On the commutator lengths of certain classes of finitely presented groups

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FINITELY PRESENTED GROUPS

H. DOOSTIE AND P. P. CAMPBELL

Received 23 June 2004; Revised 4 July 2005; Accepted 28 March 2006

For a finite groupG= X(X=G), the least positive integer MLX(G) is called the

max-imum length of G with respect to the generating set X if every element ofG may be represented as a product of at most MLX(G) elements ofX. The maximum length ofG,

denoted by ML(G), is defined to be the minimum of{MLX(G)|G= X, X=G, X=

G− {1G}}. The well-known commutator length of a groupG, denoted byc(G), satisfies the inequalityc(G)≤ML(G_{), where}_G_{is the derived subgroup of}_{G. In this paper we} study the properties of ML(G) and by using this inequality we give upper bounds for the commutator lengths of certain classes of finite groups. In some cases these upper bounds involve the interesting sequences of Fibonacci and Lucas numbers.

1. Introduction

For an abstract groupGthe commutator lengthc(G) is defined to bec(G)=Sup{λ(g)|

g∈G_}_{, where}_λ(g_{) is the minimal number of commutators of which}_g _{is the product.} This notion has been studied by many authors during the years and the results of the combinatorial methods estimate or calculate the commutator lengths of abstract groups which are mainly infinite (one may see [2,6–10], e.g.). For a finite groupG= Xwe examine the eﬀect of the generating set on the evaluation of this number by considering the following definitions, this gives us a method to calculate upper bounds forc(G). Let G= X(X=G) be a finite group. Then the following hold.

Definition 1.1. MLX(G), the maximum length ofGrelative to the generating set X, is defined to be max{λ(g)|g∈G}, whereλ(g) is the minimum number of the elements of Xof whichgis the product.

Definition 1.2. The maximum length of a groupG, denoted by ML(G), is defined to be the minimum of all numbers MLX(G), for all generating setsX(X=G) ofG.

Our notations are fairly standard, we use [x] for the integer part of the real number x, [a,b]=a−1_b−1_ab_{is defined to be the commutator of the elements}_a_and_b_{of a group,}

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 74981, Pages1–9

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the usual notationN×ϕHis used for the semidirect product of the groupNbyH, where ϕ:H→Aut(N) is a homomorphism such thathϕ=ϕhandϕh:N→Nis an element of

Aut(N), and the Reidemeister-Schreier algorithm in the form given in [1] will be used to find presentations of subgroups. In the following sections we study certain classes of finite groups for their maximal lengths and find upper bounds for the commutator lengths. The groups studied here are the dihedral groupsD2n= a,b|a2=bn=(ab)2=1,n≥3, the quaternion groupsQ2n= a,b|a2

n−1

=1, b2₌_a2n−2

,b−1_ab₌_a−1_,_n_≥_{3, the semidirect} products D2n×ϕZ2m, and Q2n×ϕZ₂_m, (n,m≥3) (where if Z₂_m= c, then ϕ:Z₂_m→

Aut(D2n) is such thatcϕ=ϕc; andϕc:D2n→D2nis defined byaϕc=aandbϕc=b−1, a similarϕexists forQ2n×ϕZ2m), the direct productsD2n×Z2m andQ2n×Z2m, and the following classes of groups:

G1=

a,b|a2₌_bn₌_abab−1_ab2_ab−1_ab−2_ab₌₁_, _n_≥_3, G2=

a,b|a2₌_bn_,_ab22_ab−12

=b2n_, _n₌_0, G3=

a,b|a2₌_bn₌_1,_abab−2_ab3_ab−2_ab−1_ab₌₁_, _n_≥_3, G4=

a,b|a2₌_b5₌_{1, [a,}_b]k_a,b3₌₁_, _k_≥_1.

(1.1)

The groupsGi (i=1, 2, 3, 4) are soluble and have been studied for their structures and orders in [3–5]. These groups are generalizations of the well-known Coxeter groups.

