Basis of Nichols Braided Lie Algebras

0.23 MB·

by  Weicai Wu

Basis of Nichols Braided Lie Algebras
7 Weicai Wu a and Jing Wang b
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0 a. Department of Mathematics, Zhejiang University
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Hangzhou 310007, P.R. China
n
a
J b. Fachbereich Mathematik und Informatik, Philipps-Universita¨t Marburg,

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1 Hans Meerweisse, 35032, Marburg, Germany

] Emails: [email protected](WC); wang [email protected] (JW)
A
Q
.
h Abstract
t
a
m We give a basis of Nichols braided Lie algebras L(V) while B(V) is Nichols
[ algebra of diagonal type with dimV = n ≥ 2 and the generalized Dynkin diagram

1 ofV withp 6= 1forall1≤ i ≤ n. Furthermore,weobtainthedimensionsofNichols
ii
v
6 braidedLiealgebras L(V)ifV is aconnected finiteCartan typewithCartan matrix
0
(a ) .
4 ij n×n
2
0
2010 Mathematics Subject Classification: 16W30, 22E60, 11F23
.
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Keywords: Nichols Lie algebra, Nichols algebra, Nichols braided Lie algebra.
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7
1
:
v 1 Introduction
i
X
r
a The question of finite-dimensionality of Nichols algebras dominates an important part of
the recent developments in the theory of (pointed) Hopf algebras(see e.g. [AHS08, AS10,
He05, He06a, He06b, WZZ15a, WZZ15b]. The interest in this problem comes from the
lifting method by Andruskiewitsch and Schneider to classify finite dimensional (Gelfand-
Kirillov)pointedHopfalgebras, whicharegeneralizationsofquantizedenveloping algebras
of semi-simple Lie algebras.
The classification of arithmetic root systems is obtained in [He05] and [He06a]. It is
proved in [WZZ15b] that Nichols algebra B(V) is finite-dimensional if and only if Nichols
braided Lie algebra L(V) is finite-dimensional.
In this paper we focus on the Nichols algebra which are finite dimension. Such finite-
dimensional Nichols algebras play a fundamental role in various subjects such as pointed
Hopf algebras and logarithmic quantum field theories.
1
Here are the main results of this paper. We give a basis of Nichols braided Lie algebras
L(V) while B(V) is Nichols algebra of diagonal type with dimV = n ≥ 2 and the
generalized Dynkin diagram of V with p 6= 1 for all 1 ≤ i ≤ n. Furthermore, we obtain
ii
the dimensions of Nichols braided Lie algebras L(V) if V is a connected finite Cartan
type with Cartan matrix (a ) .
ij n×n
2 Preliminaries
Let B(V) be the Nichols algebra generated by vector space V. Throughout this paper
braided vector space V is of diagonal type with basis x ,x ,··· ,x and C(x ⊗ x ) =
1 2 n i j
p x ⊗x , n > 1 without special announcement. Define linear map p from B(V)⊗B(V)
ij j i
to F such that p(u⊗v) = χ(deg(u),deg(v)), for any homogeneous element u,v ∈ B(V).
For convenience, p(u⊗v) is denoted by p .
uv
Denote ord(p ) the order of p with respect to multiplication. Let |u| denote
uu uu
length of homogeneous element u ∈ B(V). Let D =: {[u] | [u] is a hard super-letter },
∆+(B(V)) := {deg(u) | [u] ∈ D}, ∆(B(V)) := ∆+(B(V))∪∆−(B(V)), which is called
the root system of V. If ∆(B(V)) is finite, then it is called an arithmetic root system.
Let L(V) denote the braided Lie algebras generated by V in B(V) under Lie operations
[x,y] = yx − p xy, for any homogeneous elements x,y ∈ B(V). (L(V),[ ]) is called
yx
Nichols braided Lie algebra of V. The other notations are the same as in [WZZ15a].
Recall the dual B(V∗) of Nichols algebra B(V) of rank n in [He05, Section 1.3]. Let
y be a dual basis of x . δ(y ) = g−1 ⊗y , g ·y = p−1y and ∆(y ) = g−1 ⊗y +y ⊗1.
i i i i i i j ij j i i i i
There exists a bilinear map < ·,· >: (B(V∗)#FG)×B(V) −→ B(V) such that
< y ,uv >=< y ,u > v +g−1.u < y ,v > and < y ,< y ,u >>=< y y ,u >
i i i i i j i j
for any u,v ∈ B(V). Furthermore, for any u ∈ ⊕∞ B(V) , one has that u = 0 if and
i=1 (i)
only if < y ,u >= 0 for any 1 ≤ i ≤ n.
i
We have the braided Jacobi identity as follows:
[[u,v],w] = [u,[v,w]]+p−1[[u,w],v]+(p −p−1)v ·[u,w]. (1)
vw wv vw
[u,v·w] = p [uv]·w +v ·[uw]. (2)
wu
Throughout, Z =: {x | x is an integer}. R =: {x | x is a real number}. N =: {x | x ∈
0
Z,x ≥ 0}. N =: {x | x ∈ Z,x > 0}. F denotes the base field, which is an algebraic closed
field with characteristic zero. F∗ = F\{0}. S denotes symmetric group, n ∈ N. For any
n
set X, | X | is the cardinal of X. int(a) means the biggest integer not greater than a ∈ R.
2
3 Main results
For any homogeneous element u ∈ B(V), we denote d(u) by the subset of {x ,x ,··· ,x }
1 2 n
whichgeneratesu. Wesaythatd(u)isconnectedifforalli,j ∈ d(u)thereexistss ∈ Nand
a sequence {i = i ,i ,··· ,i = j} of elements of d(u) such that p satisfy the conditions
0 1 s ij
p p 6= 1 for all l ∈ {0,1,··· ,s−1}.
i ,i i ,i
l l+1 l+1 l
Let h ∈ {x ,x ,··· ,x },∀i ∈ N. If d(h ···h ) = {x ,··· ,x },1 ≤ j < ··· < j ≤
i 1 2 n 1 m j1 jt 1 t
n. Set d (h ···h ) := {i ∈ {1,··· ,m} | h = x } for all k ∈ {1,...,t}. It is clear
jk 1 m i jk
t
P | d (h ···h ) |= m.
jk 1 m
k=1
Lemma 3.1. If ord(p ) >| d (h ···h ) | for all r ∈ {1,...,m}, then h ···h 6= 0.
xjr,xjr jr 1 m 1 m
Proof. We show this by induction on | d(h ···h ) |= t. If x ∈ d(h ···h ), assume that
1 m k 1 m
| d (h ···h ) |= l, denote h ···h h h ···h h h ···h h h ···h
k 1 m 1 k1−1 k1 k1+1 k2−1 k2 k2+1 kl−1 kl kl+1 m
:= u x u x ···u x u for h = x ,1 ≤ i ≤ l. we know if k = 1, then u = ∅, if
1 k 2 k l k l+1 ki k 1 1
k = k +1, then u = ∅, if k = m, then u = ∅.
i i−1 i l l+1
If t = 1, we obtain h ···h 6= 0 by [He05, Lemma 1.3.3(i)].
