Selçuk J. Appl. Math. Selçuk Journal of Vol. 13. No. 1. pp. 57-67, 2012 Applied Mathematics
The Fibonacci Length over Split Extensions of Some Special Groups Aynur Yalçıner
Department of Mathematics, Science Faculty, Selçuk University, Campus, 42075, Konya, Turkiye
e-mail: ayalciner@ selcuk.edu.tr
Received Date: September 15, 2011 Accepted Date: October 11, 2011
Abstract. The main goal of this paper is to determine the Fibonacci length of split extensions over some special groups.
Key words: Fibonacci length; Fibonacci sequence; Split extension; Dihedral group; Polyhedral group.
2000 Mathematics Subject Classification: 20E22, 20F05. 1. Introduction and Preliminaries
The concept of the Fibonacci length for finite cyclic groups has been introduced by Wall ([14]), and then has been extended to abelian groups by Wilcox ([15]). After that, in [4], Campbell et. al. defined the Fibonacci orbit and the Fibonacci length of 2-generator groups and they examined the Fibonacci length of the dihedral groups. Later, this result was generalized to the power of dihedral groups in [5], and was generalized to polyhedral and binary polyhedral groups in [6]. We may refer [1, 2, 12] for some other known results on this subject. For a finitely generated group G with a generating set A = {a1, ..., an}, the Fibonacci orbit FA(G) of G with respect to the A is the sequence
(1.1) x1= a1, ..., xn= an, xi+n= n Y j=1
xi+j−1, :: (i ≥ 1).
If FA(G) is periodic, then the length of the period of the sequence is called the Fibonacci length of G with respect to the generating set A and denoted by LENA(G). We note that the Fibonacci length of a group depends on chosen the generating set and the ordering in it.
In [14], by defining “Wall numbers", it has been showed that these numbers are equivalent to the Fibonacci length of the cyclic group Znon two generators. We
recall that, for n > 1, the 2-step Wall number k(n) is the length of the minimal period of the Fibonacci numbers. Meanwhile, in [6], it has been studied the minimal period of 3-step (tribonacci ) recurrence relation for 3-generator groups. In here, we will consider the 4−step Wall number that is actually the minimal period of the integer-valued recurrence relation
ti= ti−1+ ti−2+ ti−3+ ti−4, :: t1= a, :: t2= b, :: t3= c, :: t4= d, where each entry is reduced modulo n and denoted by k(a,b,c,d)(n).
Another one of the key material in this paper is the “split extension", equiv-alently (see [3]), semi-direct products. Briefly, for any groups A and K and a homomorphism θ : A → Aut(K) given by a → θa (for all a ∈ A), the semi-direct product ([11]) G = K oθA of K by A is defined by the multiplication (a, k)(a0, k0) = (aa0, (kθ
a0)k0) on the set of all ordered pairs (a, k), where a ∈
A, k ∈ K. It is also well known that, by assuming A and K have presentations PA=< x; r > and PK =< y; s >, respectively, the group G = K oθA has a presentation (see [10])
(1.2) P =<y, x; s, r, t>,
where t = {yxλ−1yxx−1 | y ∈ y, x∈ x} and λyx ia a word on y reprsenting the element (ky)θax of K (a ∈ A, k ∈ K, x ∈ x, y ∈ y).
In this paper we first establish the Fibonacci length of the split extension of dihedral groups (Section 2). Then we determine the Fibonacci length of the split extension of polyhedral and binary polyhedral groups (Section 3). Finally, in Section 4, we present a class of groups with the property that each groups in this class has a finite Fibonacci length, although the Fibonacci length of the split extension is infinite except the direct products.
2. The Split Extension of Dihedral Groups
Recall that, for n ≥ 3, the dihedral group D2n is defined by the presentation PD2n =< a1, b1; a
2
1, bn1, (a1b1)2> .