We will use the Fibonacci and Lucas numbers:

f0= f1=1, fn= fn−1+fn−2, n≥2,

g0=2, g1=1, gn=gn−1+gn−2, n≥2, (1.2)

which are related to each other via the relationgn=fn−2+ fn. InSection 2we study the

notion of the maximum length for the mentioned direct and semidirect products, and by using the inequalityc(G)≤ML(G) which holds for every nonabelian finite groupG, we give upper bounds for the commutator lengths of the groupsGi(i=1, 2, 3, 4), in Sections

3and4.

2. The groupsD2n,Q2n,D2n×Z2m,Q2n×Z2m,D2n×ϕZ2m,Q2n×ϕZ2m

Letm,n≥3 be integers. By the definitions of the direct and semidirect products, we get the following presentations:

D2n×Z2m=

a,b,c|a2₌_bn₌_(ab)2₌_c2m₌_[a,c]₌_[b,_c]₌₁_, Q2n×Z₂_m=a,b,c|a2n−1=1,b2=a2n−2,b−1aba=c2m=[a,c]=[b,c]=1,

D2n×ϕZ2m=

a,b,c|a2₌_bn₌_(ab)2₌_c2m₌_1,_c−1_aca₌_1, _c−1_bcb₌₁_, Q2n×ϕZ₂_m=a,b,c|a2n−1=1, b2=a2n−2,b−1aba=c2m=1,c−1aca=1,c−1bcb=1,

(2.1)

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As an immediate result of the definitions we have the following.

Lemma 2.1. For every finite groupsG1= X|RandG2= Y|S,

ML{X,Y}G1×G2

≤MLXG1

+ MLYG2

. (2.2)

Proof. Obviously,G1×G2= X,Y|R,S, [X,Y], where [X,Y]= {[x,y]|x∈X, y∈Y}. There exists an element g ∈G1×G2 such that ML{X,Y}(G1×G2)=λ(g), and g = (x1x2···xm)(y1y2···yn), wherexi∈Xandyi∈Y(for [xi,yj]=1 holds for everyiand

j). Letg1=x1x2···xmandg2=y1y2···yn. The definition ofλ(g) then yields

ML{X,Y}G1×G2

=λ(g)=m+n≤λg1

+λg2

≤MLXG1

+ MLYG2

. (2.3)

Proposition 2.2. m,n≥3 are integers, then

(i) for a nonabelian metacyclic groupG,c(G)≤ |G| −1; (ii) for a metabelian groupG, where|G_{| =}_pα1

1 pα22···pkαk (herep1,p2,...are diﬀerent primes)c(G)≤(pα1

1 +···+pkαk)−k;

(iii) ML(D2n)≤[n/2] + 1 and ML(Q2n)≤2n−3+ 3;

(iv) ML(D2n×Z2m)≤[n/2] + 2mand ML(Q2n×Z2m)≤2n−3+ 2m+ 2; (v) ML(D2n×ϕZ2m)≤[n/2] + 2m;

(vi) ML(Q2n×ϕZ₂_m)≤2n−3+ 2m.

Proof. For a nonabelian metacyclic groupG, the derived subgroupGis a cyclic group, so ML(G₎_{= |}_G_{| −}_{1 and (i) follows at once.}

For a metabelian groupG, where|G_{| =}_pα1

1 pα22···p

αk

k , we may use the direct

decom-position ofGto get (ii).

We give proof of the first part of (iii), the second part is similar. The elements ofD2n are of the formaibj, wherei=0, 1 and j=0, 1,...,n−1. Consider two cases fornand for everyg∈D2ncomputeλ(g), the minimum number of the elements ofX= a,bof

whichg is the product. We see that the relations (ab)2₌_a2₌_bn₌_{1 of}_D

2n yield the relationsbn−k=b−k₌_abk_{a, for every integer}_k.

Case 1. nis even. It is easy to see that

λbn−k₌

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

k+ 2 if 1≤k≤n

2−1, n−k ifn

2 ≤k≤n−1.

(2.4)

Also,

λabk₌

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

k+ 1 if 1≤k≤n

2, n−k+ 1 ifn

2+ 1≤k≤n−1.

(2.5)

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Case 2. nis odd. In this case similar classification of the elements gives us

λbn−k₌

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

k+ 2 if 1≤k≤n−3

2 , n−k ifn−1

2 ≤k≤n−1.