1 m
If t > 1, we need to prove < yl,u x u x ···u x u >
k 1 k 2 k l k l+1
= (1+p−1)(1+p−1 +p−2)···(1+p−1 +···+p−l+1)p−1 p−1 ···p−1 u u ···u :
kk kk kk kk kk k,u1 k,u1u2 k,u1u2···ul 1 2 l+1
we show this by induction on l: If l = 1, < y ,u x u >= p−1 u u . If l > 1, then
k 1 k 2 k,u1 1 2
< y ,u x u x ···u x u >
k 1 k 2 k l+1 k l+2
= p−1 u u x ···u x u +p−1p−1 u x u u ···u x u +···
k,u1 1 2 k l+1 k l+2 kk k,u1u2 1 k 2 3 l+1 k l+2
+p−lp−1 u x u x ···x u u .
kk k,u1u2···ul+1 1 k 2 k k l+1 l+2
< yl+1,u x u x ···u x u >=< yl,< y ,u x u x ···u x u >>
k 1 k 2 k l+1 k l+2 k k 1 k 2 k l+1 k l+2
= p−1 (1+p−1)···(1+p−1 +···+p−l+1)p−1 p−1 ···p−1 u u ···u
k,u1 kk kk kk k,u1u2 k,u1u2u3 k,u1u2u3···ul+1 1 2 l+2
+p−1p−1 (1+p−1)···(1+p−1 +···+p−l+1)p−1 p−1 ···p−1 u u ···u
kk k,u1u2 kk kk kk k,u1 k,u1u2u3 k,u1u2u3···ul+1 1 2 l+2
+···
+p−lp−1 (1+p−1)···(1+p−1 +···+p−l+1)p−1 p−1 ···p−1 u u ···u
kk k,u1u2···ul+1 kk kk kk k,u1 k,u1u2 k,u1u2···ul 1 2 l+2
= (1+p−1)···(1+p−1+···+p−l+1)(1+p−1+···+p−l)p−1 ···p−1 u u ···u .
kk kk kk kk kk k,u1 k,u1u2u3···ul+1 1 2 l+2
We know | d(u u ···u ) |= t − 1, then u u ···u 6= 0 by induction hypotheses,
1 2 l+1 1 2 l+1
(1+p−1)(1+p−1+p−2)···(1+p−1+···+p−l+1) 6= 0 since ord(p ) >| d (h ···h ) |= l.
kk kk kk kk kk xk,xk k 1 m
Hence u x u x ···u x u 6= 0, completing the proof. ✷
1 k 2 k l k l+1
Lemma 3.2. Assumed that u,v,w are homogeneous elements in L(V). Let a,b,c,d,e,f
denote 1−p p ,1−p p ,1−p p ,1−p p p p ,1−p p p p ,1−p p p p ,
wv vw uw wu uv vu uv vu uw wu uv vu wv vw wv vw uw wu
respectively. If R = {r|r ∈ {uv,uw,vw} and r ∈ L(V)}, T = {t|t ∈ {a,b,c} and t 6= 0},
then (i) uvw,uwv,vwu,vuw,wuv,wvu∈ L(V) when |R| ≥ 2 and |T| =6 0. (ii) ([WZZ15b,
Lem.3.1]) uvw,uwv,vwu,vuw,wuv,wvu∈ L(V) when |T| ≥ 2.
Proof. (i) Assumed that |R| ≥ 2 and |T| =6 0. Without loss of generality, we let uv,uw ∈
L(V). If vuw ∈/ L(V), then e = 0,f = 0 by [WZZ15b, Lemma 3.2(i)] and [WZZ15a,
3
Lemma 4.12]. If a = 0, then b = 0,c = 0. this is a contradiction to |T| =6 0; If a 6= 0,
then b 6= 0,c 6= 0. Thus one obtains a contradiction to [WZZ15b, Lemma 3.1]. Then
we know vuw ∈ L(V), similarly, wuv ∈ L(V). And we find wvu,uwv ∈ L(V) and
vwu,uvw ∈ L(V) by [WZZ15b, Lemma 3.2(i)], respectively. ✷
Let b(h h ···h ) denote a set contain all nonassociative word (see [WZZ15a, Sec-
1 2 m
tion 2]) with brackets [,] on h ,h ,··· ,h , for example, b(h h h ) = {[[h ,h ],h ],
1 2 m 1 2 3 1 2 3
[[h ,h ],h ],[[h ,h ],h ],[[h ,h ],h ],[[h ,h ],h ],[[h ,h ],h ],[h ,[h ,h ]],[h ,[h ,h ]],
1 3 2 2 1 3 2 3 1 3 2 1 3 1 2 1 2 3 1 3 2
[h ,[h ,h ]],[h ,[h ,h ]],[h ,[h ,h ]],[h ,[h ,h ]]}. For i ,··· ,i are pairwise different in-
2 1 3 2 3 1 3 2 1 3 1 2 1 r
tegers, denote T the set which change all the subscript 1,2,...,r of the elements of
i1,...,ir
S to (i ,...,i ), for example, T = {(2) = (4) = (7) = id,(24),(27),(47),(247),(274)}.
r 1 r 2,4,7
It is clear |T | = |S | and T = S .
i1,...,ir r 1,...,r r
Assumed that u,v are homogeneous elements in B(V). u,v is called generalized
commutative if ∃p ∈ F∗ such that uv = pvu.
Remark 3.3. h ,h is generalizedcommutative ifp p = 1: h h = p h h ,p ∈
i j hi,hj hj,hi i j hi,hj j i hi,hj
F∗.
Lemma 3.4. (i) If d(h h ···h ) is not connected, then b(h h ···h ) = 0.
1 2 m 1 2 m
(ii) If h h ···h 6= 0 and d(h h ···h ) is not connected, then h h ···h ∈/ L(V).
1 2 m 1 2 m 1 2 m
(iii) If d(h h ···h ) is not connected, then h h ···h 6= 0 if and only if h h ···h ∈/
1 2 m 1 2 m 1 2 m
L(V).