It is clear that the D2n is the split extension of the cyclic group of order n by a cyclic group of order 2. For any generating set {a1, b1} (with has an independent ordering) of D2n, it has been showed that LEN(a1,b1)(D2n) = 6
(see [4] for the details). Also the Fibonacci length of the automorphism group of D2n has determined in [9]. Now we will present the Fibonacci length of the split extension of a dihedral group by another dihedral group.
Let us consider the dihedral groups D2n and D2m where m, n ≥ 3. Assume that the homomorphism θ : D2n−→ Aut(D2m) is defined by
a17−→ θa1, b17−→ θb1.
In here the conjugate action on D2mcan be defined by a2 θa1 7−→ a2, b2 θa1 7−→ b−12 , a2 θb1 7−→ a2, b2 θb1 7−→ b−12 .
Thus we have a semi-direct product G = D2moθD2n with a presentation (2.1)
PG=< a1, b1, a2, b2 ; a21, bn1, a22, b2m, (a1b1)2, (a2b2)2, [a1, a2], [b1, a2], a1b2a−11 = b−12 , b1b2b−11 = b−12 > .
¾
Thus the main result of this section is the following:
Theorem 2.1. Let G = D2moθD2n has a presentation as in (2.1). Then
LEN(a1,b1,a2,b2)(G) = ⎧ ⎨ ⎩ 5n 2, n ≡ 4 (mod 4) , 5n, n ≡ 2 (mod 4) , 10n, otherwise.
Proof. We need to show that the Fibonacci orbit is of the form
(2.2) xi= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ a1, i ≡ 1, −4 (mod 10) , b±1, i ≡ 2, −3 (mod 10) , a2b±2(j−3)/51 , i ≡ 3, −2 (mod 10) , b±12 b±2(j−4)(j+1)/251 , i ≡ 4, −1 (mod 10) , a2a1b∓12 b±(2j 2 −25)/25 1 , i ≡ 5, 0 (mod 10) . Using finite induction, it is easy to see that
(2.3) bα1b2b−α1 = ½
b−12 , α is odd, b2, α is even.
As in [5], to (2.2) be hold, we will use another induction on i and use the equality in (2.3). x1 = a1, x2 = b1, x3 = a2, x4 = b2, x5 = a1b1a2b2= a1a2b1b2= a2a1b−12 b1, x6 = b1a2b2a2a1b−12 b1= b1b−12 a1b−12 b1= b1a1b1= a1, x7 = a2b2a2a1b−12 b1a1= b2−1a1b−12 b1a1= a1b1a1= b−11 , x8 = b2a2a1b−12 b1a1b−11 = b2a2b2a1b1a1b−11 = a2b−21 , x9 = a2a1b−12 b1a1b−11 a2b−21 = a2a1b2−1b1a1a2b−31 = a2a1b−12 a1b−11 a2b−31 = a2b2a2b−41 = b−12 b−41 , x10 = a1b−11 a2b−21 b−12 b−41 = a1a2b−31 b−12 b−41 = a1a2b2b−31 b−41 = a2a1b2b−71 . Let k ≡ 0 (mod 10) and suppose that the result holds for all values up to k + 5.
In other words, xk+1 = a1, xk+2 = b1, xk+3 = a2b2k/51 , xk+4 = b2b(2k 2+10k)/25 1 , xk+5 = a2a1b−12 b (2k2+20k+25)/25 1 .