(2.6)

Also,

λabk₌

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

k+ 1 if 1≤k≤n−1

2 , n−k+ 1 ifn+ 1

2 ≤k≤n−1,

(2.7)

and in this case MLX(D2n)=max{λ(g)|g∈D2n} =(n+ 1)/2. Consequently, for every n≥3, ML(D2n)≤MLX(D2n)=[n/2] + 1. A similar proof for Q2n yields ML_X(Q₂n)=

2n−3+ 3, whereX= a,bis the generating set ofQ2n, and ML(Q2n)≤MLX(Q2n)=2n−3+ 3.

The inequalities of (iv) are the results ofLemma 2.1and the calculations of part (iii), where we know that ML{a,b}(D2n)=[n/2] + 1, ML{a,b}(Q2n)=2n−3+ 3, and ML_{_c_}(Z₂_m)=

2m−1.

To prove (v) we see that up to the relationsca=ac,cb=b−1_{c, and}_ba₌_ab−1_{of the} groupD2n×ϕZ2mthe 4mnelements of this group may be considered as the union of the

following sets:

S1=

ci_|₁_≤_i_≤_2m₋₁_,

S2=

ai_bj_|₀_≤_i_≤_{1, 0}_≤_j_≤_n₋₁_,

S3=

aci_|₁_≤_i_≤_2m₋₁_,

S4=

bi_cj_|₁_≤_i_≤_n₋_{1, 1}_≤_j_≤_2m₋₁_,

S5=

abi_cj_|₁_≤_i_≤_n₋_{1, 1}_≤_j_≤_2m₋₁_.

(2.8)

Since the relationsbn−k=abk_a,_bn−k₌_cbk_{c, and}_ck₌_ack_a_{hold in the group, for every}

integerk, then the minimum lengthλ(g) is definitely acceptable for everyg∈D2n×ϕZ2m, in the similar way as in (iii) and we get

maxλ(g)|g∈S1

=2m−1,

maxλ(g)|g∈S2

=1 +

_n

2

,

maxλ(g)|g∈S3

=2m,

maxλ(g)|g∈S4

=n

2

+ 2m−1,

maxλ(g)|g∈S5=

_n

2

+ 2m.

(2.9)

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The proof of (vi) is similar to the above proof and one may get the result by considering them2n+1_{elements of the group}_Q

2n×ϕZ₂_mas the union of the following sets:

S1=

ai_bj_|₀_≤_i_≤₂n−1₋_{1, 0}_≤_j_≤₁_, S2=

ai_cj_|₀_≤_i_≤₂n−1₋_{1, 1}_≤_j_≤_2m₋₁_, S3=

bci_|₁_≤_i_≤_2m₋₁_,

S4=

ai_bcj_|₁_≤_i_≤₂n−1₋_{1, 1}_≤_j_≤_2m₋₁_.

(2.10)

And this completes the proof.

3. The groupsG1andG2

The groupsG1andG2are finite and nonmetabelian soluble groups for many values ofn (see [4,5]). The following propositions are our main results on the commutator lengths of these groups.

Proposition 3.1. For everyn≥3, where g.c.d (n, 3)=1,

cG1

≤

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

gn ifn≡ ±3(mod 12),

4fn/2−1 ifn≡0(mod 12), 2gn/2 ifn≡6(mod 12).

(3.1)

Proposition 3.2. For everyn≥4,

cG2

≤

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

2n−1 ifn≡ ±1(mod 3), n+ 7

2 ifn≡3(mod 6), n+ 8

2 ifn≡0(mod 6).

(3.2)

Proof ofProposition 3.1. Letn≡ ±3(mod 12). Using the results of [5] gives us here the following presentation forG1:

G 1=

x,y|(xy)2₌_(yx)2_,_x2_,_y₌_y2_,x₌_1,_xy₌_y−fn−2xfn−3_,x1+fn−2=yfn−1. (3.3)

First we show that the relationsygn₌xgn₌_{1 hold in}G

1. The relationxy=y−fn−2xfn−3_{gives us}

xyfn−1x−1=y−fn−2fn−3x(−1+fn−3)fn−1. _(3.4)

For fn−3is odd, then [y−fn−2,x−1+fn−3_]=_{1. This relation together with the last relation of} G

1yields

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And substituting foryfn−1_yields

x(1+fn−2)2₋_f_n₋₁₍₋₁₊_f_n₋₃₎

=1. (3.6)

By an inductive method we may show thatgn=(1 +fn−2)2−fn−1(−1 + fn−3) and then xgn₌_{1. A similar proof exists for}ygn₌_{1 and since}G

1is of order 2gn(see [5]),gnis the order ofy.

