A Magnus theorem for some amalgamated products

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by  Carsten Feldkamp

A Magnus theorem for some amalgamated products
Carsten Feldkamp
7 January 18, 2017
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n Abstract
a
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A group G possesses the Magnus property if for every two elements u,v ∈G with the same normal
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1 closure,uisconjugateinGtov orv−1. WeprovetheMagnuspropertyforsomeamalgamatedproducts

] including the fundamental group of a closed non-orientable surface of genus 3. This answers a question
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G of O. Bogopolski and K. Sviridov, who obtained the analogous result for genus g >3.
.
h
t
a 1 Introduction
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[
A group G possesses the Magnus property if for every two elements u,v ∈ G with the same normal
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v closure,u isconjugateinG tov orv−1. The MagnuspropertywasnamedafterW.Magnuswhoproved
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4 the so-called Freiheitssatz (see Theorem 4.1) and the Magnus property for free groups [Mag30]. Since
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then, many mathematicians proved or disproved the Freiheitssatz and the Magnus property for certain
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0
classes of groups (see e.g. [Bog05], [BS08], [Edj89], [How81], [How04]).
.
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Let π (S+), respectively π (S−), be the fundamental group of the compact orientable, respectively
0 1 g 1 g
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non-orientable, surface of genus g. The Magnus property of π (S+) for all g was proved independently
1 1 g
v: by O. Bogopolski [Bog05] (by using algebraic methods) and by J. Howie [How04] (by using topological
i
X methods). As observed in [Bog05], there is a third, model theoretic method: two groups G , G are
1 2
r
calledelementarilyequivalentiftheirelementarytheoriescoincide: Elem(G )=Elem(G ),see[CG05].
a 1 2
It is easy to show that elementarily equivalentgroups either both possess the Magnus property,or both
do not possess it. Since the groups π (S+) for g > 2 and π (S−) for g > 4 are elementarily equivalent
1 g 1 g
to the free group on two generators,all these groups possess the Magnus property.
In [BS08], O. Bogopolskiand K. Sviridov proved the following theorem:
Theorem 1.1. [BS08, Main Theorem] Let G = ha,b,y ,...,y | [a,b]uvi, where e > 2, u,v are
1 e
non-trivial reduced words in letters y ,...,y , and u,v have no common letters. Let r,s ∈ G be two
1 e
elements with the same normal closures. Then r is conjugate to s or s−1.
As a corollary of that theorem ([BS08, Corollary 1.3]), they showed the Magnus property of π (S−)
1 g
for g > 4. Since the Magnus property trivially holds for genus 1 and 2, the authors asked, whether it
1
alsoholds for the fundamentalgroupof the non-orientablesurfaceof genus3. With ourMain Theorem,
thatprovestheMagnuspropertyforaslightlylargersubclassofone-relatorgroupsthaninTheorem 1.1,
we answer this question positively. The difficulty with genus g = 3 is essential since it is well known
that the group π (S−) = hx,y,z|x2y2z2i is not even existentially equivalent to a free group F on n
1 3 n
generators. However,in large parts, our proof follows the proof of the Main Theorem in [BS08].
Main Theorem 1.2. Let G = ha,b,y ,...,y | [a,b]ui, where n ∈ N and u is a non-trivial reduced
1 n
word in the letters y ,...,y . Then G possesses the Magnus property.
1 n
Using the isomorphism ϕ between π (S−) = hx,y,z|x2y2z2i and H = ha,b,c | [a,b]c2i defined by
1 3
e
ϕ(x)=ca−1,ϕ(y)=b−1c−1 and ϕ(z)=cbcac−1, we get the following corollary.
Corollary 1.3. The group π (S−) possesses the Magnus property.
1 3
IntheproofofTheorem1.1,theauthorsof[BS08]usedanautomorphismofGwithcertainconvenient
properties. This automorphism is in general absent for the group given in our Main Theorem. For
example, it is absent for the group π (S−). So we introduce an additional tool which we call α- and
1 3
ω-limits (see Section 3). Together with the results of [BS08,Bog05,How04] Corollary 1.3 implies the
following.
Corollary 1.4. The fundamental groups of all compact surfaces possess the Magnus property.