Now we have to show that the required result is true for the next ten entries. xk+6 = b1a2b2k/51 b2b(2k 2 +10k)/25 1 a2a1b−12 b (2k2+20k+25)/25 1 = a2b1+2k/51 b2b(2k 2+10k)/25 1 a2a1b−12 b (2k2+20k+25)/25 1 = a2b−12 b 1+2k/5 1 b (2k2+10k)/25 1 a2a1b−12 b (2k2+20k+25)/25 1 = a2b−12 a2b(2k 2+20k+25)/25 1 a1b−12 b (2k2+20k+25)/25 1 = a2b−12 a2b(2k 2 +20k+25)/25 1 b2a1b(2k 2 +20k+25)/25 1 = a2b−12 a2b−12 b (2k2+20k+25)/25 1 a1b(2k 2+20k+25)/25 1 = a1. xk+7 = a2b2k/51 b2b(2k 2+10k)/25 1 a2a1b−12 b (2k2+20k+25)/25 1 a1 = a2b2k/51 b2b(2k 2+10k)/25 1 a2a1b−12 a1b−(2k 2+20k+25)/25 1 = a2b2b2k/51 b (2k2+10k)/25 1 a2b2b−(2k 2 +20k+25)/25 1 = a2b2b(2k 2+20k)/25 1 a2b2b−(2k 2+20k+25)/25 1 = a2b2a2b(2k 2 +20k)/25 1 b2b−(2k 2 +20k+25)/25 1 = a2b2a2b2b(2k 2+20k)/25 1 b−(2k 2+20k+25)/25 1 = b−11 . xk+8 = b2b(2k 2 +10k)/25 1 a2a1b−12 b (2k2+20k+25)/25 1 a1b−11 = b2b(2k 2+10k)/25 1 a2a1b−12 a1b−2−(2k 2+20k)/25 1 = b2b(2k 2 +10k)/25 1 a2b2b−2−(2k 2 +20k)/25 1 = b2a2b(2k 2+10k)/25 1 b2b−2−(2k 2+20k)/25 1 = b2a2b2b(2k 2 +10k)/25 1 b−2−(2k 2 +20k)/25 1 = a2b−2−2k/51 . xk+9 = a2a1b−12 b (2k2+20k+25)/25 1 a1b−11 a2b−2−2k/51 = a2a1b−12 b (2k2+20k+25)/25 1 a1a2b−3−2k/51 = a2a1b−12 a1b−(2k 2+20k+25)/25 1 a2b−3−2k/51 = a2b2b−(2k 2 +20k+25)/25 1 a2b−3−2k/51 = a2b2a2b−(2k 2+20k+25)/25 1 b−3−2k/51 = b−12 b−2(k1 2+15k+50)/25.
xk+10 = a1b−11 a2b−2−2k/51 b−12 b−2(k 2+15k+50)/25 1 = a1a2b−3−2k/51 b−12 b−2(k 2 +15k+50)/25 1 = a1a2b2b−3−2k/51 b−2(k 2+15k+50)/25 1 = a2a1b2b−(2k 2 +40k+175)/25 1 . xk+11 = b−11 a2b−2−2k/51 b−12 b−2(k 2+15k+50)/25 1 a2a1b2b−(2k 2+40k+175)/25 1 = a2b−3−2k/51 b−12 b−2(k 2 +15k+50)/25 1 a2a1b2b−(2k 2 +40k+175)/25 1 = a2b−3−2k/51 b−2(k 2+15k+50)/25 1 b−12 a2a1b2b−(2k 2+40k+175)/25 1 = a2b−2(k 2 +40k+175)/25 1 a2b2a1b2b−(2k 2 +40k+175)/25 1 = b−2(k1 2+40k+175)/25a1b−(2k 2+40k+175)/25 1 = a1. xk+12 = a2b−2−2k/51 b−12 b−2(k 2+15k+50)/25 1 a2a1b2b−(2k 2+40k+175)/25 1 a1 = a2b−2−2k/51 b−2(k 2 +15k+50)/25 1 b−12 a2a1b2b−(2k 2 +40k+175)/25 1 a1 = a2b−(2k 2+40k+150)/25 1 b−12 a2a1b2a1b(2k 2+40k+175)/25 1 = a2b−(2k 2 +40k+150)/25 1 b−12 a2b−12 b (2k2+40k+175)/25 1 = a2b−(2k 2+40k+150)/25 1 a2b(2k 2+40k+175)/25 1 = b−(2k1 2+40k+150)/25b(2k1 2+40k+175)/25 = b1. xk+13 = b−12 b−2(k 2+15k+50)/25 1 a2a1b2b−(2k 2+40k+175)/25 1 a1b1 = b−12 b−2(k1 2+15k+50)/25a2a1b2a1b(2k 2 +40k+200)/25 1 = b−12 b−2(k1 2+15k+50)/25a2b−12 b (2k2+40k+200)/25 1 = b−2(k1 2+15k+50)/25b−12 a2b−12 b (2k2+40k+200)/25 1 = b−2(k1 