On systems of word equations with simple loop sets

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PROOF. Let arb be a factor of v of length |u| +1, where a and b are letters. Since both ar and rb are conjugate with u, we deduce a = b from |ar| a = |rb| a = |u| a . The claim follows. 3 Equally long primitive roots In this section we prove the result announced in the introduction. To simplify notation, we like to formulate it in the following way. Theorem 1 Let m, n be positive integers, and x 0 , . . .,x m , y 0 , . . .,y n , u 1 , . . .,u m , v 1 , . . ., v n words such that for each i, j ∈ {1, 2, . . ., m} the primitive roots of u i and u j are of equal length, and similarly for each i, j ∈ {1, 2, . . ., n} the primitive roots of v i and v j are of equal length. Let k ≥ 1 be a positive integer. If x 0 u i 1 x 1u i 2 x 2 · · ·u i m x m = y 0 v i 1 y 1v i 2 y 2 · · ·v i n y n (i = k, k + 1, k + 2) (1) then also x 0 u i 1x 1 u i 2x 2 · · ·u i mx m = y 0 v i 1y 1 v i 2y 2 · · ·v i ny n (i = 0, 1, 2, 3, . . .) . (2) PROOF. We first introduce several additional assumptions which do not harm generality. Clearly, we may suppose that the words u i , i ∈ {1, . . ., m}, and v i , i ∈ {1, 2, . . ., n}, are non-empty. Assume also that y 0 = ǫ and that either x m = ǫ or y n = ǫ. Let i ∈ {1, 2, . . ., m − 1} be such that x i is empty. We then may suppose that u i and u i+1 do not commute, since otherwise we merge them by writing (u i u i+1 ) j instead of u j iu j i+1. We say that two words u and v, at least one of which is nonempty, are marked if they do not begin with the same symbol. Let i ∈ {1, 2, . . ., m} be such that x i is nonempty. The reasoning below verifies that we may consider, without loss of generality, only cases in which u i and x i , are marked. Suppose that z is the longest nonempty prefix of x i , which is also a prefix of u i x i . Let x ′ i−1 = x i−1z, u ′ i = z−1 u i z, and x ′ i = z−1 x i . It is not difficult to see that x ′ i−1, x ′ i and u ′ i are well defined, and for any j the word x 0 u j 1x 1 u j 2x 2 · · ·u j mx m 4
does not change if we substitute x i−1 , x i , and u i by x ′ i−1 , x′ i , and u′ i , respectively. Repeating the procedure finitely many times we shall obtain the markedness. Analogously we assume that for each i ∈ {1, 2, . . ., n − 1} such that y i = ǫ, the words v i and v i+1 do not commute and that for each j ∈ {1, 2, . . ., n} such that y j ≠ ǫ, the words v j and y j are marked. The proof of the theorem will now proceed by induction with respect to the number m + n. Suppose that m + n ≤ 2. An obvious length argument yields that m = n = 1, x 1 = ǫ, |x 0 | = |y 1 |, and |u 1 | = |v 1 |. From equalities x 0 u i 1 = v i 1y 1 (i = k, k + 1) (3) one obtains that v 1 and u 1 are conjugate over x 0 u k 1 = vk 1 y 1. Lemma 1 now easily implies that (2) holds. Suppose that m + n > 2. We distinguish two main cases: 1 ◦ |x 0 | > |v k 1 |; 2 ◦ |x 0 | ≤ |v k 1|. Consider the first case. If |y 1 | > 0 then the words x 0 , v 1 and y 1 begin with the same symbol, and v 1 , y 1 are not marked, which is against our assumptions. x 0 v k 1 y 1 v k 1 v 1 Let therefore y 1 = ǫ. If |x 0 u k 1 | ≥ min{|vk+1 1 |, |v1 k v 2|} then the words v 1 and v 2 are comparable, i.e., one of them is a prefix of the other. Since the primitive roots of v 1 and v 2 are equally long, they coincide, and v 1 and v 2 commute, again a contradiction with the global assumption. x 0 u k 1 v k 1 v 1 v k 1 v 2 Suppose, on the other hand, that |x 0 u k 1| < min{|v k+1 1 |, |v k 1v 2 |}. 5
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On systems of word equations with simple loop sets

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