Main Lemmas

Before we embark on the proof of Proposition V.1.2, we shall establish some preliminary lemmas.

Now consider an arbitrary factor groupHofG_∞equipped with the presentation

H=hA ∪T |R₁∪R₂∪Ri (V.6)

whereRis a finite subset inG_∞. ThenH∼=G_∞/hhRiiG∞. Note that ifR=R₃,H=G, ifR=/0,H=G_∞which are two major examples we concern. We have following lemmas forH.

Lemma V.3.1. Let H be a factor group of G_∞equipped with presentation (V.6) and w be a word in(T ∪T⁻¹)^∗such that|w|=n, then

w=Hw¯

p

∏

i=1

b^u_iⁱ

where p6n²,b_i∈A^±1,u_i∈F,Tr(θ(ui))⊂B_n. In addition, the cost of converting LHS to RHS is bounded by n². Proof. Since ¯w=t₁^mi. . .t_k^mk for somem₁, . . . ,m_k∈Zsuch that∑^k_i=1|mi|6n, to move each letter inwto the desired place, it will cost at most n commutators of the form[ti,tj],16i< j6k. By the discussion in Section II.3 and Section II.4, in total, we need at mostn²such commutators. That is,

w=w¯

p

∏

i=1

[ti1,ti2]^εiu′i,whereu^′_i∈F,p6n²,16i₁<i₂6k,εi∈ {±1}.

Moreover, since the length ofwis bounded byn, Tr(θ(u^′_i))6n.

By applying relations in{ai j= [ai,aj]|16i<j6n} ptimes we immediately have

w=Hw¯

p

∏

i=1

b^u_iⁱ,Tr(θ(ui))6n.

The cost of relations is bounded byp6n².

In particular, forw∈Fsuch thatπ(w) =1, it costs at mostn²relations inHto convert it to a product of conjugates of elements inA.

Lemma V.3.2. Let H be a factor group of G_∞ equipped with presentation (V.6) then there exists a constant K only depends onR₁∪R₂such that

Area([a,b^u])6Kⁿ,∀a,b∈A,kθ(u)k<n.

Proof. LetF ={θ(supp(λ))|λ∈Λ}thenF is a finite colletction of finite sets. By the choice ofΛ,F satisfies the assumptions of Lemma III.3.1.

By Lemma III.3.1, each x∈B_r+ε(r) can be taken from B_r by F for r>R. Recall that R is defined to be max{D,D²/2C,D²/(4kC−4)} andε(r) =C−D²/2r, whereC,Dare purely determined byΛhenceR₁∪R₂ as we stated in Lemma III.3.5.

According to our choice ofR, we note thatε(r)>ε(2kD²/(4kC−4)) =_2k¹ forr>R. LetK1be the constant which is large enough such that f(n)6K₁ⁿforn6R, andK₂be the constant

K₂:=max

λ∈Λ{

∑

u∈F¯

|λ(u)|}+2.

Since eachλ has finite support,K₂is well-defined. Now letK:=max{K1,K₂^2k}.

Suppose forn>R, Area([a,b^u])6Kⁿ,∀a,b∈A,kθ(u)k<n. We then prove our lemma by induction. Let us first consider the caser=n+_2k¹. Fix somev∈F¯satisfyingkθ(v)k<r. Sinceε(n)>_2k¹,B_n+1

2k can be taken fromB_nby F. Then there isλ ∈Λwithθ(supp(λv))ˆ ⊂B_nby the definition of “taken from”.

Therefore we have two cases depending onλ∈C(A)orC(A^∗). Firstly assume thatλ∈Λ∩C(A). Then by applying the commutator formula[x,yz] = [x,y]^x−1zx[x,z], we obtain

[a,b^v] =G[a,

∏

u∈F¯

(b^λ(u)ˆ)^uv] =

∏

u∈F¯

[a,b^λ(u)uvˆ ]^h(u),

where theh(u)’s are certain elements inHwhich need not concern us. Note that in the first equality above, we apply relations in (V.3) twice to replacebby∏_u∈F¯b^λ(y)u. Since supp(λ)⊂B¯_Dwe havekθ(u)k<D<_2kⁿ.Additionally, we havekθ(v)k<n+_2k¹ andkθ(uv)k<n. It meets all assumptions of Proposition III.3.4 (c). Note thatHis a factor group ofH_nwhich we defined in Section III.3. Then[a,b^λ(u)uvˆ ]is conjugate inHto[a,b^λ(u)uvˆ ], the area of which is bounded by|λ(u)|K¯ ⁿ. It follows that

Area([a,b^v])62+

∑

u∈F¯

Area([a,b^λ(u)uvˆ ])62+

∑

u∈F¯

|λ(u)|Kˆ ⁿ6K2Kⁿ.

Repeating this process 2ktimes, we obtain that

Area([a,b^v])6K₂^2kKⁿ6Kⁿ⁺¹, forv∈F¯,kθ(v)k<n+1.

Ifλ∈Λ∩C(A^∗), the only different is that

[a,b^v] = [a^v−1,b]^v−1 = [

∏

u∈F¯

(a^λ(u)ˆ)^uv−1,b]^v−1.

Similarly we obtain that

Area([a,b^v])6K₂^2kKⁿ6Kⁿ⁺¹.

Furthermore, Lemma V.3.2 allows us to estimate the cost to commute two conjugates of elements inA. Since the normal closure ofA inHis abelian, this lemma provides a tool to estimate the cost of converting words inhhAiiH, in particular, inG. Also Lemma V.3.2 reveals how much metabelianness costs in a finitely presented metabelian group.

We will discuss this topic further in Section VI.1.