2+15k+50)/25a2b(2k 2+40k+200)/25 1 = a2b4+(10k/25)1 . xk+14 = a2a1b2b−(2k 2+40k+175)/25 1 a1b1a2b4+(10k/25)1 = a2a1b2a1b(2k 2 +40k+175)/25 1 b1a2b4+(10k/25)1 = a2b−12 b 1+(2k2+40k+175)/25 1 a2b4+(10k/25)1 = a2b−12 a2b1+(2k 2 +40k+175)/25 1 b 4+(10k/25) 1 = b2b(2k 2+50k+300)/25 1 . xk+15 = a1b1a2b4+(10k/25)1 b2b(2k 2 +50k+300)/25 1 = a1a2b1b4+(10k/25)1 b2b(2k 2+50k+300)/25 1 = a1a2b5+(10k/25)1 b2b(2k 2 +50k+300)/25 1 = a1a2b−12 b (2k2+60k+425)/25 1 .
to be smallest positive integer such that
xLEN +3 = a2b2LEN/51 = a2,
xLEN +4 = b2b2LEN (LEN +5)/251 = b2. This means that
LEN = ⎧ ⎨ ⎩ 5n 2 , n ≡ 0 (mod 4) , 5n, n ≡ 2 (mod 4) , 10n, otherwise. Hence the result.
3. The Split Extension of Binary Polyhedral and Polyhedral Groups It is known that, for l, m, n > 1, the binary polyhedral group B(l, m, n) and polyhedral group T (l, m, n) are defined by the presentations
PB(l,m,n)=< a, b, c; al= bm= cn= abc > and
PT (l,m,n)=< a, b, c; al, bm, cn, abc >,
respectively. We note that, for l = m = 2, the polyhedral group T (2, 2, n) simply becomes dihedral group. Also, the binary polyhedral group B(2, m, n) is an extension of the cyclic group Z2 by the group T (2, m, n) whenever the group T (2, m, n) is finite ([7]). We next note that Fibonacci lengths of binary polyhedral and polyhedral groups can be found in [6]. In this section we will examine the Fibonacci length of these groups under split extensions.
3.1. Case I: (Binary) polyhedral by finite cyclic groups
Let us consider two different split extensions of the binary polyhedral group by a cyclic group of order n and the polyhedral group by a cyclic group of order n, respectively.
Firstly, let us give our attention to the product B(m, 2, 2) oθZn, where (3.1) θ : Zn−→ Aut(T (m, 2, 2)), r 7−→ θa, r 7−→ θb, r 7−→ θc. By considering the 4-step Wall number k(n), we have the following result. Theorem 3.1. Let G be the group B(m, 2, 2) oθZn, where θ is defined as in (3.1). Then LEN(a,b,c,r)(G) = k(n).
Proof. By (1.2), G has a presentation
and then it is easy to see that the Fibonacci orbit can be written as the form: xi= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ arti, i ≡ 1 (mod 10) , brti, i ≡ 2 (mod 10) , crti, i ≡ 3 (mod 10) , rti, i ≡ 4 (mod 10) , c2rti, i ≡ 5 (mod 10) , c2bcrti, i ≡ 6 (mod 10) , cbcrti, i ≡ 7 (mod 10) , crti, i ≡ 8 (mod 10) , rti, i ≡ 9 (mod 10) , rti, i ≡ 0 (mod 10) , where ti= ti−1+ ti−2+ ti−3+ ti−4, t1= t2= t3= 0, t4= 1.
