ON THE COMMUTATOR LENGTHS OF CERTAIN CLASSES OF FINITELY PRESENTED GROUPS

For a finite group G= 〈X〉 (X = G), the least positive integer MLX(G) is called the maximum length of G with respect to the generating set X if every element of G may be represented as a product of at most MLX(G) elements of X . The maximum length of G, 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 group G, denoted by c(G), satisfies the inequality c(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.


Introduction
For an abstract group G the commutator length c(G) is defined to be c(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][7][8][9][10], e.g.).For a finite group G = X we 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 for c(G).Let G = X (X = G) be a finite group.Then the following hold.Definition 1.1.ML X (G), the maximum length of G relative to the generating set X, is defined to be max{λ(g) | g ∈ G}, where λ(g) is the minimum number of the elements of X of which g is the product.Definition 1.2.The maximum length of a group G, denoted by ML(G), is defined to be the minimum of all numbers ML X (G), for all generating sets X (X = G) of G.
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, the usual notation N × ϕ H is used for the semidirect product of the group N by H, where ϕ : H → Aut(N) is a homomorphism such that hϕ = ϕ h and ϕ h : N → N is 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 groups and the following classes of groups: The groups G i (i = 1,2,3,4) are soluble and have been studied for their structures and orders in [3][4][5].These groups are generalizations of the well-known Coxeter groups.
We will use the Fibonacci and Lucas numbers: which are related to each other via the relation g n = f n−2 + f n .In Section 2 we study the notion of the maximum length for the mentioned direct and semidirect products, and by using the inequality c(G) ≤ ML(G ) which holds for every nonabelian finite group G, we give upper bounds for the commutator lengths of the groups G i (i = 1,2,3,4), in Sections 3 and 4.

The groups
Let m,n ≥ 3 be integers.By the definitions of the direct and semidirect products, we get the following presentations: where ϕ is the homomorphism defined in the last section.
H. Doostie and P. P. Campbell 3 As an immediate result of the definitions we have the following.
Lemma 2.1.For every finite groups There exists an element , where x i ∈ X and y i ∈ Y (for [x i , y j ] = 1 holds for every i and j).Let Proof.For a nonabelian metacyclic group G, the derived subgroup G is a cyclic group, so ML(G ) = |G | − 1 and (i) follows at once.
For a metabelian group G, where |G | = p α1 1 p α2 2 ••• p αk k , we may use the direct decomposition of G to get (ii).
We give proof of the first part of (iii), the second part is similar.The elements of D 2n are of the form a i b j , where i = 0,1 and j = 0,1,...,n − 1.Consider two cases for n and for every g ∈ D 2n compute λ(g), the minimum number of the elements of X = a,b of which g is the product.We see that the relations (ab (2.4) Also, Case 2. n is odd.In this case similar classification of the elements gives us (2.6) Also, and in this case ML X (D 2n ) = max{λ(g) | g ∈ D 2n } = (n + 1)/2.Consequently, for every The inequalities of (iv) are the results of Lemma 2.1 and the calculations of part (iii), where we know that ML {a,b} To prove (v) we see that up to the relations ca = ac, cb = b −1 c, and ba = ab −1 of the group D 2n × ϕ Z 2m the 4mn elements of this group may be considered as the union of the following sets: Since the relations b n−k = ab k a, b n−k = cb k c, and c k = ac k a hold in the group, for every integer k, then the minimum length λ(g) is definitely acceptable for every g ∈ D 2n × ϕ Z 2m , in the similar way as in (iii) and we get (2.9) H. Doostie and P. P. Campbell 5 The proof of (vi) is similar to the above proof and one may get the result by considering the m2 n+1 elements of the group Q 2 n × ϕ Z 2m as the union of the following sets: (2.10)And this completes the proof.

