On the Fibonacci length of powers of dihedral groups - CiteSeerX

On the Fibonacci length of powers of dihedral groups C. M. Campbell∗ , P. P. Campbell∗ , H. Doostie† and E. F. Robertson∗ ∗ Mathematical Institute, University of St Andrews, North Haugh, St Andrews, KY16 9SS, Scotland. † Mathematics Department, University for Teacher Education, 49 Mofateh Avenue, Tehran, 15614, Iran. E-mail: [email protected], [email protected], [email protected], [email protected] Abstract. For a finitely generated group Qn G = hAi, where A = {a1 , a2 , . . . , an }, the sequence xi = ai , 1 ≤ i ≤ n, xi+n = j=1 xi+j−1 , i ≥ 1, is called the Fibonacci orbit of G with respect to the generating set A, denoted FA (G). If FA (G) is periodic we call the length of the period of the sequence the Fibonacci length of G with respect to i , i > 1 A, written LENA (G). In this paper we examine the Fibonacci lengths of D2m where D2m is the dihedral group of order 2m. 2001 Mathematics Subject Classification: 20D06 Keywords: Group, Fibonacci length, Fibonacci sequence

1

Introduction

The idea of forming a sequence of group elements based on a Fibonacci-like recurrence relation was initially introduced by Wall in [14] and later developed by other authors, see [1], [7], [16]. This idea was then refined in [3] and [5] to that of a Fibonacci orbit and Fibonacci length of a given group G = hAi where A = {a1 , a2 , . . . , an } as follows: Definition 1.1 For the finitely generated group G = hAi, where A = {a1 , . . . , an }, the Fibonacci orbit of G with respect to theQgenerating set A, written FA (G), is the sequence x1 = a1 , . . . , xn = an , xi+n = nj=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, written LENA (G). When it is clear 1

C. M. Campbell, P. P. Campbell, H. Doostie and E. F. Robertson

2

which generating set is being investigated we will write LEN (G) for LENA (G). It is clear that FA (G) is periodic if G is an epimorphic image of a Fibonacci group; see [13] for more information on Fibonacci groups. As can be seen from the above definitions the Fibonacci length of a group depends in general on the chosen generating set. It was noted in [3] that, for any generating pair {a, b} ∈ D2m , LEN{a,b} (D2m ) = 6, where D2m is the dihedral group of order 2m. In this paper we intend to investigate this phenomenon further i by calculating the Fibonacci length of D2m , for any integer i, with respect to the natural generating set. We use the natural generating set for D2m , as in [4], 2 defined as satisfying h a1 , b1 | a21 = 1, bm 1 = 1, (a1 b1 ) = 1 i. This is extended to the direct product by using the following well known method of construction: if G = h X | R i and N = h Y | S i then G × N = h X, Y | R, S, [X, Y ] i where [X, Y ] = {[x, y] : x ∈ X, y ∈ Y }, see [9]. In the above we have used relations and relators as appropriate and will continue to do so in this paper. When rewriting words we use the convention found in [11], namely the use of underscores to highlight the subwords which are replaced in passing from one word to the next. Over the years there has been interest in particular types of presentations for direct powers of groups, with one strand of this interest motivated by questions of Wiegold in [15]; see for example [2] and [4]. Other strands are discussed in i [10] and [12]. In this paper we find the structure of the Fibonacci orbit of D∞ i and use this to find the Fibonacci length of D2m . We show that for i > 1 the i on the natural generating set no longer remains constant Fibonacci length of D2m but varies with i and m. We also examine the Fibonacci length of a restricted i set of D2m , for odd m, on a different generating set. The results in this paper were suggested by data from computer programs written in the computational algebra system GAP [8].

2

i The Fibonacci Orbit of D∞ , i≥2

i D∞ has a presentation with 2i generators and 2i2 relations: i = h a1 , b1 , a2 , b2 , . . . , ai , bi | a2l = (al bl )2 = 1, D∞ [aj , ak ] = [aj , bk ] = [bj , bk ] = 1, 1 ≤ l ≤ i, 1 ≤ j < k ≤ i i.