Again for simplicity, let us denote the Fibonacci length by LEN. Since we require LEN to be the smallest positive integer such that
xLEN +1 = artLEN +1, xLEN +2 = brtLEN+2, xLEN +3 = crtLEN +3, xLEN +4 = rtLEN+4,
this means that LEN = k(n), where k(n) is the 4-step Wall number of the positive integer n.
Hence the result.
By considering the polyhedral group T (m, 2, 2), we also have the following con-sequence of Theorem 3.1.
Corollary 3.1. Let G = T (m, 2, 2) oθZn. Then LEN(a,b,c,r)(G) = k(n). Proof. We note that G has a presentation
PG=< a, b, c, r; am, b2, c2, abc, rn, rar−1= a−1, [r, b], [r, c] > . Since c2= 1 in the proof of Theorem 3.1, the result can be seen directly. Moreover the next theorem presents a similar result by considering the binary polyhedral group B(2, m, 2).
Theorem 3.2. Let G = B(2, m, 2) oθZn. Then LEN(a,b,c,r)(G) = k(n), where k(n) denotes the 4-step Wall number.
Proof. Suppose that θ : Zn−→ Aut(B(2, m, 2)) is defined by r 7−→ θa, r 7−→ θb, r 7−→ θc.
Therefore G has a presentation
PG=< a, b, c, r; a2= bm= c2= abc, rn, rbr−1= b−1, [r, a], [r, c] > . Using the sequence in (1.1), the proof can be seen quite similarly as in the proof of Theorem 3.1.
Remark 3.1. As a result of Theorem 3.2, the Fibonacci length of G = T (2, m, 2) oθZn is k(n).
Since the direct product of any two groups is actually as a special case of the split extension of these groups by the meaning of θ is the identity homohorphism, one can also investigate the Fibonacci length on direct products.
Lemma 3.1. Let G be a finitely generated group with a generating set A = {a1, ..., aj}, and let LEN(G) = s. Then LEN(a1,...,an,r)(G × Zn) = k(n),
where k(n) denotes the (j + 1)-step Wall number. Proof. The members of the Fibonacci orbit are
x1= a1, · · · , xj = aj, xj+1= r, xj+2= wj+2(a1, · · · , aj, r), · · · · · · , xi−1= wi−1(a1, · · · , aj, r),
xi= a1rti, · · · , xi+j = ajrti+j, xi+j+1= rti+j+1, · · · where
ti= ti−1+ ti−2+ · · · + ti−j−1, t1= t2= · · · = tj = 0, tj+1= 1. Let us denote LEN(a1,...,an,r)(G × Zn) by LEN. Since we need the smallest
nontrivial integer LEN such that
xLEN +1= a1rtLEN +1, xLEN +2= a2rtLEN+2...xLEN +j+1= rtLEN+j+1, we have LEN = k(n).
The following result is an easy consequence of Lemma 3.1.
Corollary 3.2. Let G1 = B(m, 2, 2) × Zn, G2 = B(2, m, 2) × Zn, G3 = T (m, 2, 2) × Zn and G4= T (2, m, 2) × Zn. Then, for 1 ≤ i ≤ 4,
LEN (Gi) = k(n).
LEN (G1) = k(n), LEN (G2) = k(n), LEN (G3) = k(n), LEN (G4) = k(n), where k(n) denotes the 4-step Wall number of n.
3.2. Case II: (Binary) polyhedral by (binary) polyhedral groups In this part, let us again consider two different split extensions of the binary polyhedral group by a binary polyhedral group, and the polyhedral group by a polyhedral group, respectively.
Let us take the binary polyhedral groups B(n, 2, 2) and B(m, 2, 2), and consider the homomorphism θ : B(n, 2, 2) −→ Aut(B(m, 2, 2)) defined by
a17−→ θa1, b17−→ θb1, c17−→ θc1.
The action on B(m, 2, 2) can be defined by a2 θa1 7−→ a−12 , b2 θa1 7−→ b2, c2 θa1 7−→ c2, a2 θb1 7−→ a−12 , b2 θb1 7−→ b2, c2 θb1 7−→ c2, a2 θc1 7−→ a−12 , b2 θc1 7−→ b2, c2 θc1 7−→ c2.
