s ¨ ny 1Ž X n , . . . , X ny2 , X 1 .. Proof of the Claim. If none of these is a proper power, then the assertion follows immediately from Lemma 6. So assume that one of these is a proper power. Then, in view of the induction hypothesis of Lemma 13

ON CERTAIN C-TEST WORDS FOR FREE GROUPS

537

and Lemmas 8 and 11, the other two also have to be proper powers. Here, to avoid a contradiction of the non-triviality of these words, all of these words must have the same type, either ŽB2. or ŽB4.. The treatment of ŽB2. is the same as in the Theorem Ž n s 4.; if all of these words are of type ŽB4., then we have X 1 s X ny2 s X ny1 s X n / 1, which proves the claim. The proof of Lemma 13 is complete. Proof of the Theorem. The additional property that ¨ nŽ X 1 , . . . , X n . s 1 if and only if the subgroup ² X 1 , . . . , X n : of Fm is cyclic follows immediately from definition Ž1.4. of ¨ nŽ x 1 , . . . , x n . and Lemma 13. Now supposing 1 / ¨ nŽ X 1 , . . . , X n . s ¨ nŽ Y1 , . . . , Yn ., we want to prove that there exists a word Z g Fm such that Yi s ZX i Zy1

for all i s 1, . . . , n,

that is, we want to show that ¨ nŽ x 1 , . . . , x n . is a C-test word. From here on, repeat the proof of the Theorem Ž n s 4., replacing references to Lemmas 10, 11 Ž n s 4., and 13 Ž n s 4. and the Theorem Ž n s 3. by references to Lemma 13, Lemmas 8 and 11, Lemma 13, and the induction hypothesis of the Theorem, respectively. A situation different from that of the Theorem Ž n s 4. can possibly occur only in Case II.2.2, which we reconsider below. Case II.2.2. Both ¨ ny 1Ž X ny1 , X 2 , X 3 , . . . , X ny2 , X n . and ¨ ny1Ž X n , X 2 , X 3 , . . . , X ny2 , X 1 . are proper powers. In view of Lemma 13, ¨ ny 1Ž X ny1 , X 2 , . . . , X ny2 , X n . & ¨ ny1Ž X n , X 2 , . . . , X ny 2 , X 1 . has 16 possible types, of which it suffices to consider the types involving ŽB4., namely, ŽB1. & ŽB4., ŽB2. & ŽB4., ŽB3. & ŽB4., ŽB4. & ŽB4., ŽB4. & ŽB1., ŽB4. & ŽB2., and ŽB4. & ŽB3., since the consideration for the remaining types is the same as in the proof of the Theorem Ž n s 4.. However, none of these types except for ŽB4. & ŽB4. can actually occur, for these types except for ŽB4. & ŽB4., together with the hypothesis of Case II.2 that ² ¨ ny 1Ž X ny1 , X 2 , . . . , X ny2 , X n ., ¨ ny1Ž X n , X 2 , . . . , X ny2 , X 1 .: is cyclic, yield a contradiction to Lemma 8. If ŽB4. & ŽB4. occurs, then the equality corresponding to the second one of Ž4.10., namely, 1 / u Ž ¨ ny 1 Ž X ny1 , X 2 , . . . , X ny2 , X n . , ¨ ny1 Ž X n , X 2 , . . . , X ny2 , X 1 . . s Su Ž ¨ ny 1 Ž Yny1 , Y2 , . . . , Yny2 , Yn . , ¨ ny1 Ž Yn , Y2 , . . . , Yny2 , Y1 . . Sy1 forces ¨ ny 1Ž Yny1 , Y2 , . . . , Yny2 , Yn . & ¨ ny1Ž Yn , Y2 , . . . , Yny2 , Y1 . to have type ŽB4. & ŽB4. as well. Thus, we have X 1 s X ny2 s X ny1 s X n / 1 and

Y1 s Yny2 s Yny1 s Yn / 1,

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DONGHI LEE

which together with X ny 1 s W2 Yny1Wy1 2 ,

X 2 s W2 Y2Wy1 2 ,...,

X ny2 s W2 Yny2Wy1 2 ,

X n s W2 YnWy1 2 , corresponding to the equalities on the second line of Ž4.15., implies the desired result. This completes Case II.2.2.

6. PROOF OF COROLLARY 2 We are now in a position to prove Corollary 2. Proof of Corollary 2. Let ␾ and ␺ be endomorphisms of Fm with non-cyclic images. Take u1 s ¨ m Ž x 1 , x 2 , . . . , x m .

and

u 2 s ¨ mq1 Ž x my1 , x my2 , . . . , x 1 , x m , x 1 . .

Supposing ␾ Ž u i . s ␺ Ž u i ., for i s 1, 2, we want to prove ␾ s ␺ . By Corollary 1, the equality ␾ Ž u1 . s ␺ Ž u1 . implies that

Ž 6.1.

␾ s ␶S ( ␺ ,

where ² S, ␺ Ž u1 .: is cyclic.