Lemma V.3.3. Let H be a factor group of G_∞ equipped with presentation (V.6) and K be the same constant in Lemma V.3.2. Then in H we have

Area(a^ua^−¯u)6(2K)ⁿ,∀a∈A,u∈F,Tr(u)⊂B_n

Proof. We prove it by an induction onn. Suppose fori6n, the result holds. Then for the casen+1, we writeu=u^′t_s^±1 then Tr(u)⊂B_n+1,Tr(u^′)⊂B_n.

a^u=a^u′ts±1= (a^u¯′)^ts±1ν1,

where Area(ν1)6(2K)ⁿby our inductive assumption. Write ¯u^′=t₁^m1. . .t_k^mk, we claim that

¯ u^′t_s^±1=u

m

∏

j=1

c^α_j^j wherecj∈ {[ts,tl]^±1|16s<l6k},αj∈F¯,m=

k i=s+1

∑

|mi|6n.

We need to be really careful here. Let us first consider the case that the exponent oftsis 1. We assumes<k, otherwise

it is trivial. Note that ifm_k>0

t_k^mkt_s=t_k^mk−1t_st_k[ts,t_k]⁻¹=t_k^mk−2t_st_k²[ts,t_k]^−tk[ts,t_k]⁻¹

=t_k^mk−3t_st_k³[ts,t_k]^−tk2[ts,tk]^−tk[ts,t_k]⁻¹ ...

=tst_k^mk[ts,t_k]^−tkmk−1. . .[ts,t_k]^−tk[ts,tk]⁻¹.

Ifm_k<0, we have

t_k^mkt_s=t_k^mk+1t_st_k⁻¹[ts,tk]^ts=t_k^mk+2t_st_k⁻²[ts,t_k]^tst−1k [ts,tk]^ts

=t_k^mk+3tst_k⁻³[ts,t_k]^tst−2k [ts,tk]^tstk−1[ts,tk]^ts−1 ...

=t_st_k^mk[ts,t_k]^tstkmk+1. . .[ts,t_k]^tstk−1[ts,t_k]^ts.

Repeating this process, we then prove the claim for the case that the exponent oft_sis 1.

On the other hand, if the exponent oft_sis−1, then similarly, consider ifm_k>0

t_k^mkt_s⁻¹=t_k^mk−1t_s⁻¹t_k[ts,tk]^tk=t_k^mk−2t_s⁻¹t_k²[ts,tk]^tk2[ts,tk]^ts

=t_k^mk−3t_s⁻¹t_k³[ts,t_k]^tk3[ts,t_k]^tk2[ts,t_k]^tk ...

=t_s⁻¹t_k^mk[ts,t_k]^{tmkk −1}. . .[ts,tk]^tk2[ts,tk]^tk,

and ifm_k<0

t_k^mkt_s⁻¹=t_k^mk+1t_s⁻¹t_k⁻¹[ts,tk]⁻¹=t_k^mk+2t_s⁻¹t_k⁻²[ts,tk]^−tk−1[ts,tk]⁻¹

=t_k^mk+3t_s⁻¹t_k⁻³[ts,tk]^−tk−2[ts,tk]^−t−1k [ts,t_k]⁻¹ ...

=t_s⁻¹t_k^mk[ts,t_k]^−tkmk+1. . .[ts,t_k]^−tk−1[ts,tk]⁻¹.

Again by repeating this process, the claim is proved. Thus by induction onk, we can movetsto the desired place.

Now we have

a^u=a^u¯′ts±1ν₁= (

m

∏

j=1

c^α_j^j)⁻¹a^u¯(

m

∏

j=1

c^α_j^j)ν₁. Apply relations from{ai j= [ai,a_j]|16i<j6n}2mtimes, we have that

a^u= (

m

∏

j=1

d^α_j^j)⁻¹a^u¯(

m

∏

j=1

d^α_j^j)ν2ν1

wheredj∈A^±1and Area(ν₂)62mby our disccusion.

Next we need to commutea^u¯andd^α_j^j for j=1, . . . ,mto the left and estimate the cost. Note that[a^u¯,d^α_j^j] is conjugate to[a,d^α_j^j(u)¯−1]. From the computation above,αjis either a tail of ¯uor a tail of ¯umultiplied byt_s^±1. Therefore (u)¯ ⁻¹,αj,αj(u)¯ ⁻¹satisfy the assumption of Proposition III.3.4 (b). Thus[a,d^α_j^j(u)¯−1]is conjugate to[a,d^α_j^ju−1]. Since kθ(αju⁻¹)k6n+1, the area of[a,d^α_j^ju−1], by Lemma V.3.2, is bounded byKⁿ⁺¹.

Applying[a,d^α_j^ju−1]toa^uandd^α_j^j for j=1, . . . ,m, we can commute alld^α_j^j to the left such that it cancels with d^−α_j ^j. Then we finally have

a^u=a^u¯ν3ν2ν1, where

Area(ν3)6mKⁿ⁺¹ In total, the cost of convertinga^utoa^u¯is bounded by

Area(ν3ν2ν1)6Area(ν3) +Area(ν2) +Area(ν1)6(2K)ⁿ+2m+mKⁿ⁺¹6(2K)ⁿ⁺¹.

Note that we use the fact thatm6nand we can chooseK≫1.

Lemma V.3.3 provides a method for us to “organize” the exponent of a conjugate. In particular, combining all three lemmas introduced this section (Lemma V.3.1, Lemma V.3.2 and Lemma V.3.3), we are able to convert any word in hhAiiHto its ordered form. This forms the foundation of converting the word problem in groupGto the membership problem of a submodule in the freeT-module generated byA.