Proof. (i) We show this by induction on |h h ···h |. ∀w ∈ b(h h ···h ), then ∃u ∈
1 2 m 1 2 m
b(h h ···h ),v ∈ b(h h ···h ), τ ∈ S ,k ∈ {1,...,m} such that w =
τ τ τ τ τ τ m
(1) (2) (k) (k+1) (k+2) (m)
[u,v]. If d(u) or d(v) is not connected, then u = 0 or v = 0 by induction hypotheses,
so w = 0. If d(u) and d(v) are connected, we know p p = 1 for any h ∈ d(u),
hi,hj hj,hi i
h ∈ d(v) since d(h h ···h ) is not connected.
j 1 2 m
(1) For ∀h ∈ d(v), if p p = 1 for l ∈ {1,...,k}, then [u,h ] = 0. We
j τ ,j j,τ j
(l) (l)
show this by induction on |u|. If k = 1, it is clear. If k > 1, we set u = [u ,u ],
1 2
[[u ,u ],h ] = [u ,[u ,h ]]+ p−1 [[u ,h ],u ] +(p −p−1 )u ·[u ,h ] = 0 by (1) and
1 2 j 1 2 j u2hj 1 j 2 hju2 u2hj 2 1 j
induction hypotheses.
(2)Ifp p = 1forl ∈ {1,...,k},t ∈ {k+1,...,m}, then[u,h ···h ] =
τ ,τ τ ,τ τ τ
(l) (t) (t) (l) (k+1) (m)
h ·[u,h ···h ] = ··· = h ···h ·[u,h ] = 0 by (2).
τ(k+1) τ(k+2) τ(m) τ(k+1) τ(m−1) τ(m)
(3) We know v = Pk h ···h for some k ∈ F∗,τ ∈ T , then
i τiτ(k+1) τiτ(m) i i τ(k+1),...,τ(m)
i
[u,v] = [u,Pk h ···h ] = Pk [u,h ···h ] = 0.
i τiτ(k+1) τiτ(m) i τiτ(k+1) τiτ(m)
i i
✷
(ii) and (iii) are clear by (i).
Corollary 3.5. If u and v are homogeneous elements in B(V), d(u)∩d(v) = ∅, p p = 1
ij ji
for any x ∈ d(u), x ∈ d(v), then the following conditions are equivalent: (i) u 6= 0,v 6= 0.
i j
(ii) uv 6= 0. (iii) uv ∈/ L(V).
4
Proof. We know (ii)⇐⇒ (iii) by Lemma 3.4. (ii)=⇒ (i) is clear. (ii)⇐= (i): assume
| u |= k, then ∃y ,...,y ∈ d(u) such that < y ···y ,u >∈ F∗, then < y ···y ,uv >
i1 ik i1 ik i1 ik
=< y ···y ,u > v 6= 0. ✷
i1 ik
Now we assumed that p 6= 1 for all i ∈ {1,...,n}.
ii
Proposition 3.6. (i) Ifh ···h ∈ L(V)and p p 6= 1forsomek ∈ {1,...,m−
1 m−1 hk,hm hm,hk
1} or h ···h ∈ L(V) and p p 6= 1 for some l ∈ {2,...,m}, then h ···h and
2 m hl,h1 h1,hl 1 b
h ···h ∈ L(V) for some b ∈ {1,...,m} and p p 6= 1 for some a ∈ {1,...,b},
b+1 m ha,hc hc,ha
c ∈ {b+1,...,m}.
(ii) Assumed that h ···h 6= 0 or h ···h 6= 0 for some b ∈ {1,...,m}. If
1 b b+1 m
h ···h ,h ···h ∈ L(V) and p p 6= 1 for some a ∈ {1,...,b},c ∈ {b +
1 b b+1 m ha,hc hc,ha
1,...,m}, then there exist τ ∈ T ,p ∈ F∗ such that h ···h = ph ···h with
1,...,m 1 m τ(1) τ(m)
h ···h ∈ L(V),p p 6= 1 for some k ∈ {1,...,m − 1} or with
τ(1) τ(m−1) hτ(k),hτ(m) hτ(m),hτ(k)
h ···h ∈ L(V),p p 6= 1 for some l ∈ {2,...,m}.
τ(2) τ(m) hτ(l),hτ(1) hτ(1),hτ(l)
(iii) If h ···h ∈ L(V) and p p 6= 1 for some k ∈ {1,...,m − 1} or
1 m−1 hk,hm hm,hk
h ···h ∈ L(V) and p p 6= 1 for some l ∈ {2,...,m}, then h ···h ∈ L(V).
2 m hl,h1 h1,hl 1 m
(iv) Assumed that h ···h 6= 0. If h ···h ∈ L(V), then there exist τ ∈ T ,p ∈
1 m 1 m 1,...,m
F∗ such thath ···h = ph ···h with 0 6= h ···h ∈ L(V),p p
1 m τ(1) τ(m) τ(1) τ(m−1) hτ(k),hτ(m) hτ(m),hτ(k)
6= 1 for some k ∈ {1,...,m−1} or with 0 6= h ···h ∈ L(V),p p 6= 1
τ(2) τ(m) hτ(l),hτ(1) hτ(1),hτ(l)
for some l ∈ {2,...,m}.
Proof. (i) Let a = k,b = m−1,c = m or a = 1,b = 1,c = l.
We show (ii), (iii) and (iv) by induction on the length of |h h ···h | = m.
1 2 m
Assume m = 2. (ii) is clear. If h 6= h , then (iii),(iv) follows from [WZZ15a, Lemma
1 2
4.12],[WZZ15a, Lemma 5.2], respectively. If h = h , then (iii),(iv) follows from[WZZ15a,
1 2
Lemma 4.3] , [He05, Lemma 1.3.3(i)], respectively.
Now let m > 2.
(ii) If b = m − 1, let τ = id,k = a,p = 1; if b = 1, let τ = id,l = c,p = 1. Now
assumed that 1 < b < m−1. We denote p˜ := p p for ∀u,v ∈ B(V).
u,v u,v v,u
(1) Assumed that h ···h 6= 0, we proceed by induction over b.
1 b
We know there exist τ ∈ T ,p ∈ F∗ such that h ···h = p h ···h with (a)
1,...,b 1 1 b 1 τ(1) τ(b)
0 6= h ···h ∈ L(V),p p 6= 1 for some k ∈ {1,...,b − 1} or
τ(1) τ(b−1) hτ(k1),hτ(b) hτ(b),hτ(k1) 1
with (b) 0 6= h ···h ∈ L(V),p p 6= 1 for some l ∈ {2,...,b} by
τ(2) τ(b) hτ(l1),hτ(1) hτ(1),hτ(l1) 1
h ···h ∈ L(V) and induction hypotheses of (iv).