We will startthe proofof our maintheoremby recapitulating the notationof[BS08, Section 3]with
some small alterations. In Section 3 we introduce α- and ω-limits, α-ω-length ||·|| and suitable
α,ω
elements. In Section 4 we give the proof for the case, where the α-ω-length of suitable elements is
positive (the considerationof this case is close to that in [BS08]). In Section 5 we complete the proof in
the remaining case.
2 Reduction to a new group and left/right bases
WedenotethenormalclosureofanelementginagroupGbyhhgii andtheexponentsumofanelement
G
g ∈Ginaletter xbyg . Note thatthis sumiswell-definedifallrelationsofGconsideredaswordsofa
x
free grouphaveexponentsum 0 inx. Our maintheoremcanbe deduced fromthe following proposition
in the same way as in the Main Theorem of [BS08] from [BS08, Proposition 2.1]. Therefore we leave
this argumentation out.
Proposition 2.1. Let H = hx,b,y ,...,y | [xk,b]ui, where k,n ≥ 1 and u is a non-trivial reduced
1 n
word in the letters y ,...,y . Further, let r,s ∈ H\{1} with r = 0. Then hhrii = hhsii implies that
1 n x H H
r is conjugate to s or s−1.
We briefly summarise the concept of left and right bases from [BS08]. For a given group G and an
element g ∈G we denote by g the element x−igxi for i∈Z.
i
2
Let H be as in Proposition 2.1. We consider the homomorphism ϕ:H →Z which sends x to 1 and
b,y ,y ,...,y to 0. For each i ∈ Z let Y = {b ,y ,y ,...,y }, where y := x−iy xi. Using the
1 2 n i i 1,i 2,i n,i j,i j
rewriting process of Reidemeister-Schreier, we obtain that the kernel N of ϕ has the presentation
N =h[Yi | biui =bi+k (i∈Z)i.
i∈Z
The group N is the free product of the free groups G = hY | i (i ∈ Z) with amalgamation, where
i i
G and G are amalgamated over a cyclic group Z that is generated by b u in G and by b in
i i+k i+k i i i i+k
G . This gives N =N ∗···∗N for all t∈Z, where
i+k t t+k−1
N =... ∗ G ∗ G ∗ G ∗ ...
l l−k l l+k
Zl−k Zl Zl+k Zl+2k
for l ∈Z.
Proposition 2.2. For every i∈Z, the group N is free with basis
B(i)={b ,b ,...,b } ∪{y |16m6n,j ∈Z}.
i i+1 i+k−1 m,j
Proof. Since N =N ∗···∗N , it suffices to show that each N has the basis
i i+k−1 l
B :={b }∪{y |16m6n,t∈Z}.
l l m,l+tk
We have N = N , where N = G ∗ ... ∗ G ∗ ... ∗ G . Using Tietze trans-
l S l,p l,p l−pk l l+pk
p∈N Zl−(p−1)k Zl Zl+k Zl+pk
formations, one can show that N is free with basis B := {b}∪{y | 1 6 m 6 n,−p6 t 6p}.
l,p l,p l m,l+tk
Since B ⊂B , the group N is free with basis B = B .
l,p l,p+1 l l S l,p
p∈N
Notation. Let G = hG ,G ...,G i, G = hG | l > ii and G = hG | l 6 ii for all i,j ∈ Z
i,j i i+1 j i,∞ l −∞,i l
with i6j. Further, we use the
b-left basis B+(i) := {b ,b ,...,b } ∪ {y |16m6n,j >i} of G and the
i i+1 i+k−1 m,j i,∞
b-right basis B−(i) := {b ,b ,...,b } ∪ {y |16m6n,j 6i} of G .
i−k+1 i−k+2 i m,j −∞,i
Remark 2.3. Let r ∈ N be written as a word in the letters b ,y (1 6 m 6 n,j ∈ Z). For an
j m,j
arbitrary i ∈ Z we describe how to rewrite r in the basis B+(i). First, replace each letter b of r by
j
b u , if j > i+k−1, and by b u−1, if j < i, and reduce. If the resulting word contains b that
j−k j−k j+k j j
do not belong to B+(i), we repeat this procedure. After finitely many steps, we will obtain the desired
form of r. Analogously, we can rewrite r in the basis B−(i).