The groups G 1 and G 2
The groups G 1 and G 2 are finite and nonmetabelian soluble groups for many values of n (see [4,5]).The following propositions are our main results on the commutator lengths of these groups.
Proposition 3.1.For every n ≥ 3, where g.c.d (n,3 Proof of Proposition 3.1.Let n ≡ ±3(mod 12).Using the results of [5] gives us here the following presentation for G 1 : And substituting for y fn−1 yields By an inductive method we may show that ) and then x gn = 1.A similar proof exists for y gn = 1 and since G 1 is of order 2g n (see [5]), g n is the order of y.Now consider the relations [x 2 , y] = [y 2 ,x] = 1 and conclude that the words xy −1+gn (or y −1+gn x) are the words of largest length; that is, c(G 1 ) ≤ ML(G 1 ) ≤ ML {x,y} (G 1 ) = g n .
Let n ≡ 6(mod12).In this case G 1 may be presented as This group is of order 2(−2 + g n ) (see [5]).First we show that the relations x gn/2 = y gn/2 = 1 hold in G 1 .Combining the last two relations of G 1 yields x fn−4− f2 = y fn−3+ f1 .Again combining this relation with x 1+ fn−3 = y −1+ fn−2 gives us x fn−5+ f3 = y fn−4− f2 .We repeat this method and get the relations For i = (n − 4)/2 and i = (n − 6)/2 we get x t = y t and x t = y 2t , respectively, where t = f n/2 + f (n−4)/2 .Consequently, y t = 1 holds in G 1 .Now it is easy to see that t = g n/2 .The word yx −1+gn/2 y −1+gn/2 x is one of the words of maximum length.So c(G 1 ) ≤ ML(G 1 ) ≤ ML {x,y} (G 1 ) = 2g n/2 as required in this case.The remained case is n ≡ 0(mod12).In this case G 1 has the previous case's presentation.We first show that the relations hold in G 1 .As well as in the last case we get the relations If we let i = −2 + n/2 then, after a simplification we get Raising both sides to the power 3 yields x 3 fn/2−1 = y 9 fn/2−1 .Also for the value i = −3 + n/2 we get x 3 fn/2−1 = y 4 fn/2−1 .Consequently, y 5 fn/2−1 = 1, and these relations together with the relations [x 2 , y] = [y 2 ,x] = 1 show that the word H. Doostie and P. P. Campbell 7 is of the largest length in G 1 .However, by considering the relation x fn/2−1 = y 3 fn/2−1 , this word will be reduced to as required.This completes the proof.
Proof of Proposition 3.2.For every n ≡ ±1(mod 3), G 2 is a metabelian group and as a result of the computations of [4], G 2 is a cyclic group of order 2n.So c(G 2 ) ≤ 2n − 1 comes from the results of Section 2.
Let n ≡ 3(mod6).Using the Todd-Coxeter coset enumeration algorithm gives us the presentation It is easy to show the validity of the relations x n = y n and x 2n = y 2n = 1 in G 2 .Let n=6k + 3. We claim that the words w 1 = x 2 y 3k+1 xy 2 and w 2 = y 2 x 3k+1 yx 2 in G 1 are of the largest length 3k + 5. Indeed, the relations x 3 y 3 = 1, xyx = y n−1 , and yxy = x n−1 show that any word with maximal length could not contain the subwords x 2 y 2 and y 2 x 2 .The remained words which have to be examined are indeed w 3 = y 3k+1 xy 2 x and w 4 = x 3k+1 yx 2 y.These words are of length 3k + 4, for we have Similarly, w 4 is of length 3k + 4. To complete the proof we now show that w 1 and w 2 are of length 3k + 5.By using the relations we get w 1 = x 2 y 3k (yxy)y = x 2 y 3k x 6k+2 y = x 2 y 3k x 3k x 3k+2 y = x 3k+4 y, (3.16) and in a similar way, w 2 will be reduced to y 3k+4 x.So c(G 2 ) ≤ ML(G 2 ) ≤ ML {x,y} (G 2 ) = 3k + 5.
Let n ≡ 0(mod6).We use the Todd-Coxeter coset enumeration algorithm to find a presentation for G 2 .In two different cases, n ≡ 0(mod12) and n ≡ 6(mod12), we get the following presentations: .17) respectively, where R 1 = y 2+n/2 x 3 y −1 xy −1 x 3 and R 2 = yxy −1 xyx 3 y −1 x.The largest power of x in every word of G 2 is equal to 3, however, the largest power of y is n/2 + 1, for R 1 = 1 yields y 2+n/2 = xyx 3 yx.(3.18)This relation also gives us the relation (xy) 2 = y 2+n/2 x.The relation R 2 = 1 yields yxy n−1 xy = x 3 yx and hereby we deduce that the only words of maximum length must be among the following words: Obviously the words w 1 and w 2 are of length 4 + n/2, however w 4 is of length 5, for we see that Proof.The subgroup