The elements of the group defined by the above presentation can be written qi−1 ri−1 in the normal form aqi i ai−1 . . . aq11 bri i bi−1 . . . br11 where the ql are either 0 or 1. Throughout this paper we will always reduce elements of the Fibonacci orbit to this normal form. In order to elucidate the main proof of this section we start by considering 2 the Fibonacci type sequence of elements of D∞ as x1 = a1 , x2 = b1 , x3 = a2 , x4 = b2 , xn = xn−4 xn−3 xn−2 xn−1 , (n ≥ 5).


3

We have: 2 Lemma 2.1 Every element of the Fibonacci orbit {xj } of D∞ = ha1 , b1 , a2 , b2 i may be represented by:  a1 , j ≡ 1, −4 mod 10    ±1  , j ≡ 2, −3 mod 10   b1 ±2(j−3)/5 a2 b 1 , j ≡ 3, −2 mod 10 xj =  ±1 ±2(j−4)(j+1)/52   , j ≡ 4, −1 mod 10 b2 b1    ±1 ±(2j 2 −52 )/52 , j ≡ 5, 0 mod 10 a2 a1 b 2 b 1

where the positive exponent is chosen for the first value of j and the negative exponent is chosen for the second value of j. Proof. The assertion may be proved by induction on j. We first prove the anchor step of the induction. Since ai bai = b−1 , i odd, we have x 1 = a1 , x 2 = b1 , x 3 = a2 , x 4 = b2 , x 5 = a2 a1 b 2 b 1 , x 6 = b 1 a2 b 2 a1 b 1 a2 b 2 = a1 , x7 = a2 b2 a1 b1 a2 b2 a1 = a1 b1 a1 = b−1 1 , −1 −2 x8 = b2 a1 b1 a2 b2 a1 b1 = b2 a2 b2 .a1 b1 a1 b−1 1 = a2 b 1 , −2 −3 −1 −4 x9 = a1 b1 a2 b2 a1 b−1 1 a2 b1 = a2 b2 a2 .a1 b1 a1 b1 = b2 b1 , −2 −1 −4 −1 −7 x10 = a1 b−1 1 a2 b 1 b 2 b 1 = a2 a1 b 2 b 1 . Now let k ≡ 0 mod 10 and assume that the result holds for all values up to k + 5, namely xk+1 xk+2 xk+3 xk+4 xk+5

= a1 , = b1 , 2((k+3)−3)/5 2k/5 = a2 b 1 = a2 b 1 , 2((k+4)−4)((k+4)+1)/25 2k(k+5)/25 = b2 b1 = b2 b1 , 2 (2(k+5) −25)/25 (2k2 +20k+25)/25 = a1 a2 b1 b 2 = a2 a1 b 2 b 1 .

We now prove that the next ten entries have the required form and thus complete the induction. 2k/5

2k(k+5)/25

(2k2 +20k+25)/25

xk+6 = b1 a2 b1 b2 b1 a2 a1 b 2 b 1 1+2k/5+2k2 /25+10k/25 (2k2 +20k+25)/25 = b1 a1 b 1 a2 b 2 a2 b 2 = a1 .


2k/5

4

(2k2 +20k+25)/25

2k(k+5)/25

xk+7 = a2 b1 b2 b1 a2 a1 b 2 b 1 a1 2k/5+2k2 /25+10k/25 (2k2 +20k+25)/25 = b1 a1 b1 a1 a2 b 2 a2 b 2 −1 = b1 . (2k2 +20k+25)/25

2k(k+5)/25

xk+8 = b2 b1 a2 a1 b2 b1 a1 b−1 1 2 2 2k /25+10k/25 (2k +20k+25)/25 −1 = b1 a1 b 1 a1 b 1 b 2 a2 b 2 −2((k+8)−3)/5 = a2 b 1 . (2k2 +20k+25)/25

−2((k+8)−3)/5

xk+9 = a2 a1 b2 b1 a1 b−1 1 a2 b 1 (2k2 +20k+25)/25 −2k/5−3 a1 b 1 a2 b 2 a2 = a1 b 1 −1 −2((k+9)−4)((k+9)+1)/25 = b2 b1 .