α ω
3 - and -limits
We keepusingthe notationintroducedinSection2. An arbitraryelementr∈N\{1}canbe written in
manywaysasareducedwordinlettersb ,y . Forexample, b u u−1b−1 =b b−1 =b u−1u b−1 .
i m,i 0 0 1 1 k k+1 2k k k+1 2k+1
We give an algorithm which finds a word r∗ representing r such that the smallest index of letters used
3
inr∗ ismaximumpossible. We callsuchindexthe α-limit of r anddenoteitbyα . Inotherwords,α
r r
is the largest index such that r is an element of G . The following algorithm rewrites an arbitrary
αr,∞
wordr ∈N\{1}into the presentationr∗ ofr writteninbasisB+(α )(see Lemma3.2). Fora wordr in
r
the alphabet {b |i∈Z}∪{y |16m6n,i∈Z}, let min(r) denote the minimal index of letters of r.
i m,i
Algorithm 3.1. Let r ∈ N\{1}. Suppose that r is given as a finite word in letters b ,b ,... and
i i+1
y ,y ,... (1 6 m 6 n). In particular, r ∈ G . Let r[0] be the reduced word representing r in
m,i m,i+1 i,∞
basis B+(i). Increasing i if necessary, we may assume that i=min(r[0]).
(1) Let r[1] be the word obtained from r[0] by replacement of each occurrence of the letter b by
i
b u−1 followed by free reduction. Then r[1] presents r in the following basis of G :
i+k i i,∞
{b ,b ,...,b }∪{y |16m6n,i6l}. (⋆)
i+1 i+2 i+k m,l
(1a) If r[1] does not contain a letter y (16m6n), we reset r[0]:=r[1], i:=min(r[1]), and go
m,i
back to (1). Clearly, the new i is larger than the old one.
(1b) If r[1] contains a letter y (16m6n), the algorithm ends with r∗ =r[0] and α =i.
m,i r
Lemma 3.2. The output r∗ of Algorithm 3.1 coincides with the word representing r in Basis B+(α ),
r
where α is the α-limit of r.
r
Proof. We shall prove that the algorithm ends and that the integer assigned to α by the algorithm is
r
really the α-limit of r.
The algorithm ends since the length of r[0] decreases with each iteration of (1) which does not
terminate the algorithm. This can be verified in the following way: If we arrive in (1a), then the word
r[1] does not contain a letter y (16m6n). Recall that r[1] was obtained from r[0] by replacing all
m,i
b with b u−1 and reducing the resulting word. The fact that there is no letter y (16m6n) left
i i+k i m,i
in r[1] means that each b in r[0] occurred in a subword of the form b u . Therefore r[1] is shorter than
i i i
r[0].
Finally, we show that r is not an element of G , where i as in (1b). Observe that
i+1,∞
G =G ∗hy |16m6ni.
i,∞ i+1,∞ m,i
By (1b), r, written in basis (⋆), uses a letter y (16m6n). Therefore r∈/ G and α =i.
m,i i+1,∞ r
Corollary 3.3. Let r be an element from N\{1} given as a word in some b-left basis. Starting
Algorithm 3.1 with r, we either get the same presentation or a presentation of shorter length.
Proof. In the proofof Lemma 3.2, we alreadyshowed that the length of r[0] in Algorithm 3.1 decreases
with each iteration of (1) which does not terminate the algorithm. Moreover, the last iteration of
Algorithm 3.1 does not change the current presentation r[0].
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Analogously, we can find a presentation r of a word r ∈N\{1} such that the largest used index of
∗
letters in r is minimum possible. We call this index ω-limit of r and denote it by ω . In other words,
∗ r
ω is the smallest index such that r is an element of G . The algorithm to find r can be received
r −∞,ωr ∗
from Algorithm 3.1 by “mirroring” this algorithm using the b-right basis of G and replacements of
−∞,i
b by b u . We define the α-ω-length of r by ||r|| := ω −α +1. Note that the α-ω-length of
i i−k i−k α,ω r r
a non-trivial element can be non-positive.
Examples.
(i) Letr =b u y b u−1andk =3. Todetermineα ,wewriter =b u y b u−1 =b y b u−1
−2 −2 1,0 4 1 r −2 −2 1,0 4 1 1 1,0 4 1
and get α = 0. For ω we have r = b u y b u−1 = b u y b u u−1 = b u y b
r r −2 −2 1,0 4 1 −2 −2 1,0 1 1 1 −2 −2 1,0 1
=b u y b u . Thus, ω =0 and ||r|| =1. Note that r ∈/ G =G .
