On systems of word equations with simple loop sets

Note that the previous argument would not work for k = 1.
The only remaining possibility is that α is a factor of ψ(u k+1
j ), shorter than
ψ(u 2 j ).
Then it is easy to see that
The proof is now complete.
|l k+1 | α − |l k | α = |l k+2 | α − |l k+1 | α = 1.
Equality (13) yields
|ϕ(l i )| p = ∑
ϕ(α)=p
|l i | α
for i = k, k + 1, k + 2. In this point we shall exploit the requirement (ii) in
the definition of p. The condition guarantees that there is at least one word
α ∈ H + satisfying ϕ(α) = p, which is a factor of l k+2 and is neither a factor
of l k nor of l k+1 . That implies, together with (15), that
|ϕ(l k+2 )| p − |ϕ(ϕ(l k+1 )| p > |ϕ(l k+1 )| p − |ϕ(l k )| p . (16)
Confronting the last inequality with the structure of the right side of our
equations we get the following claim.
Lemma 6 The primitive root of ϕ(v 1 ) is conjugate with t.
PROOF. Let α be a word from F(r k ) ∪ F(r k+1 ) satisfying ϕ(α) = p. In a
similar manner as in Lemma 5 one can show that for our α the equality
|r k+1 | α − |r k | α = |r k+2 | α − |r k+1 | α . (17)
holds. Then from (16) we deduce that there must exist at least one factor α of
r k+2 , which is neither a factor of r k nor of r k+1 , such that ϕ(α) = p. Such an
α necessarily contains the factor ψ(v k+1 ). The Periodicity Lemma concludes
the proof.
Now, it can be intuitively clear that there cannot exist a loop, the primitive
root of which is not conjugate with t. A proof of this fact is given in the
following lemma.
Lemma 7 For any i ∈ {1, 2, . . ., m} the primitive root of ϕ(u i ) is conjugate
with t.
13