−2((k+9)−4)((k+9)+1)/25

−2((k+8)−3)/5

xk+10 = a1 b−1 b−1 2 b1 1 a2 b 1 2 −2k/5−7−2k /25−15k/25 = a1 b 1 a2 b−1 2 2 −(2(k+10) −25)/25 = a2 a1 b−1 b . 2 1 −2((k+8)−3)/5

−(2k2 +40k+175)/25

−2((k+9)−4)((k+9)+1)/25

xk+11 = b−1 b−1 a2 a1 b−1 1 a2 b1 2 b1 2 b1 −10k/25−7−2k2 /25−6k/5 −2k2 /25−8k/5−7 −1 −1 = b1 a1 b 1 a2 b 2 a2 b 2 = a1 . −2((k+8)−3)/5

−2((k+9)−4)((k+9)+1)/25

−(2k2 +40k+175)/25

b−1 a2 a1 b−1 xk+12 = a2 b1 2 b1 2 b1 −2k/5−2−2k2 /25−6k/5−4 −2k2 /25−8k/5−7 −1 = b1 a1 b 1 a1 a2 b2 a2 b−1 2 = b1 . −2((k+9)−4)((k+9)+1)/25

−(2k2 +40k+175)/25

a2 a1 b−1 xk+13 = b−1 2 b1 2 b1 −2k2 /25−6k/5−4 −2k2 /25−8k/5−7 −1 = b1 a1 b 1 a1 b1 b−1 2 a2 b 2 2((k+13)−3)/5 = a2 b 1 . −(2k2 +40k+175)/25

a1

a1 b 1

2(k+10)/5

xk+14 = a2 a1 b−1 a1 b 1 a2 b 1 2 b1 −(2k2 +40k+175)/25 1+2(k+10)/5 = a1 b 1 a1 b 1 a2 b−1 2 a2 2((k+14)−4)((k+14)+1)/25 = b2 b1 . 2(k+10)/5

2((k+14)−4)((k+14)+1)/25

xk+15 = a1 b1 a2 b1 b2 b1 1+2(k+10)/5+2((k+14)−4)((k+14)+1)/25 = a1 b1 a2 b 2 (2(k+15)2 −25)/25 = a2 a1 b2 b1 .

In the same manner as the previous case we examine the Fibonacci sequence


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3 of D∞ i.e.

x 1 = a1 , x 2 = b 1 , x 3 = a2 , x 4 = b 2 , x 5 = a3 , x 6 = b 3 , x n =

6 Y

xn−7+j , (n ≥ 7).

j=1

3 Lemma 2.2 Every element of the Fibonacci orbit {xj } of D∞ = ha1 , b1 , a2 , b2 , a3 , b3 i may be represented by:  j ≡ 1, −6 mod 14  1,  a±1   , j ≡ 2, −5 mod 14 b  1   ±2(j−3)/7   , j ≡ 3, −4 mod 14   a2 b1 ±2(j−4)(j+3)/7 2 ±1 b2 b1 , j ≡ 4, −3 mod 14 xj =  ±2(j−5)/7 ±4(j−5)(j+2)(j+9)/(3×73 )   b1 , j ≡ 5, −2 mod 14 a3 b 2    ±1 ±2(j+1)(j−6)/72 ±2(j−6)(j+1)(j+8)(j+15)/(3×74 )  b3 b2 b1 , j ≡ 6, −1 mod 14     a a a b±1 b±(2j 2 −72 )/72 b±(2(j/7)4 +8(j/7)3 +4(j/7)2 −8(j/7)−3)/3 , j ≡ 7, 0 mod 14 3 2 1 3 2 1

where the positive exponent is chosen for the first value of j and the negative exponent is chosen for the second value of j. Proof. As in Lemma 2.1, it is sufficient to compute x1 , x2 , . . . , x14 and the result follows by induction on j. We now seek to generalize the above to obtain a normal form for the Fibonacci i , i ≥ 2. We will need the following result in our calculations. orbit of D∞ Lemma 2.3 For n ≥ 3 the following polynomial identity holds: 2+