Therefore, by (1.2), we have a semi-direct product G = B(m, 2, 2) oθB(n, 2, 2) with a presentation (3.3) PG=< a1, b1, c1, a2, b2, c2 ; an1= b21= c12= a1b1c1, am2= b22= c22= a2b2c2, [a1, b2], [a1, c2], [b1, b2], [b1, c2], [c1, b2], [c1, c2], a1a2a−11 = a−12 , b1a2b−11 = a−12 , c1a2c−11 = a−12 > . ⎫ ⎬ ⎭
Theorem 3.3. Let G has a presentation as defined in (3.3). Then LEN(a1,b1,c1,a2,b2,c2)(G) = 14.
Proof. We prove this by a direct calculation. We have the sequence a1, b1, c1, a2, b2, c2, c21c22, c21b1c1, c1b1c1, c1, b2c2c22c21, c2b2c2, c2, c21, a1, b1, c1, a2, b2, c2, ...
and the Fibonacci length is 14.
As a consequence of this theorem, we can give the following result. Corollary 3.3. Let G = T (m, 2, 2) oθT (n, 2, 2). Then
LEN(a1,b1,c1,a2,b2,c2)(G) = 14.
Now, let us consider the semi-direct product B(2, m, 2) oθB(2, n, 2), where (3.4) θ : B(2, n, 2) −→ Aut(B(2, m, 2)), a17−→ θa1, b17−→ θb1, c17−→ θc1.
Thus we obtain the following result.
Theorem 3.4. Let the group G be as defined in (3.4). Then LEN(a1,b1,c1,a2,b2,c2)(G) = 14.
Proof. We note that G has a presentation
PG=< a1, b1, c1, a2, b2, c2 ; a21= bn1= c12= a1b1c1, a22= bm2= c22= a2b2c2, [a1, a2], [a1, c2], [b1, a2], [b1, c2], [c1, a2], [c1, c2], a1b2a−11 = b−12 , b1b2b−11 = b−12 , c1b2c−11 = b−12 > . ⎫ ⎬ ⎭ Then the members of the Fibonacci orbit are
a1, b1, c1, a2, b2, c2, c21c22, b1c1, c1a1, b1a1, b2c2a21, c2a2, b2a2, a21, a1, b1, c1, a2, b2, c2, · · ·
Hence the Fibonacci length is 14, as required.
We further obtain the following result as a consequence of Theorem 3.4. Corollary 3.4. Let G = T (2, m, 2) oθT (2, n, 2). Then
LEN(a1,b1,c1,a2,b2,c2)(G) = 14.
4. The Split Extension of Extended Hecke Groups
In this section we mainly reveal that while the Fibonacci length of extended Hecke groups are finite, the Fibonacci length of the split extension of those are not.
We remind that the extended Hecke group H(λq) is defined by a presentation ([13])
PH(λq)=< a, b, c | a
2, bq, c2, (ac)2, (cb)2> .
As a special case, if q = 3, then H(λ3) is called the extended modular group and denoted by Γ. By [8], the extended Hecke group is actually the split extension of the Hecke group by a cyclic group of order 2.
Theorem 4.1. The Fibonacci length of the extended Hecke group is 8. Proof. Since the members of the Fibonacci orbit are
a, b, c, abc, bab−1, b−1, cb−2, bac, a, b, c, · · · , the required length is clearly 8.
We then obtain the following result as a quick consequence of Lemma 3.1 and Theorem 4.1 for the direct products.
Corollary 4.1. Let G = H(λq) × Zn. Then LEN(a,b,c,r)(G) = k(n).
In fact, for the case θ is not equal to identity homomorphism, a simple calcula-tion shows that the split extension H(λq)oθZnhas an infinite Fibonacci length. Since the memebers of the Fibonacci orbit are as in the following:
a, b, c, r, abcr, babr2, ab2abr4, acb2abab2abr8, ...
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