τ(1) τ(b)
(a) . If there exists j ∈ {b+1,...,m} such that p˜ 6= 1, then h h ···h ∈
1 1 hτ(b),hj1 τ(b) b+1 m
L(V) by induction hypotheses of (iii), and p˜ 6= 1 for k ∈ {1,...,b − 1}. It is
hτ(k1),hτ(b) 1
proved since |h ···h | = b−1 < b and induction hypotheses.
τ(1) τ(b−1)
(a) . If all j ∈ {b + 1,...,m} such that p˜ = 1, then h h ···h =
2 1 hτ(b),hj1 τ(b) b+1 m
p h ···h h forsomep ∈ F∗ byRemark3.3. Ontheotherhand, weknowp˜ 6=
2 b+1 m τ(b) 2 hτ(k1),hc
5
1 for some k ∈ {1,...,b − 1} by p˜ 6= 1 , then h ···h h ···h ∈
1 ha,hc τ1τ(1) τ1τ(b−1) τ1(b+1) τ1(m)
L(V), h ···h h ···h = p h ···h h ···h for some τ ∈
τ(1) τ(b−1) b+1 m 3 τ1τ(1) τ1τ(b−1) τ1(b+1) τ1(m) 1
T ,p ∈ F∗ by the induction hypotheses of (ii) and (iii). We know there
τ(1),...,τ(b−1),b+1,...,m 3
exists k ∈ {τ(1),...,τ(b−1),b+1,...,m} such that p˜ 6= 1 since p˜ 6= 1.
2 hτ(b),hτ1(k2) hτ(b),hτ(k1)
We obtain h ···h = p h ···h h ···h = p p h ···h h ···h h
1 m 1 τ(1) τ(b) b+1 m 1 2 τ(1) τ(b−1) b+1 m τ(b)
= p p p h ···h h ···h h
1 2 3 τ1τ(1) τ1τ(b−1) τ1(b+1) τ1(m) τ(b)
= p p p h ···h h ···h h byh ···h = h ···h
1 2 3 τ1τ(1) τ1τ(b−1) τ1τ(b+1) τ1τ(m) τ1τ(b) τ1(b+1) τ1(m) τ1τ(b+1) τ1τ(m)
since τ ∈ T and h = h since τ ∈ T . Set
1,2,...,b τ1τ(b) τ(b) 1 τ(1),τ(2),...,τ(b−1),b+1,...,m
τ′ = τ1τ1(1) ······ τ1τb(−b−11) τ1τ(bb+1) ······ τm1τ−(m1) τ1mτ(b) !, then we obtain hτ′(1)···hτ′(m−1) ∈
L(V),p˜hτ′(k),hτ′(m) 6= 1 and h1···hm = p1p2p3hτ′(1)···hτ′(m) for some τ′ ∈ T1,...,m,k ∈
{1,...,m−1},p p p ∈ F∗.
1 2 3
(b) . If there exists j ∈ {2,...,b},j ∈ {b+1,...,m} such that p˜ 6= 1, then
1 1 2 hτ(j1),hj2
h ···h h ···h ∈ L(V), and h ···h h ···h
τ1τ(2) τ1τ(b) τ1(b+1) τ1(m) τ(2) τ(b) b+1 m
= p h ···h h ···h for some τ ∈ T ,p ∈ F∗ by the in-
4 τ1τ(2) τ1τ(b) τ1(b+1) τ1(m) 1 τ(2),...,τ(b),b+1,...,m 4
ductionhypotheses of(ii)and(iii). Weknowthereexistsl ∈ {τ(2),...,τ(b),b+1,...,m}
2
such that p˜ 6= 1 since p˜ 6= 1. We obtain h ···h
hτ(1),hτ1(l2) hτ(1),hτ(l1) 1 m
= p h ···h h ···h = p p h h ···h h ···h
1 τ(1) τ(b) b+1 m 1 4 τ(1) τ1τ(2) τ1τ(b) τ1(b+1) τ1(m)
= p p h h ···h h ···h by h ···h = h ···h
1 4 τ1τ(1) τ1τ(2) τ1τ(b) τ1τ(b+1) τ1τ(m) τ1(b+1) τ1(m) τ1τ(b+1) τ1τ(m)
since τ ∈ T and h = h since τ ∈ T . Set τ′ = τ τ, then we
1,...,b τ1τ(1) τ(1) 1 τ(2),...,τ(b),b+1,...,m 1
obtain hτ′(2)···hτ′(m) ∈ L(V),p˜hτ′(l),hτ′(1) 6= 1 and h1···hm = p1p4hτ′(1)···hτ′(m) for some
τ′ ∈ T ,l ∈ {2,...,m},p p ∈ F∗.
1,...,m 1 4
(b) . If all j ∈ {2,...,b},l ∈ {b+1,...,m} such that p˜ = 1, then
2 1 2 hτ(j1),hj2
h ···h h ···h = p h ···h h ···h for some p ∈ F∗ by Remark 3.3.
τ(2) τ(b) b+1 m 5 b+1 m τ(2) τ(b) 5
On the other hand, we know p˜ 6= 1 by p˜ 6= 1, then h h ···h ∈ L(V)
hτ(1),hc ha,hc τ(1) b+1 m
by induction hypotheses of (iii). We know h ···h = p h ···h h ···h =
1 m 1 τ(1) τ(b) b+1 m
p p h h ···h h ···h and p˜ 6= 1. It is proved since |h ···h | =
1 5 τ(1) b+1 m τ(2) τ(b) hτ(l1),hτ(1) τ(2) τ(b)
b−1 < b and induction hypotheses.
(2) Assumed that h h ···h 6= 0, we proceed by induction over m−b.
b+1 b+2 m
We know there exist τ ∈ T ,p ∈ F∗ such that h ···h = p h ···h
b+1,...,m 6 b+1 m 6 τ(b+1) τ(m)
with (c) 0 6= h ···h ∈ L(V),p p 6= 1 for some k ∈ {b +
τ(b+1) τ(m−1) hτ(k1),hτ(m) hτ(m),hτ(k1) 1
1,...,m − 1} or with (d) 0 6= h ···h ∈ L(V),p p 6= 1 for
τ(b+2) τ(m) hτ(l1),hτ(b+1) hτ(b+1),hτ(l1)
some l ∈ {b+2,...,m} by h ···h ∈ L(V) and induction hypotheses of (iv).