+

n−1 X 2j−2 m(m + 1)(m + 2) . . . (m + j − 3) (j − 2)! j=3

n X 2j−2 (m − 1)m(m + 1) . . . (m + j − 4) (j − 2)! j=3

=

2n−2 m(m + 1)(m + 2) . . . (m + n − 3) (n − 2)!

with the convention that when n = 3, the first term in the first summation on the left hand side is zero. Proof. We use induction on n. When n = 3 the result holds since 2 + 0 + 1!2 (m − 1) = 2m, i.e. 1!2 m = 2m. When n = 4 the result also holds since 2+

2 2 22 22 m + (m − 1) + (m − 1)m = 2m(m + 1), i.e. m(m + 1) = 2m(m + 1). 1! 1! 2! 2!


6

Now assume that the result holds for all values less than n. Then n−1 n X X 2j−2 2j−2 2+ m(m + 1) . . . (m + j − 3) + (m − 1)m . . . (m + j − 4) (j − 2)! (j − 2)! j=3 j=3 n−2 n−1 X X 2j−2 2j−2 = 2+ m(m + 1) . . . (m + j − 3) + (m − 1)m . . . (m + j − 4) (j − 2)! (j − 2)! j=3 j=3

+

2n−3 2n−2 m(m + 1) . . . (m + n − 4) + (m − 1)m . . . (m + n − 4). (n − 3)! (n − 2)!

By the inductive hypothesis this is equal to 2n−3 2n−3 m(m + 1) . . . (m + n − 4) + m(m + 1) . . . (m + n − 4) (n − 3)! (n − 3)! + =

2n−2 (m − 1)m . . . (m + n − 4) (n − 2)!

2n−2 2n−2 m(m + 1) . . . (m + n − 4) + (m − 1)m . . . (m + n − 4) (n − 3)! (n − 2)! =

as required.

2n−2 m(m + 1) . . . (m + n − 3) (n − 2)!

Now we make the following observations about the Fibonacci orbits {xj } of i the groups D∞ = ha1 , b1 , a2 , b2 , . . . , ai , bi i (where i ∈ N). Lemma 2.4 The exponent of bl in xj is zero if j ≡ 1, 2, . . . , 2l − 1 mod (2i + 1), where l ∈ {1, 2, . . . , i}. Proof. This is easy to see if we look at the ‘structure’ of the entries of the i Fibonacci orbit {xj } of D∞ . For j ∈ {1, 2, . . . , 2i} we have a(j+1)/2 , if j is odd, xj = bj/2 , if j is even,

C. M. Campbell, P. P. Campbell, H. Doostie and E. F. Robertson and x2i+1 =

2i Y

7

xd .

d=1

Now it is easy to see that the next term of the Fibonacci orbit to contain an al will be 2l+2i−1 Y x2l+2i = xd . d=2l

Likewise the next bl occurs in x2l+2i+1 =

2l+2i Y

xd .

d=2l+1

The final stage of the induction follows by using an argument analogous to that above. Remark It can easily be shown that the exponents of bl can be calculated once one knows the exponents of b1 since the exponents of bl ‘lag’ behind the exponents of b1 . This holds because all bl ’s initially have exponent one and from Lemma 2.4 above it can be seen when the exponents of bl are nonzero. As an example of this 3 case where, when j ≡ 3 mod 14, the power of b1 is 2(j − 3)/7, ’lag’ see the D∞ and when j ≡ 5 mod 14, the power of b2 is 2(j − 5)/7. 3 These results are best illustrated by looking at the D∞ case (given separately in Lemma 2.2). It can be used as an example in the following proof to elucidate concepts. In this case the orbit is