−2 −2 1,0 −2 −2 r α,ω αr,ωr 0
(ii) Let r = b b−1 and k = 4. Clearly, α = 5. To determine ω , we write r = b b−1 = b u−1b−1
5 6 r r 5 6 5 2 2
=b u u−1b−1 =b u u−1u−1b−1. Thus, ω =2 and ||r|| =−2.
1 1 2 2 1 1 2 −2 −2 r α,ω
Lemma 3.4. Let r ∈ N. Then α = α +j and ω = ω +j for all i,j ∈ Z. In particular,
ri+j ri ri+j ri
||r || =||r || for all i,j ∈Z.
i α,ω j α,ω
Proof. Indeed, r can be obtained from r by increasing the (second) indices of all letters by j.
i+j i
Notation. Let r ∈N. For the following two lemmata, let r(i) be the presentationof r written in basis
B(i).
Lemma 3.5. Let r ∈N\{1}. The following statements are equivalent:
(1) For some i∈Z, the word r(i) begins with a positive power of a b-letter.
(2) For all i∈Z, the word r(i) begins with a positive power of a b-letter.
Proof. The word r(i+1) can be obtained from r(i), by replacement of each occurrence of b in r(i) by
i
b u−1 followed by free reduction. The new letter b does not lie in {b ,...,b }. That prevents
i+k i i+k i i+k−1
cancellationbetweenb-lettersinr(i+1). Thereforer(i)startswithapositiveexponentofab-letterifand
only if r(i+1) starts with a positive exponent of a b-letter. This proves the equivalence (1)⇔(2).
Corollary 3.6. For every r ∈N\{1} there exists a conjugate r of r such that r(i) is cyclically reduced
e e
for each i∈Z.
Proof. Using conjugation,we mayassumethatr(0) iscyclicallyreduced. Ifr(0)containsonly y-letters,
we are done with the element r represented by r(0). Suppose that r(0) contains a b-letter. Let r be
e e
the element represented by a cyclic permutation of r(0) which either starts with a positive power of a
b-letter,orendswithanegativepowerofab-letter(butnotboth). ByLemma3.5,thisr hasthedesired
e
property.
Definition 3.7. Let r ∈N\{1}. Any element r as in Lemma 3.6 is called suitable conjugate for r.
e
5
Remark 3.8. (dual structure of N)Denoteb′ :=b u , y′ :=y−1 (16m6n)andG′ :=G .
i −i −i m,i m,−i i −i
We call the elements b′,y′ (1 6 m 6 n) dual to b ,y (1 6 m 6 n) and the subgroup G′ dual to
i m,i i m,i i
G . Expressing b-letters and y-letters via their dual, we obtain b =b′ u′ , y =y′−1 (1 6m 6n),
i i −i −i m,i m,−i
where u′ (j ∈Z) is the wordobtained fromu by replacing eachletter y with y′−1 . That justifies
−j j m,j m,−j
the terminology.
Observe that the old relations b u = b (i ∈ Z) can be rewritten in dual letters as b′u′ = b′
i i i+k i i i+k
(i ∈ Z). Thus, the relations preserve their form. Moreover, we have G′ = hb′,y′ | 1 6 m 6 ni
i i m,i
that repeats the form G = hb ,y | 1 6 m 6 ni. Other dual objects can be defined analogously: For
i i m,i
example,G′ :=hG′,G′ ...i=G andG′ :=h...,G′ ,G′i=G . Weusethefollowing
i,∞ i i+1 −∞,−i −∞,i i−1 i −i,∞
bases of N (i∈Z):
B(i)′ ={b′,b′ ,...,b′ } ∪{y′ |16m6n,j ∈Z}.
i i+1 i+k−1 m,j
along with the
b′-left basis B+(i)′ := {b′,b′ ,...,b′ } ∪ {y′ |16m6n,j >i} of G′ and the
i i+1 i+k−1 m,j i,∞
b′-right basis B−(i)′ := {b′ ,b′ ,...,b′} ∪ {y′ |16m6n,j 6i} of G′ .
i−k+1 i−k+2 i m,j −∞,i
For r ∈ N\{1}, let α′ be the largest index i such that r is an element of G′ and let ω′ be the
r i,∞ r
smallest index i such that r is an element of G′−∞,i. We denote ||r||α′,ω′ :=ωr′ −α′r+1.