Here, it is sufficient to show S s 1 to draw the desired result. By Ž6.1., the other equality ␾ Ž u 2 . s ␺ Ž u 2 . yields that

␺ Ž u 2 . s ␾ Ž u 2 . s S␺ Ž u 2 . Sy1 , so that ² S, ␺ Ž u 2 .: is cyclic. Then S s 1 follows obviously from the following CLAIM. ² ␺ Ž u1 ., ␺ Ž u 2 .: is non-cyclic. Proof of the Claim. Putting X i s ␺ Ž x i ., for all i s 1, 2, . . . , m, the claim is equivalent to

² ¨ m Ž X1 , X 2 , . . . , X m . , ¨ mq1Ž X my1 , X my2 , . . . , X1 , X m , X1 .: is non-cyclic. Since ␺ has a non-cyclic image, ² X 1 , X 2 , . . . , X m : is non-cyclic. This implies by the Theorem that ¨ m Ž X1 , X 2 , . . . , X m . / 1

and

¨ mq1 Ž X my1 , X my2 , . . . , X 1 , X m , X 1 . / 1.

We treat the two cases separately. Case I. ¨ mq 1Ž X my1 , X my2 , . . . , X 1 , X m , X 1 . is not a proper power.

ON CERTAIN C-TEST WORDS FOR FREE GROUPS

539

In this case, ¨ mŽ X 1 , X 2 , . . . , X m . cannot be of form ŽB4., for if ¨ m Ž X 1 , X 2 , . . . , X m . were of form ŽB4. then X 1 s X my1 s X m / 1, which forces ¨ mq 1Ž X my1 , X my2 , . . . , X 1 , X m , X 1 . to be a proper power of form ŽB4., a contradiction. Then the claim follows from Lemmas 8 and 11. Case II. ¨ mq 1Ž X my1 , X my2 , . . . , X 1 , X m , X 1 . is not a proper power. This case can occur only when m G 3, for if ¨ 3 Ž X 1 , X 2 , X 1 . were a proper power, then we should have X 2 s 1; hence ² X 1 , X 2 : is cyclic, contrary to our assumption that ␺ has a non-cyclic image. It then follows from Lemma 13 that ¨ mq 1Ž X my1 , X my2 , . . . , X 1 , X m , X 1 . has one of four types ŽB1. ᎐ ŽB4.. Of these types, ŽB1. cannot occur, for ŽB1. implies X my 1 s X my2 s ⭈⭈⭈ s X 1 s 1, contrary to the fact that ² X 1 , X 2 , . . . , X m : is non-cyclic. For a similar reason, ŽB2. cannot occur, either. On the other hand, ŽB4. cannot occur when m s 3, for ŽB4. together with m s 3 implies X 2 s X 3 s X 1 / 1, contrary to ² X 1 , X 2 , X 3 : being non-cyclic. When m G 4, ŽB4. implies that X my 1 s X m s X 1 / 1, so that ¨ mŽ X 1 , X 2 , . . . , X m . has type ŽB4. as well. Then the claim follows from Lemma 11. It remains to consider type ŽB3.. If ¨ mq 1Ž X my1 , X my2 , . . . , X 1 , X m , X 1 . has type ŽB3., then we have that ¨ mq 1 Ž X my1 , X my2 , . . . , X 1 , X m , X 1 .

¡ ŽX ¢¨ Ž X

¨ s~

Ž 6.2.

2 2

and

X m . 576⭈400

Ž my 2.

X my1 . 576⭈400

Ž my 2.

my1 , m,

,

for m even

,

for m odd,

X my 2 s X my3 s ⭈⭈⭈ s X 1 s 1.

Then Ž6.2. forces ¨ mŽ X 1 , X 2 , . . . , X m . to have type ŽB1.; hence

¡¨ Ž X

¨ m Ž X1 , X 2 , X m . s

~

¢

2

m , X my1

. 960⭈400

Ž my 3.

,

for m even

¨ 2 Ž X my1 , X m .

960⭈400 Ž my 3.

,

for m odd.

Now, by way of contradiction, suppose the contrary of the claim. It then follows that

² ¨ 2 Ž Xmq , Xm . , ¨ 2 Ž Xm , Xmy1 .: is cyclic, 1

and so by Lemmas 1 and 3 that 1 F ¨ 2 Ž X my1 , X m . s ¨ 2 Ž X m , X my1 . .

540

DONGHI LEE

Applying Lemma 2 to this equality yields the existence of T g Fm such that

Ž 6.3.

X my 1 s TX m Ty1

and

X m s TX my1Ty1 .

Combining these equalities, we have that both ² X my 1 , T : and ² X m , T : are cyclic. Here, if T / 1, then ² X my 1 , X m : is cyclic; if T s 1, then X my 1 s X m by Ž6.3.. This together with Ž6.2. yields a contradiction of the fact that ² X 1 , X 2 , . . . , X m : is non-cyclic, which cimpletes the proof of the claim. The proof of Corollary 2 is completed.

ACKNOWLEDGMENTS The author expresses her sincere thanks to Professor Sergei V. Ivanov for his valuable comments and kind encouragement. The author also thanks the referee for many helpful comments.

REFERENCES 1. G. Baumslag, A. G. Myasnikov, and V. Shpilrain, Open problems in combinatorial group theory, Contemp. Math. 250 Ž1999., 1᎐27. 2. S. V. Ivanov, On certaln elements of free groups, J. Algebra 204 Ž1998., 394᎐405. 3. A. G. Kurosh, ‘‘The Theory of Groups, I, II,’’ 2nd ed., Chelsea, New York, 1960. 4. R. C. Lyndon and P. E. Schupp, ‘‘Combinatorial Group Theory,’’ Springer-Verlag, New YorkrBerlin, 1977. 5. R. C. Lyndon, Cohomology theory of groups with a single defining relation, Ann. of Math. 52 Ž1962., 650᎐665.