Z = (a1 , b1 , a2 , b2 , a3 , b3 , a3 a2 a1 b3 b2 b1 , −2 −1 −4 −2 −4 −1 −4 −8 −1 −7 −19 a1 , b−1 1 , a2 b1 , b2 b1 , a3 b2 b1 , b3 b2 b1 , a3 a2 a1 b3 b2 b1 , a1 , b1 , . . .). So the Fibonacci orbit behaves as if it is in ‘layers’ of length 2i + 1 where in alternate layers the exponents of bl are all positive and all negative. Since one layer’s ‘structure’ depends only on the previous layer, a proof by induction will only need the first layer to be proved to anchor the induction. Also we note here that it is unnecessary to know the general form of the k(2i + 1)th entry (k ∈ N) in the Fibonacci orbit since this will always be the product of all the previous terms w1 i wi−1 in its layer and so this entry has a known shape, namely ai ai−1 . . . a1 bw i bi−1 . . . b1 where wl is the sum of the exponents of bl in the layer. We are now ready to give the main result of this section.


8

i Proposition 2.5 The exponents of b1 in the Fibonacci orbit {xj } of D∞ are given in the table below

0,

j ≡ 1, 2i + 2 mod (4i + 2)

±1,

j ≡ 2, 2i + 3 mod (4i + 2)

Q n−2 ±An ( n−3 , j ≡ n, 2i + 1 + n mod (4i + 2) d=0 (j − n + d(2i + 1)))/(2i + 1) Pj−1

j ≡ 2i + 1, 0 mod (4i + 2)

d=j−2i zd ,

where An = 2n−2 /(n − 2)!, zr is the exponent of b1 in xr , n ≡ j mod (2i + 1) so 3 ≤ n ≤ 2i and the positive forms of elements in the orbit are chosen if 1 ≤ j mod (4i + 2) ≤ 2i + 1; otherwise choose the (second) negative forms. Proof. Let the exponent of b1 in the xr entry of the Fibonacci orbit be zr . Assume that the result is true for all values of j such that 1 ≤ j < k. There are several cases to examine:

Case 1. k ≡ 1, 2i + 2 mod (4i + 2) In this case we have z

b1k−2i

+...+zk−2

z

a1 b1k−2i

+...+zk−2

= a1 .

Case 2. k ≡ 2, 2i + 3 mod (4i + 2) Here we have z

b1k−2i

+...+zk−3

z

a1 b1k−2i−2

+...+zk−3

0+zk−2i−1

a1 = a1 b1

−zk−2i−1

a1 = b 1

= b±1 1 .

Case 3. k ≡ n, 2i + n + 1 mod (4i + 2), 3 ≤ n ≤ 2i + 1 When we are trying to calculate the exponent of b1 for the kth entry in the Fibonacci orbit we need only concentrate on the terms in a1 and b1 . The exponent of b1 and a1 in the kth entry in the Fibonacci orbit is (

x−1 Y

x−1 Y

bz1l )(a1

l=k−2i

l=x−2i

bz1l )a1 (

k−1 Y

bz1l )

l=x+1

where x = (2i + 1)bk/(2i + 1)c. (Note the first bracket is from the layer below that of xk , the second bracket is the last entry in the lower layer). The above sum can be simplified using the group relations as follows (

x−1 Y

l=k−2i

bz1l )(a1

x−1 Y

l=x−2i

bz1l )a1 (

k−1 Y

l=x+1

k−2i−1 Y

bz1l ) = a1 (

l=x−2i

bz1l )a1 (

k−1 Y

l=x+1

bz1l ),


=

9

k−2i−1 Y

k−1 Y −zl ( b1 )( bz11 ). l=x−2i l=x+1

Thus the exponent of b1 is given by k−1 X

l=x+1

zl −

k−2i−1 X

zl

l=x−2i

= ±(0 + 1 + A3 ((x + 3) − 3)/(2i + 1) + . . . +Ak−x−1

hQ

k−x−4 ((k d=0

i − 1) − (k − x − 1) + d(2i + 1)) /(2i + 1)k−x−3 )

−(∓(0 + 1 + A3 ((x − 2i + 2) − 3)/(2i + 1) + . . . +Ak−x

hQ

k−x−3 ((k d=0

i − 2i − 1) − (k − x) + d(2i + 1)) /(2i + 1)k−x−2 )