Lemma 3.9. For any non-trivial r ∈N the following statements are valid.
(1) We have α′r =−ωr, ωr′ =−αr, and ||r||α′,ω′ =||r||α,ω.
(2) Suppose that r written in a basis B(i) is cyclically reduced and begins with a positive power of a
b-letter. Then r written in the basis B(−i−k+1)′ is cyclically reduced and begins with a positive
power of a b′-letter.
Proof. Both statements can be verifiedstraightforwardby using the relations b =b′ u′ , y =y′−1
i −i −i m,i m,−i
(16m6n).
r α ω
4 Proof of Proposition 2.1 for with positive - -length
e
4.1 Properties of r with positive α-ω-length
e
We use the following version of Magnus’ Freiheitssatz:
Theorem 4.1 (Magnus’ Freiheitssatz (cf. [Mag30])). Let F be a free group on a basis X, and let
g be a cyclically reduced word in F with respect to X, containing a letter x ∈ X. Then the subgroup
generated by X\{x} is naturally embedded into the group F/hhgii .
F
By abuse of notation, we write A/hhaii instead of A/hhaii , where A is a group and a∈A.
A
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Corollary 4.2. Let r ∈N be a suitable element. Then we get the embeddings G ֒→N/hhrii and
αre+1,∞
e e
G ֒→N/hhrii.
−∞,ωre−1
e
Proof. By the definition of α-limits, r written in basis B+(αre) contains at least one letter in Yαre, and
e
by Definition 3.7, the suitable element r is cyclically reduced in this basis. We extend B+(αre) to a free
e
basis of N by adding the letters {ym,i | 1 6 m 6 n,i < αre}. The presentation of r will not change by
e
that. Now, G ֒→ N/hhrii follows immediately from Theorem 4.1. The other embedding follows
αre+1,∞
e
analogously.
Lemma 4.3. Suppose that r ∈N satisfies ||r|| >1. Then r, written in the basis B+(α ) of G ,
α,ω r αr,∞
contains at least one letter y with l>ω .
m,l r
Proof. By assumption, α 6ω . To the contrary, suppose that
r r
r ∈hb ,b ,...,b ,y ,y ,...,y |16m6ni.
αr αr+1 αr+k−1 m,αr m,αr+1 m,ωr−1
Then, using relations b =b u , we obtain
i i−k i−k
r ∈hb ,...,b ,y ,y ,...,y |16m6ni.
αr−k αr−1 m,αr−k m,αr−k+1 m,ωr−1
Hence r ∈G . A contradiction.
−∞,ωr−1
4.2 The structure of some quotients of N
This section and the next one are very similar to [BS08, Section 4 and 5], but due to some important
changes we cannot skip them. Our aim in this subsection is to present N/hhr ,r ,...,r ii as an
i i+1 j
e e e
amalgamated product. We denote w =b u .
i i i
Lemma 4.4. Let r ∈ N be a suitable element with ||r|| > 1, and let i,j be two integers with i 6 j.
α,ω
e e
We denote s=αrej and t=ωrej −1. Then we have:
(1) N/hhri,ri+1,...,rjii ∼= G−∞,t/hhri,ri+1,...,rj−1ii ∗ Gs,∞/hhrjii.
e e e e e e Gs,t e
w =b ,
t−k+1 t+1
...
w =b
t t+k
(2) G naturally embeds into N/hhr ,r ,...,r ii.
s+1,∞ i i+1 j
e e e
Beforewe give a formalproof,we consider anillustratedexample. This will help to visualisea lot of
technical details in the formulation of lemma.
Example. Let k = 4. We consider the element r = r = b y y b u and the integers i = −1 and
0 4 2,1 1,3 0 0
e e
j =2.