Now consider the relations [x2_,_y]₌_[y2_,x]₌_{1 and conclude that the words}_xy−1+gn

(ory−1+gnx) are the words of largest length; that is,c(G

1)≤ML(G1)≤ML{x,y}(G1)=gn. Letn≡6(mod 12). In this caseG1may be presented as

x,y|(xy)2₌_(yx)2_,_x2_,_y₌_y2_,x₌_1,_x1+fn−3y−1+fn−2_,x−1+fn−2=yfn−1

. (3.7)

This group is of order 2(−2 +gn) (see [5]). First we show that the relationsxgn/2=ygn/2= 1 hold in G1. Combining the last two relations ofG1 yields xfn−4−f2=yfn−3+f1. Again combining this relation withx1+fn−3=y−1+fn−2_{gives us}xfn−5+f3=yfn−4−f2_{. We repeat this} method and get the relations

xfn−2−i−(−1)ifi₌yfn−1−i+(−1)ifi−1_, i=2, 3,.... _(3.8)

Fori=(n−4)/2 andi=(n−6)/2 we get xt₌_yt _and_xt₌_y2t_{, respectively, where}_t₌ fn/2+f(n−4)/2. Consequently, yt=1 holds inG1. Now it is easy to see thatt=gn/2. The word yx−1+gn/2y−1+gn/2x_{is one of the words of maximum length. So}c(G

1)≤ML(G1)≤ ML{x,y}(G1)=2gn/2as required in this case.

The remained case isn≡0(mod 12). In this caseG1has the previous case’s presenta-tion. We first show that the relations

xfn/2−1=y3fn/2−1_, y5fn/2−1=₁ _(3.9)

hold inG1. As well as in the last case we get the relations

xfn−2−i−(−1)ifi₌yfn−1−i+(−1)ifi−1_, i=2, 3,.... _(3.10)

If we leti= −2 +n/2 then, after a simplification we get

xfn/2−1=y3fn/2−1. _(3.11)

Raising both sides to the power 3 yieldsx3fn/2−1=y9fn/2−1_{. Also for the value}i= −_{3 +}n/2 we getx3fn/2−1=y4fn/2−1_{. Consequently,} y5fn/2−1=_{1, and these relations together with the} relations [x2_,_y]₌_[y2_,x]₌_{1 show that the word}

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is of the largest length inG1. However, by considering the relationxfn/2−1=y3fn/2−1, this word will be reduced to

y−1+2fn/2−1x−1+2fn/2−1yx, _(3.13)

thenc(G1)≤ML(G1)≤ML{x,y}(G1)=4fn/2−1, as required. This completes the proof.

Proof ofProposition 3.2. For everyn≡ ±1(mod 3),G2is a metabelian group and as a re-sult of the computations of [4],G2is a cyclic group of order 2n. Soc(G2)≤2n−1 comes from the results ofSection 2.

Letn≡3(mod 6). Using the Todd-Coxeter coset enumeration algorithm gives us the presentation

G 2=

x,y|(xy)2₌_yn_{, (yx)}2₌_xn_,_x3_y3₌₁_. _(3.14)

It is easy to show the validity of the relationsxn=yn_and_x2n₌_y2n₌_{1 in}_G

2. Letn=6k+ 3. We claim that the wordsw1=x2y3k+1xy2andw2=y2x3k+1yx2in G1 are of the largest length 3k+ 5. Indeed, the relationsx3_y3₌_1,_xyx₌_yn−1_{, and}_yxy₌_xn−1_{show that any} word with maximal length could not contain the subwordsx2_y2_and_y2_x2_{. The remained} words which have to be examined are indeedw3=y3k+1xy2xandw4=x3k+1yx2y. These words are of length 3k+ 4, for we have

w3=y3k(yxy)yx=y3kx6k+2yx=x3k+2yx. (3.15)

Similarly,w4is of length 3k+ 4. To complete the proof we now show thatw1andw2are of length 3k+ 5. By using the relations we get

w1=x2y3k(yxy)y=x2y3kx6k+2y=x2

y3k_x3k_x3k+2_y₌_x3k+4_y, _(3.16)

and in a similar way,w2will be reduced toy3k+4x. Soc(G2)≤ML(G2)≤ML{x,y}(G2)= 3k+ 5.

