On systems of word equations with simple loop sets

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