UNIQUENESS OF ROOTS UP TO CONJUGACY FOR SOME ...

10 EON-KYUNG LEE AND SANG-JIN LEE
Claim. Without loss of generality, we may assume that {z 2 , . . . , z m } = {2, . . . , m} (i.e. π α0 (i) = i
for all 1 ≤ i ≤ m) and each of the other cycles of π α0 is of the form (i + r i , . . . , i + 2, i + 1) for
some i ≥ m and r i ≥ 2.
Proof of Claim. Choose an r-permutation θ such that θ(1) = 1, θ({z 2 , . . . , z m }) = {2, . . . , m} and
each cycle (of length ≥ 2) of θπ α0 θ −1 is of the form (i+r i , . . . , i+2, i+1). Note that θπ α0 θ −1 fixes
each point of {1, . . . , m}. Let ζ 0 be an r-braid whose induced permutation is θ, and let ζ = 〈ζ 0 〉 n .
Since ζ 0 is 1-pure, ζ is also 1-pure. Applying Lemma 3.1 (i) to ζ and (α, β, P, k), it suffices to show
that Theorem 1.6 is true for (ζαζ −1 , ζβζ −1 , π ζ (P ), k). Note that R ext (ζαζ −1 ) = ζ ∗ R ext (α) =
C ζ0∗n is standard and that (ζαζ −1 ) ext = ζ 0 α ext ζ0 −1 = ζ 0 β ext ζ0 −1 = (ζβζ −1 ) ext . □
Using the above claim, we assume that π α0 (i) = i for all 1 ≤ i ≤ m and each of the other cycles
of π α0 is of the form (i + r i , . . . , i + 2, i + 1) for some i ≥ m and r i ≥ 2. Then n, α and β are as
follows:
n = (n 1 , . . . , n m , n m+1 , . . . , n m+1 , . . . , n s , . . . , n s ),
} {{ } } {{ }
r m+1
r s
α = 〈α 0 〉 n (α 1 ⊕ · · · ⊕ α m ⊕ (α m+1,1 ⊕ · · · ⊕ α m+1,rm+1 ) ⊕ · · · ⊕ (α s,1 ⊕ · · · ⊕ α s,rs )) n ,
β = 〈α 0 〉 n (β 1 ⊕ · · · ⊕ β m ⊕ (β m+1,1 ⊕ · · · ⊕ β m+1,rm+1 ) ⊕ · · · ⊕ (β s,1 ⊕ · · · ⊕ β s,rs )) n .
By raising the power k large enough, we may assume that the lengths r i of the cycles of π α0
are all divisors of k. Let k = r i p i for m < i ≤ s. Then
where
α k = 〈α0〉 k n (α1 k ⊕ · · · ⊕ αm k ⊕ (˜α p m+1
m+1,1 ⊕ · · · ⊕ ˜αp m+1
m+1,r m+1
) ⊕ · · · ⊕ (˜α p s
s,1 ⊕ · · · ⊕ ˜αps s,r s
)) n ,
β k = 〈α0〉 k n (β1 k ⊕ · · · ⊕ βm k ⊕ ( ˜β p m+1
m+1,1 ⊕ · · · ⊕ ˜β p m+1
ps
m+1,r m+1
) ⊕ · · · ⊕ ( ˜β s,1 ⊕ · · · ⊕ ˜β p s
s,r s
)) n ,
˜α i,j = α i,j−ri +1α i,j−ri +2 · · · α i,j−1 α i,j ,
˜β i,j = β i,j−ri+1β i,j−ri+2 · · · β i,j−1 β i,j
for m < i ≤ s and 1 ≤ j ≤ r i . Hereafter we regard the second index j of (i, j) as being taken
modulo r i . Since α k = β k , one has
α k i = βk i for 1 ≤ i ≤ m,
˜α p i
i,j = ˜β p i
i,j for m < i ≤ s and 1 ≤ j ≤ r i .
Recall that α and β are P -pure, hence α i and β i are P n,i -pure for 1 ≤ i ≤ m by Lemma 2.15.
Recall also that the induced permutation of α 0 fixes no point i > m, hence P n,i = ∅ for i > m.
From now on, we will construct an n-braid γ such that β = γαγ −1 . It will be of the form
γ = (γ 1 ⊕ · · · ⊕ γ m ⊕ (γ m+1,1 ⊕ · · · ⊕ γ m+1,rm+1 ) ⊕ · · · ⊕ (γ s,1 ⊕ · · · ⊕ γ s,rs )) n ,
where γ 1 is 1-unlinked and P n,1 -straight, and γ i is P n,i -straight for 2 ≤ i ≤ m. Then γ is 1-
unlinked by Lemma 2.15 (iii) because γ 1 is 1-unlinked. And γ is P -straight by Lemma 2.15 (ii)
because γ i is P n,i -straight for 1 ≤ i ≤ m and P n,i = ∅ for i > m.
Note that α k 1 = β k 1 and that 1 ∈ P n,1 because 1 ∈ P . By the induction hypothesis on the braid
index, there exists a P n,1 -straight, 1-unlinked n 1 -braid γ 1 with β 1 = γ 1 α 1 γ −1
1 .
Let 2 ≤ i ≤ m. Note that α k i = βk i . If P n,i = ∅, there is an n i -braid γ i such that β i = γ i α i γ −1
i
by [Gon03]. Suppose P n,i ≠ ∅. Then there is an n i -braid ζ i with 1 ∈ π ζi (P n,i ). Since α i and β i are