On systems of word equations with simple loop sets

comparable, and since primitive roots of u 1 and u 2 are equally long, they
commute.
d u k 1 u 2 1
d u k+1
1
u 2
z 1 d u k
1
z 1 d u k
1
z 2
The second main case was |x 0 | ≤ |v k 1|. Recall that we consider the system of
equations
x 0 u i 1 x 1u i 2 x 2 · · ·u i m x m = v i 1 y 1v i 2 y 2 · · ·v i n y n (i = k, k + 1, k + 2) (6)
where k ∈ N + and either x m = ǫ or y n = ǫ . If
|u k+2
1 | ≥ |u 1 | + |v 1 | − 1 and |v k+2
1 | − |x 0 | ≥ |u 1 | + |v 1 | − 1 (7)
then, by the Periodicity Lemma, the primitive roots of u 1 and v 1 are conjugate.
Clearly they are conjugate over x 0 .
x 0 u
k+2
1
v k+2
1
Now the number n + m can be decreased by eliminating u 1 (if |u 1 | > |v 1 |) or
v 1 (if |v 1 | > |u 1 |) or both (if |u 1 | = |v 1 |), and we are through by induction.
Let us be more more rigorous. Using Lemma 1, let t = t 1 t 2 be a primitive
word, and q, r and s positive integers such that u 1 = (t 1 t 2 ) q , v 1 = (t 2 t 1 ) r and
x 0 = t 2 (t 1 t 2 ) s . Suppose that q + s ≥ r (the opposite case being similar). Now
(1) allows us to deduce that
t 2 (t q+s−r ) i x 2 u i 2 · · ·ui m x m = y 1 v i 2 y 2 · · ·v i n y n (i = k, k + 1, k + 2) . (8)
By induction, we deduce that
t 2 (t q+s−r ) i x 2 u i 2 · · ·ui m x m = y 1 v i 2 y 2 · · ·v i n y n (i = 0, 1, 2, . . .) (9)
is true. Obviously, also
t 2 (t q+s ) i x 2 u i 2 · · ·ui m x m = ((t 2 t 1 ) r ) i y 1 v i 2 y 2 · · ·v i n y n (i = 0, 1, 2, . . .) (10)
holds, and we are done.
Assume that (7) does not hold because of
|v 1 | > |u k+1
1 | + 1. (11)
If x 1 ≠ ǫ, the words u 1 and x 1 are not marked. Suppose that x 1 = ǫ. If
|v k+1
1 | ≥ |x 0 | + |u k+2
1 |
7