1 τ(b+1) τ(m)
(c) . If there exists j ∈ {b+1,...,m−1},j ∈ {1,...,b} such that p˜ 6= 1, then
1 1 2 hτ(j1),hj2
h ···h h ···h ∈ L(V),andh ···h h ···h = p h ···h
τ1(1) τ1(b) τ1τ(b+1) τ1τ(m−1) 1 b τ(b+1) τ(m−1) 7 τ1(1) τ1(b)
h ···h for some τ ∈ T ,p ∈ F∗ by the induction hypothe-
τ1τ(b+1) τ1τ(m−1) 1 1,...,b,τ(b+1),...,τ(m−1) 7
ses of (ii) and (iii). We know there exists k ∈ {1,...,b,τ(b+1),...,τ(m−1)} such that
2
p˜ 6= 1 since p˜ 6= 1. We obtain h ···h
hτ(m),hτ1(k2) hτ(m),hτ(k1) 1 m
= p h ···h h ···h h = p p h ···h h ···h h
6 1 b τ(b+1) τ(m−1) τ(m) 6 7 τ1(1) τ1(b) τ1τ(b+1) τ1τ(m−1) τ(m)
6
= p p h ···h h ···h h byh ···h = h ···h since
6 7 τ1τ(1) τ1τ(b) τ1τ(b+1) τ1τ(m−1) τ1τ(m) τ1(1) τ1(b) τ1τ(1) τ1τ(b)
τ ∈ T and h = h since τ ∈ T . Set τ′ = τ τ,
b+1,b+2,...,m τ1τ(m) τ(m) 1 1,2,...,b,τ(b+1),τ(b+2),...,τ(m−1) 1
thenweobtainhτ′(1)···hτ′(m−1) ∈ L(V),p˜hτ′(k),hτ′(m) 6= 1andh1···hm = p6p7hτ′(1)···hτ′(m)
for some τ′ ∈ T ,k ∈ {1,...,m−1},p p ∈ F∗.
1,...,m 6 7
(c) . If all j ∈ {b + 1,...,m − 1},j ∈ {1,...,b} such that p˜ = 1, then
2 1 2 hτ(j1),hj2
h ···h h ···h = p h ···h h ···h for some p ∈ F∗ by Remark
1 b τ(b+1) τ(m−1) 8 τ(b+1) τ(m−1) 1 b 8
3.3. On the other hand, we know p˜ 6= 1 by p˜ 6= 1, then h ···h h ∈
hτ(m),ha ha,hc 1 b τ(m)
L(V) by induction hypotheses of (iii). We know h ···h = p h ···h h ···h =
1 m 6 1 b τ(b+1) τ(m)
p p h ···h h ···h h and p˜ 6= 1. Then |h h ···h | =
6 8 τ(b+1) τ(m−1) 1 b τ(m) hτ(k1),hτ(m) τ(b+1) τ(b+2) τ(m−1)
m−b−1 < m−b and induction hypotheses give the proof.
(d) . If there exists j ∈ {1,...,b} such that p˜ 6= 1, then h ···h h ∈
1 1 hτ(b+1),hj1 1 b τ(b+1)
L(V) by induction hypotheses of (iii), and p˜ 6= 1 for l ∈ {b+2,...,m}. It is
hτ(l1),hτ(b+1) 1
proved since |h ···h | = m−b−1 < m−b and induction hypotheses.
τ(b+2) τ(m)
(d) . If all j ∈ {1,...,b} such that p˜ = 1, then we have h ···h h =
2 1 hτ(b+1),hj1 1 b τ(b+1)
p h h ···h forsome p ∈ F∗ by Remark3.3. Ontheother hand, we know p˜ 6=
9 τ(b+1) 1 b 9 hτ(l1),ha
1forsomel ∈ {b+2,...,m}byp˜ 6= 1, thenh ···h h ···h ∈ L(V),
1 ha,hc τ1(1) τ1(b) τ1τ(b+2) τ1τ(m)
h ···h h ···h = p h ···h h ···h forsomeτ ∈ T ,
1 b τ(b+2) τ(m) 10 τ1(1) τ1(b) τ1τ(b+2) τ1τ(m) 1 1,...,b,τ(b+2),...,τ(m)
p ∈ F∗ by the induction hypotheses of (ii) and (iii). We know there exists l ∈
10 2
{1,...,b,τ(b + 2),...,τ(m)} such that p˜ 6= 1 since p˜ 6= 1. We
h ,h h ,h
τ(b+1) τ1(l2) τ(b+1) τ(l1)
obtain h ···h = p h ···h h ···h = p p h h ···h h ···h
1 m 6 1 b τ(b+1) τ(m) 6 9 τ(b+1) 1 b τ(b+2) τ(m)
= p p p h h ···h h ···h
6 9 10 τ(b+1) τ1(1) τ1(b) τ1τ(b+2) τ1τ(m)
= p p p h h h h ···h by h ···h = h ···h since
6 9 10 τ1τ(b+1) τ1τ(1) τ1τ(b) τ1τ(b+2) τ1τ(m) τ1(1) τ1(b) τ1τ(1) τ1τ(b)
τ ∈ T and h = h since τ ∈ T . Set
b+1,...,m τ1τ(b+1) τ(b+1) 1 1,...,b,τ(b+2),...,τ(m)
τ′ = τ1τ(b1+1) τ1τ2(1) ······ τb1+τ(b1) τ1τb(+b+22) ······ τ1τm(m) !,thenweobtainhτ′(2)···hτ′(m) ∈ L(V),
p˜hτ′(l),hτ′(1) 6= 1 and h1···hm = p6p9p10hτ′(1)···hτ′(m) for some τ′ ∈ T1,...,m,l ∈ {2,...,m},
p p p ∈ F∗.
6 9 10
(iii) Assumed that h h ···h ∈/ L(V). For convenience, set τ0 = τ = id,τ ∈
1 2 m 0 k
T or T or ··· or T , τk = τ τ ···τ
τk−1(1),...,τk−1(m−k) τk−1(2),...,τk−1(m−k+1) τk−1(k+1),...,τk−1(m) k k−1 1
for k ≥ 1. We need to prove Assertion(j−i) holds for all j−i ∈ {0,...,m−3}. Denote
A = {1,...,m}−{τm−j+i−2(i),...,τm−j+i−2(j)} for all i ≤ j ∈ {1,...,m}
ij
Assertion(j−i) : 0 6= hτm−j+i−2(i)···hτm−j+i−2(j) ∈ L(V), p˜hr,ht = 1 for ∀r 6= t ∈ Aij.
p˜ 6= 1, and p˜ = 1 for some k ∈ {i,...,j}, all r ∈ A .
hτm−j+i−2(k),hr hτm−j+i−2(i)···hτm−j+i−2(j),hr ij
h1···hm = p1···pm−j+i−2hτm−j+i−2(1)···hτm−j+i−2(m) for some p1,...,pm−j+i−2 ∈ F∗.
We prove this by induction on j −i.