7
Algorithm 3.1 applied to r gives αre0 = 1. Its “mirrored” version gives ωre0 = 3. In particular,
e
||r||α,ω =3. Furthermore, we have s=αre2 =3 and t=ωre2 −1=4. Then Lemma 4.4 (1) says that
e
N/hhr−1,r0,r1,r2ii ∼= G−∞,4/hhr−1,r0,r1ii ∗ G3,∞/hhr2ii.
e e e e e e e G3,4 e
w = b ,
1 5
w = b
2 6
w = b ,
3 7
w = b
4 8
InFigure1, the wordr =r is picturedby the (partiallydashed)line crossingthe blocksG ,...,G
0 0 4
e e
sincer usesletterswithindicesfromthesegment[0,4]. Wecanrepresentr bythewordb y y b that
4 2,1 1,3 4
e e
uses letters with indices from the segment [1,4], and we can represent r by the word b u y y b u
0 0 2,1 1,3 0 0
e
that uses letters with indices fromthe segment[0,3]. To visualise the fact that αre0 =1 and ωre0 =3, we
draw a continuous line crossing the blocks G ,G ,G .
1 2 3
Figure 1: Illustration to Lemma 4.4 for k = 4, r = b4y2,1y1,3b0u0, i = −1, and j = 2
e
Proof. First, we prove that (1) implies (2). By Corollary 4.2, we have G ֒→ G /hhr ii, and if
s+1,∞ s,∞ j
e
(1) holds, then we have G /hhr ii ֒→N/hhr ,r ,...,r ii. The composition of these two embeddings
s,∞ j i i+1 j
e e e e
gives (2). Now we prove (1) for fixed i by induction on j.
Base of induction. For j =i we shall show
N/hhrjii ∼= G−∞,t ∗ Gs,∞/hhrjii. (4.1)
e Gs,t e
w =b ,
t−k+1 t+1
...
w =b
t t+k
It suffices to show the following claim.
8
Claim. Let P be the subgroup of N generated by G ∪{b ,...,b }. Then
s,t t+1 t+k
(a) P embeds into G and G /hhr ii.
−∞,t s,∞ j
e
(b) The abstractamalgamatedproduct in the right side of (4.1) is canonically isomorphic to N/hhr ii.
j
e
Proof of the claim. (a) Clearly, P embeds into G . So, we show that P embeds into G /hhr ii.
−∞,t s,∞ j
e
NotethatP =hY ∪···∪Y ∪{b ,...,b }i=hb ,b ,...,b ,y ,y ,...,y |16m6ni.
s t t+1 t+k s s+1 s+k−1 m,s m,s+1 m,t
Thus, the group P is generated by the set
{b ,b ,...,b ,y ,y ,...,y |16m6n}. (4.2)
s s+1 s+k−1 m,s m,s+1 m,t
This set is a part of the free basis B+(s) of G . By Definition 3.7, r written in B+(s) is cyclically
s,∞ j
e
reduced. Further, by statement(1) of Lemma 4.3, the element r written in B+(s) contains at leastone
j
e
letter y with 1 6 m 6 n and ℓ > t+1. In particular, this letter does not lie in the set (4.2). By
m,ℓ
Magnus’ Freiheitssatz (Theorem 4.1), P embeds into G /hhr ii.
s,∞ j
e
Now, the amalgamated product in (4.1) is well defined. It is easy to check that the groups written
in (4.1) are isomorphic by finding a common presentation. This completes the base of induction.
Inductive step i,j →i,j+1. We need to show the formula, where s=αrej and t=ωrej −1:
N/hhri,ri+1...,rj+1ii ∼= G−∞,t+1/hhri,ri+1,...,rjii ∗ Gs+1,∞/hhrj+1ii. (4.3)
e e e e e e Gs+1,t+1 e
w =b ,
t−k+2 t+2
...
w =b
t+1 t+k+1
Let P be the subgroup of N generated by G and the set {b ,b ,...,b }. First, we
s+1,t+1 t+2 t+3 t+k+1
prove that P canonically embeds into both factors. As above P embeds into G /hhr ii. So we
s+1,∞ j+1
e
show that P embeds into G /hhr ,r ,...,r ii using the following commutative diagram:
−∞,t+1 i i+1 j
e e e
(cid:31)(cid:127) (cid:31)(cid:127) by (2)
P(cid:15)(cid:15)(cid:127)_ ❯ ❯ ❯ ❯ ❯ ❯ ❯ ❯//G❯s+❯1,❯∞❯ϕ❯ ❯ ❯ ❯ ❯ ❯ ❯ ❯// N❯/❯h**hrei,rei(cid:31)+?OO 1,...,rejii
G // G /hhr ,r ,...,r ii
−∞,t+1 −∞,t+1 i i+1 j
e e e
LetϕbethecompositionofthecanonicalembeddingofthesubgroupP intoG andthecanonical
−∞,t+1
homomorphismfromG tothefactorgroupG /hhr ,r ,...,r ii. Itremainstoprovethatϕ
−∞,t+1 −∞,t+1 i i+1 j
e e e
isanembedding. ConsideringP asasubgroupofG andusingstatement(2)fori,j (recallthat(1)
s+1,∞
implies (2)), we have an embedding of P into N/hhr ,r ,...,r ii. Since the diagram is commutative,
i i+1 j
e e e
ϕ:P →G /hhr ,r ,...,r ii is an embedding. Again, it is easy to check that the groups in (4.3)
−∞,t+1 i i+1 j
e e e
are isomorphic.