Letn≡0(mod 6). We use the Todd-Coxeter coset enumeration algorithm to find a presentation forG2. In two diﬀerent cases,n≡0(mod 12) andn≡6(mod 12), we get the following presentations:

G 2=

x,y|x2_,_y₌_x,_y3₌_x4₌_yn₌_R

1=R2=1, (xy)n/2=x2

, G

2=

x,y|x2_,_y₌_x,_y3₌_x4₌_yn₌_R

1=R2=1, (xy)n/2=yn/2

,

(3.17)

respectively, whereR1=y2+n/2x3y−1xy−1x3andR2=yxy−1xyx3y−1x. The largest power ofxin every word ofG2is equal to 3, however, the largest power ofyisn/2 + 1, forR1=1 yields

y2+n/2₌_xyx3_yx. _(3.18)

This relation also gives us the relation (xy)2₌_y2+n/2_{x. The relation}_R

2=1 yieldsyxyn−1xy

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the following words:

w1=x3y1+n/2, w2=y1+n/2x3, w3=x3y1+n/2xy, w4=xyx3y1+n/2. (3.19)

Obviously the wordsw1andw2are of length 4 +n/2, howeverw4is of length 5, for we see that

w4=

xyx3_y_yn/2₌_y2+n/2_x−1_yn/2₌_y2+n/2_x3_yn/2₌_yn+2_x3₌_y2_x3_. _(3.20)

Finallyw3is also of length 4 +n/2, for

w3=yn/2x3yxy=yn/2yxyn−1xyy=ynyxy−1+n/2xy2=yxy−1+n/2xy2. (3.21)

Consequently,c(G2)≤ML(G2)≤ML{x,y}(G2)=4 +n/2. This completes the proof.

4. The groupsG3andG4

These groups are examples of metacyclic groups and we have the following result.

Proposition 4.1. For everyn≥3, the groupsG3, and for everyk≥1, the groupsG4 are finite. Moreover,c(G3)≤(2n−3−(−1)n)/3 andc(G4)≤k(k3+ 2k2+ 4k+ 3).

Proof. The subgroupH= xi=[a−1,b−1]b−i: 0≤i≤n−1ofG3 is indeed the derived subgroup of G3 (one may easily check that |G3:H| = |G3:G3| =2nand H⊆G3). A presentation forHmay be given as

H=x0,...,xn−1|xixi2+1=1,xn−1x20=1, 0≤i≤n−2

. (4.1)

Obviously this group is finite and cyclic of order (2n−(−1)n)/3, andc(G3)≤ML{x,y}(G3)=

|G

3| −1=(2n−3−(−1)n)/3 is a result ofSection 2. In a similar way we consider the sub-group

K=a−1_,b_,_b−1_,ab−i_{: 0}_≤_i_≤₃ _(4.2)

ofG4.Kis the derived subgroup ofG4and may be presented by K=a1,...,a5|a1a2a3a4a5=1,aki =ai+1ai+2, 1≤i≤5

, (4.3)

(where indices are reduced modulo 5). An almost easy simplification of the relations shows thatKis a cyclic group of orderk4_{+ 2k}3_{+ 4k}2_{+ 3k}_{+ 1, and}_c(G

4)≤ML{x,y}(G4)=

|G

4| −1=k(k3+ 2k2+ 4k+ 3) is a result ofSection 2. This completes the proof.

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H. Doostie: Mathematics Department, Teacher Training University, 49 Mofateh Avenue, Tehran 15614, Iran

E-mail address:[email protected]

P. P. Campbell: Institute of Mathematics, St. Andrews University, St. Andrews, KY16 9SS, Scotland, UK

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