Step 1. Assertion(m−3) holds:
Ifh ···h ∈ L(V),thenp˜ = 1by[WZZ15a,Lemma4.12]. Ifh ···h =
1 m−1 h1···hm−1,hm 1 m−1
0, this is a contradiction to h h ···h ∈/ L(V), then h h ···h 6= 0. We know
1 2 m 1 2 m−1
there exist τ ∈ T , p ∈ F∗ such that h ···h = p h ···h with (a)
1 1,...,m−1 1 1 m−1 1 τ1(1) τ1(m−1)
7
0 6= h ···h ∈ L(V),p˜ 6= 1 for some k ∈ {1,...,m−2} or with (b)
τ1(1) τ1(m−2) hτ1(k),hτ1(m−1)
0 6= h ···h ∈ L(V),p˜ 6= 1 for some l ∈ {2,...,m − 1} by induction
τ1(2) τ1(m−1) hτ1(l),hτ1(1)
hypotheses of (iv). We know h = h .
τ1(m) m
(a) If p˜ 6= 1, then p˜ 6= 1 since p˜ = 1. We obtain
hτ1(m−1),hm hτ1(1)···hτ1(m−2),hm h1···hm−1,hm
h ···h h h ∈ L(V) by Lemma 3.2(iii). It is a contradiction to
τ1(1) τ1(m−2) τ1(m−1) m
p h ···h h = h ···h ∈/ L(V). Then p˜ = 1 and p˜ =
1 τ1(1) τ1(m−1) m 1 m hτ1(m−1),hm hτ1(1)···hτ1(m−2),hm
1. We obtain p˜ 6= 1 for some k ∈ {1,...,m − 2} since p˜ 6= 1. Then
hτ1(k1),hm 1 hk,hm
h ···h h ∈ L(V) by induction hypotheses. pp h ···h h h
τ1(1) τ1(m−2) m 1 τ1(1) τ1(m−2) m τ1(m−1)
= p h ···h h h = h ···h for some p ∈ F∗ by Remark 3.3, then
1 τ1(1) τ1(m−2) τ1(m−1) m 1 m
p˜ = 1by[WZZ15a,Lemma4.12],weobtainp˜ =
hτ1(m−1),hτ1(1)···hτ1(m−2)hm hτ1(m−1),hτ1(1)···hτ1(m−2)
1.
(b) If p˜ 6= 1, then p˜ 6= 1 since p˜ = 1. We obtain
hτ1(1),hm hτ1(2)···hτ1(m−1),hm h1···hm−1,hm
h ···h h h ∈ L(V) by Lemma 3.2(iii). It is a contradiction to
τ1(1) τ1(m−2) τ1(m−1) m
p h ···h h = h ···h ∈/ L(V). Then p˜ = 1 and p˜ =
1 τ1(1) τ1(m−1) m 1 m hτ1(1),hm hτ1(2)···hτ1(m−1),hm
1. We obtain p˜ 6= 1 for some k ∈ {2,...,m − 1} since p˜ 6= 1. Then
hτ1(k1),hm 1 hk,hm
h ···h h ∈ L(V)byinductionhypotheses. p h h ···h h = h ···h ,
τ1(2) τ1(m−1) m 1 τ1(1) τ1(2) τ1(m−1) m 1 m
thenp˜ = 1by[WZZ15a,Lemma4.12],weobtainp˜ =
hτ1(1),hτ1(2)···hτ1(m−1)hm hτ1(1),hτ1(2)···hτ1(m−1)
1.
If h ···h ∈ L(V), then p˜ = 1 by [WZZ15a, Lemma 4.12]. If h ···h = 0,
2 m h2···hm,h1 2 m
this isa contradictiontoh h ···h ∈/ L(V), thenh ···h 6= 0. Weknow thereexist τ ∈
1 2 m 2 m 1
T , p ∈ F∗ such that h ···h = p h ···h with (c) 0 6= h ···h ∈
2,...,m 1 2 m 1 τ1(2) τ1(m) τ1(2) τ1(m−1)
L(V),p˜ 6= 1 for some k ∈ {2,...,m − 1} or with (d) 0 6= h ···h ∈
hτ1(k),hτ1(m) τ1(3) τ1(m)
L(V),p˜ 6= 1 for some l ∈ {3,...,m} by induction hypotheses of (iv). We know
h ,h
τ1(l) τ1(2)
h = h .
τ1(1) 1
(c) If p˜ 6= 1, then p˜ 6= 1 since p˜ = 1. We obtain
hτ1(m),h1 hτ1(2)···hτ1(m−1),h1 h2···hm,h1
h h ···h ∈ L(V) by Lemma 3.2(iii). It is a contradiction to p h h ···h =
1 τ1(2) τ1(m) 1 1 τ1(2) τ1(m)
h ···h ∈/ L(V). Then p˜ = 1 and p˜ = 1. We obtain p˜ 6= 1
1 m hτ1(m),h1 hτ1(2)···hτ1(m−1),h1 hτ1(l1),h1
for some l ∈ {2,...,m−1} since p˜ 6= 1. Then h h ···h ∈ L(V) by induc-
1 hl,h1 1 τ1(2) τ1(m−1)
tion hypotheses. p h h ···h h = h ···h , then p˜ = 1
1 1 τ1(2) τ1(m−1) τ1(m) 1 m hτ1(m),h1hτ1(2)···hτ1(m−1)
by [WZZ15a, Lemma 4.12], we obtain p˜ = 1.
hτ1(m),hτ1(2)···hτ1(m−1)
(d) If p˜ 6= 1, then p˜ 6= 1 since p˜ = 1. We obtain
hτ1(2),h1 hτ1(3)···hτ1(m),h1 h2···hm,h1
h h ···h ∈ L(V) by Lemma 3.2(iii). It is a contradiction to p h h ···h =
1 τ1(2) τ1(m) 1 1 τ1(2) τ1(m)
h ···h ∈/ L(V). Then p˜ = 1 and p˜ = 1. We obtain p˜ 6= 1
1 m hτ1(2),h1 hτ1(3)···hτ1(m),h1 hτ1(l1),h1
for some l ∈ {3,...,m} since p˜ 6= 1. Then h h ···h ∈ L(V) by induction
1 hl,h1 1 τ1(3) τ1(m)
hypotheses. pp h h h ···h = p h h ···h = h ···h for some p ∈
1 τ1(2) 1 τ1(3) τ1(m) 1 1 τ1(2) τ1(m) 1 m
F∗ by Remark 3.3, then p˜ = 1 by [WZZ15a, Lemma 4.12], we obtain
hτ1(2),h1hτ1(3)···hτ1(m)
p˜ = 1. It is proved.