Finally, we need a “mirrored” version of Lemma 4.4:
9
Lemma 4.5. Let r ∈ N be a suitable element with ||r|| > 1, and let i,j be two integers with i 6 j.
α,ω
e e
We denote s=αrei +1 and t=ωrei. Then we have:
(1) N/hhri,ri+1,...,rjii ∼= G−∞,t/hhriii ∗ Gs,∞/hhri+1,ri+1,...,rjii.
e e e e Gs,t e e e
w =b ,
s−k s
...
w =b
s−1 s+k−1
(2) G naturally embeds into N/hhr ,r ,...,r ii.
−∞,t−1 i i+1 j
e e e
Proof. Due to the dual structure noticed in Remark 3.8, the statements of Corollary 4.2, Lemma 4.3
and Lemma 4.4 also hold for the dual objects. By rewriting the dual version of Lemma 4.4 with the
help of Lemma 3.9, we get the desired statement.
4.3 Conclusion of the proof for r with positive α-ω-length
e
By assumption of Proposition 2.1, we have two elements r,s ∈ H with the same normal closure and
r = 0. Thus, r,s are elements of N. By Lemma 3.6, we can choose conjugates r,s of r,s such that
x
e e
r and s are suitable elements. Normal closures are invariant under conjugation. So, without loss off
e e
generality, we can replace r,s by r,s. In this section we assume ||r|| > 1. Since the normal closures
α,ω
e e e
of r and s in H are equal, the normal closures of R := {r | i ∈ Z} and S := {s | i ∈ Z} in N are
i i
e e e e
equal. In particular, s is trivial in N/hhRii. Thus, there are indices i,j ∈ Z such that s is trivial in
0 0
e e
N/hhr ,r ,...,r ii. We choose a pair i,j with this property and j−i minimal. By Lemma 4.4 (1), we
i i+1 j
e e e
have
N/hhri,ri+1,...,rjii ∼= G−∞,ωrej−1/hhri,ri+1,...,rj−1ii A∗ Gαrej,∞/hhrjii (4.4)
e e e e e e e
for some subgroup A.
Lemma 4.6. Suppose that ||r|| >1. Then
α,ω
e
(1) ωes0 >ωrej >ωrei,
(2) αes0 6αrei 6αrej,
(3) ||s|| >||r|| .
α,ω α,ω
e e
Proof. (1) We prove ωes0 > ωrej. Suppose the contrary. Then s0 ∈ G−∞,ωrej−1. Since s0 is trivial
e e
in N/hhr ,r ,...,r ii, it is also trivial in the left factor of the amalgamated product (4.4). There-
i i+1 j
e e e
fore, s is trivial in N/hhr ,r ,...,r ii. This contradicts the minimality of j −i. The inequality
0 i i+1 j−1
e e e e
ωrej >ωrei follows by Lemma 3.4 since j >i. Inequalities (2) canbe provedanalogouslywith the help of
Lemma 4.5 (1). Inequality (3) follows straightforwardfrom (1) and (2).
Since ||s|| > ||r|| > 1, we have ||r|| > ||s|| by symmetry and therefore ||s|| = ||r|| .
α,ω α,ω α,ω α,ω α,ω α,ω
e e e e e e
By Lemma 3.4, this is equivalentto ||s || =||r || . This andthe firsttwostatements of Lemma 4.6
0 α,ω i α,ω
e e
implythatαes0 =αrei =αrej,andωrej =ωrei =ωes0,inparticular,j =i. Therefore,s0 istrivialinN/hhriii,
e e
andthe index iis determinedby αes0 =αrei. Onecanproveinthe samewaythatri is trivialinN/hhs0ii.
e e
10