h ,h ···h
τ1(2) τ1(3) τ1(m)
Step 2. Assumed that Assertion(j−i) hold, we prove Assertion(j−i−1) holds,
8
j −i ≤ m−3:
We know 0 6= hτm−j+i−2(i)···hτm−j+i−2(j) ∈ L(V), then there exists
τm−j+i−1 ∈ Thτm−j+i−2(i),...,hτm−j+i−2(j),pm−j+i−1 ∈ F∗ suchthathτm−j+i−2(i)···hτm−j+i−2(j) =
pm−j+i−1hτm−j+i−1(i)···hτm−j+i−1(j) with (1◦) 0 6= hτm−j+i−1(i)···hτm−j+i−1(j−1) ∈ L(V),
p˜ 6= 1 for some k ∈ {i,...,j −1} or with (2◦)
hτm−j+i−1(k),hτm−j+i−1(j)
0 6= hτm−j+i−1(i+1)···hτm−j+i−1(j) ∈ L(V), p˜hτm−j+i−1(l),hτm−j+i−1(i) 6= 1 for some l ∈ {i +
1,...,j} by induction hypotheses of (iv). We obtain h ···h
1 m
= p1···pm−j+i−2pm−j+i−1hτm−j+i−1(1)···hτm−j+i−1(m). For convenience, h′k := hτm−j+i−1(k)
for all k ∈ {1,...,m}. We know τm−j+i−1(t) = t for all t ∈ A by the definition of
ij
τm−j+i−1.
(1◦) If there exists r ∈ Aij such that p˜h′α,hr = 1 for all α ∈ {i,...,j−1}, then p˜h′j,hr 6= 1
by Assertion(j−i), we know p˜h′i···h′j,hr 6= 1, it is a contradiction to Assertion(j−i).
Thus exists α ∈ {i,...,j −1} such that p˜h′α,hr 6= 1 for all r ∈ Aij, we obtain
h′ ···h′ h′ ···h′ ∈ L(V) by induction hypotheses.
1 i−1 i j−1
Set {r ,··· ,r ,l ,··· ,l } = {τm−j+i−1(j +1),...,τm−j+i−1(m−1)} with r < ··· <
1 β 1 η 1
rβ,l1 < ··· < lη such that p˜h′j,hrk 6= 1 and p˜h′j,hlξ = 1 for all k ∈ {1,...,β}, all
ξ ∈ {1,...,η}. Then h′ ···h′ h′ ···h′ ∈ L(V) for all r ∈ {1,...,η} and h′h′ ···h′ ∈
1 j−1 l1 lr j r1 rt
L(V)forallt ∈ {1,...,β}byinductionhypotheses. Ifp˜h′,h′ 6= 1,thenp˜h′···h′ h′ ···h′ ,h′ 6=
j m 1 j−1 l1 lη m
1 and p˜h′jh′r1···h′rβ,h′m 6= 1 by Assertion(j−i). Then h′1···h′j−1h′l1 ···h′lηh′jh′r1···h′rβh′m ∈
L(V) by Lemma 3.2(iii), h ···h = p ···p h′ ···h′
1 m 1 m−j+i−1 1 m
= p ···p p′h′ ···h′ h′ ···h′ h′h′ ···h′ h′ for some p′ ∈ F∗ by Remark 3.3. It
1 m−j+i−1 1 1 j−1 l1 lη j r1 rβ m 1
is a contradiction. We obtain p˜h′j,h′m = 1. We obtain h′1···h′j−1h′l1 ···h′lηh′m ∈ L(V) by
induction hypotheses. If β ≥ 1, then p˜h′jh′r1···h′rβ−1,h′rβ 6= 1 and p˜h′1···h′j−1h′l1···h′lηh′m,h′rβ 6= 1
by Assertion(j−i). then h′ ···h′ h′ ···h′ h′ h′h′ ···h′ h′ ∈ L(V) by Lemma
1 j−1 l1 lη m j r1 rβ−1 rβ
3.2(iii). h h ···h = p ···p p′h′ ···h′ h′ ···h′ h′ h′h′ ···h′ h′ for some
1 2 m 1 m−j+i−1 2 1 j−1 l1 lη m j r1 rβ−1 rβ
p′ ∈ F∗ by Remark 3.3. It is contradiction. Then β = 0. η = m − j − 1, i.e.
2
p˜h′,h′ = 1 for all r ∈ {j + 1,...,m}. So p˜h′···h′ ,h′ = 1 for all r ∈ {j + 1,...,m} by
j r i j−1 r
Assertion(j−i). Assumed that there exists θ ∈ {1,...,i− 1} such that p˜h′,h′ 6= 1. If
θ j
p˜h′j,h′1···h′θ−1h′θ+1···h′i−1h′i···h′j−1h′j+1···h′m = 1, then p˜h′j,h′1···h′j−1h′j+1···h′m 6= 1, h′1···h′j−1h′j+1···h′m ∈
L(V) is clear by induction hypotheses. then h′ ···h′ h′ ···h′ h′ ∈ L(V) by [WZZ15a,
1 j−1 j+1 m j
Lemma 4.12]. h h ···h = p ···p p′h′ ···h′ h′ ···h′ h′ for some p′ ∈ F∗ by
1 2 m 1 m−j+i−1 3 1 j−1 j+1 m j 3
Remark 3.3. It is contradiction. If p˜h′,h′···h′ h′ ···h′ h′···h′ h′ ···h′ 6= 1,
j 1 θ−1 θ+1 i−1 i j−1 j+1 m
h′ ···h′ h′ ···h′ h′ ···h′ h′ ···h′ ∈ L(V)isclearbyinductionhypotheses. Then
1 θ−1 θ+1 i−1 i j−1 j+1 m
h′h′ ···h′ h′ ···h′ h′ ···h′ h′ ···h′ h′ ∈ L(V) by Lemma 3.2(iii). h h ···h =
θ 1 θ−1 θ+1 i−1 i j−1 j+1 m j 1 2 m
p ···p p′h′h′ ···h′ h′ ···h′ h′ ···h′ h′ ···h′ h′ for some p′ ∈ F∗ by Re-
1 m−j+i−1 4 θ 1 θ−1 θ+1 i−1 i j−1 j+1 m j 4
mark 3.3. It is contradiction. Then p˜h′,h′ = 1 for all θ ∈ {1,...,i − 1}, we obtain
θ j
p˜h′θ,h′i···h′j−1 = 1forallθ ∈ {1,...,i−1}byAssertion(j−i). Weknowh′1···h′j−1h′j+1···h′m
∈ L(V) by induction hypotheses. If p˜h′j,h′i···h′j−1 6= 1, then h′1···h′j−1h′j+1···h′mh′j ∈ L(V)
9
by [WZZ15a, Lemma 4.12], h h ···h = p ···p p′h′ ···h′ h′ ···h′ h′ for
1 2 m 1 m−j+i−1 5 1 j−1 j+1 m j
some p′5 ∈ F∗ by Remark 3.3. It is a contradiction. Then p˜h′j,h′i···h′j−1 = 1.
(2◦) If there exists r ∈ Aij such that p˜h′α,hr = 1 for all α ∈ {i+1,...,j}, then p˜h′i,hr 6= 1
by Assertion(j−i), we know p˜h′i···h′j,hr 6= 1, it is a contradiction to Assertion(j−i).
Thus exists α ∈ {i+1,...,j} such that p˜h′α,hr 6= 1 for all r ∈ Aij, we obtain
h′ ···h′h′ ···h′ ∈ L(V) by induction hypotheses.
i+1 j j+1 m
Set {r ,··· ,r ,l ,··· ,l } = {τm−j+i−1(2),...,τm−j+i−1(i − 1)} with r < ··· <
1 β 1 η 1
rβ,l1 < ··· < lη such that p˜h′i,hrk 6= 1 and p˜h′i,hlξ = 1 for all k ∈ {1,...,β}, all
ξ ∈ {1,...,η}. Then h′ ···h′ h′ ···h′ ∈ L(V) for all r ∈ {1,...,η} and h′ ···h′ h′ ∈
l1 lr i+1 m r1 rt i
L(V) for all t ∈ {1,...,β}by induction hypotheses. If p˜h′,h′ 6= 1, thenp˜h′ ···h′ h′ ···h′ ,h′ 6=
i 1 l1 lη i+1 m 1
1 and p˜h′r1···h′rβh′i,h′1 6= 1 by Assertion(j−i). Then h′1h′r1···h′rβh′ih′l1···h′lηh′i+1···h′m ∈
L(V) by Lemma 3.2(iii), h ···h = p ···p h′ ···h′
1 m 1 m−j+i−1 1 m
= p ···p p′h′h′ ···h′ h′h′ ···h′ h′ ···h′ for some p′ ∈ F∗ by Remark 3.3. It
1 m−j+i−1 1 1 r1 rβ i l1 lη i+1 m 1
is a contradiction. We obtain p˜h′i,h′1 = 1. We obtain h′1h′l1 ···h′lηh′i+1···h′m ∈ L(V) by
induction hypotheses. If β ≥ 1, then p˜h′ ···h′ h′,h′ 6= 1 and p˜h′h′ ···h′ h′ ···h′ ,h′ 6= 1 by
r2 rβ i r1 1 l1 lη i+1 m r1
Assertion(j−i). then h′ h′ ···h′ h′h′h′ ···h′ h′ ···h′ ∈ L(V) by Lemma 3.2(iii).
r1 r2 rβ i 1 l1 lη i+1 m
h h ···h = p ···p p′h′ h′ ···h′ h′h′h′ ···h′ h′ ···h′ for some p′ ∈ F∗ by
1 2 m 1 m−j+i−1 2 r1 r2 rβ i 1 l1 lη i+1 m 2
Remark3.3. It iscontradiction. Thenβ = 0. η = i−2, i.e. p˜h′,h′ = 1forallr ∈ {1,...,i−
i r
1}. So p˜h′ ···h′,h′ = 1 for all r ∈ {1,...,i−1} by Assertion(j−i). Assumed that there
i+1 j r
exists θ ∈ {j + 1,...,m} such that p˜h′,h′ 6= 1. If p˜h′,h′···h′ h′ ···h′h′ ···h′ h′ ···h′ = 1,
θ i i 1 i−1 i+1 j j+1 θ−1 θ+1 m
then p˜h′i,h′1···h′i−1h′i+1···h′m 6= 1, h′1···h′i−1h′i+1···h′m ∈ L(V) is clear by induction hypothe-
ses. then h′h′ ···h′ h′ ···h′ ∈ L(V) by [WZZ15a, Lemma 4.12]. h h ···h =
i 1 i−1 i+1 m 1 2 m
p ···p p′h′h′ ···h′ h′ ···h′ for some p′ ∈ F∗ by Remark 3.3. It is contradic-
1 m−j+i−1 3 i 1 i−1 i+1 m 3
tion. Ifp˜h′i,h′1···h′i−1h′i+1···h′jh′j+1···h′θ−1h′θ+1···h′m 6= 1,h′1···h′i−1h′i+1···h′jh′j+1···h′θ−1h′θ+1···h′m ∈
L(V)isclearbyinductionhypotheses. Thenh′h′ ···h′ h′ ···h′h′ ···h′ h′ ···h′ h′
i 1 i−1 i+1 j j+1 θ−1 θ+1 m θ
∈ L(V) by Lemma 3.2(iii). h h ···h
1 2 m
= p ···p p′h′h′ ···h′ h′ ···h′h′ ···h′ h′ ···h′ h′ for some p′ ∈ F∗ by
1 m−j+i−1 4 i 1 i−1 i+1 j j+1 θ−1 θ+1 m θ 4
Remark 3.3. It is contradiction. Then p˜h′,h′ = 1 for all θ ∈ {j + 1,...,m}, we obtain
θ i
p˜h′θ,h′i+1···h′j = 1forallθ ∈ {j+1,...,m}byAssertion(j−i). Weknowh′1···h′i−1h′i+1···h′m
∈ L(V) by induction hypotheses. If p˜h′i,h′i+1···h′j 6= 1, then h′ih′1···h′i−1h′i+1···h′m ∈ L(V)
by[WZZ15a, Lemma4.12],h h ···h = p ···p p′h′h′ ···h′ h′ ···h′ forsome
1 2 m 1 m−j+i−1 5 i 1 i−1 i+1 m
p′5 ∈ F∗ by Remark 3.3. It is a contradiction. Then p˜h′i,h′i+1···h′j = 1.
Step3. WeobtainAssertion(0)byStep1and2. i.e. p˜ = 1andp˜ 6=
hτm−2(i),hr hτm−2(i),hr
1 for all r ∈ A . It is a contradiction.
ii
(iv)Weknowd(h h ···h )isconnectedbyLemma3.4(iii). Ifh ···h orh ···h ∈
1 2 m 1 m−1 2 m
L(V), it is clear. Now we assumed that h ···h ∈/ L(V) and h ···h ∈/ L(V).
1 m−1 2 m
Step 1. Let the largest integer j − i+ 1 such that h h ···h ∈ L(V) and
τ1(i) τ1(i+1) τ1(j)
h h ···h = p h h ···h for some p ∈ F∗, all τ ∈ T , it is clear
1 2 m−1 1 τ1(1) τ1(2) τ1(m−1) 1 1 1,2,...,m−1
10