So we want to show that n−1 n X X 2l−2 2l−2 2+ m(m + 1) . . . (m + l − 3) + (m − 1)m(m + 1) . . . (m + l − 4) (l − 2)! (l − 2)! l=3 l=3

=

2n−2 m(m + 1) . . . (m + n − 3) (n − 2)!

where ≡ k mod (2i + 1), m = bk/(2i + 1)c and 2 < n < 2i + 1 and Pn−1 nl−2 m(m+1) . . . (m+l −3)/(l −2)!) is zero if n = 3. Now the result follows l=3 (2 by using Lemma 2.3.

Case 4. k ≡ 2i + 1, 0 mod (4i + 2) Here there is nothing to prove.

3

i The Fibonacci Length of D2m , i≥2

i In order to give an expression for the Fibonacci length of D2m , i ≥ 2, we require the following definitions and lemmas. i i Definition In D2m let M inLEN (D2m ) = 2m(2i + 1)/(4, m).


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i i i Lemma 3.1 In D2m , i ≥ 2, M inLEN (D2m ) divides LEN (D2m ), and the quoi i tient LEN (D2m )/M inLEN (D2m ) only involves odd prime divisors of m. i Proof. We first note that M inLEN (D2m ) is the smallest possible Fibonacci i length since we must have (4i + 2) | LEN (D2m ) and, from the third entry in i i the Fibonacci orbit of D2m , we also have m | (2LEN (D2m )/(2i + 1)). Thus i i M inLEN (D2m ) divides LEN (D2m ). i Now let l = M inLEN (D2m ). By Proposition 2.5 for 3 ≤ n ≤ 2i we have

zl+n =

2n−2 l(l + (2i + 1)) . . . (l + (n − 3)(2i + 1)) . (n − 2)!(2i + 1)n−2

If l = l/(2i + 1) = 2m/(4, m) the above becomes zl+n =

2n−2 l(l + 1) . . . (l + (n − 3)) . (n − 2)!

Now zl+n is obviously an integer. Since zl+n is the power of b1 in the Fibonacci i )+n , 3 ≤ n ≤ 2i. It may occur that orbit we require that m divides zLEN (D2m m 6 |zl+n because powers of primes from the factorization of m that also occur in the numerator of zl+n may be factored out by the (n − 2)! denominator. Let these α α ‘missing’ primes be pj j . . . pαr r . Now multiplying l by a factor q less than pj j . . . pαr r will not be sufficient. For m will not divide zql+n since the denominator (n − 2)! will be the same as in the zl+n case and the numerator will still be a product of 2n−2 and n − 2 consecutive integers but this time starting at 2qm/(4, m). Thus α α (n − 2)! will still factor out pj j . . . pαr r and since q is less than pj j . . . pαr r we still α have m 6 |zql+n . If we let q = pj j . . . pαr r then we will have m|zql+n . i Definition Let πm,i ∈ N be defined as satisfying the equation LEN (D2m ) = i πm,i M inLEN (D2m ).

We now give a property of πm,i that will be used in a later lemma. Lemma 3.2 For any fixed m the sequence (πm,i )∞ i=2 is monotonically increasing. i Proof. Let l = LEN (D2m ) = 2mπm,i (2i + 1)/(4, m) and l = l/(2i + 1). From Proposition 2.5 for 3 ≤ n ≤ 2i we have

2n−2 l(l + 1) . . . (l + (n − 3)) (n − 2)! n−2 2 (2mπm,i /(4, m))((2mπm,i /(4, m)) + 1) . . . ((2mπm,i /(4, m)) + (n − 3)) . = (n − 2)!

zl+n =

As i increases, and m remains the same, the number of entries in the layers of the Fibonacci orbit increases as does (2i − 2)!. This gives the possibility that the


11

number and size of primes in the prime decomposition of (2i − 2)! will increase and as a consequence so will πm,i . Note if πm,i+1 6> πm,i then since (2i)! > (2i−2)!, πm,i+1 = πm,i . Now we show a multiplicative property of πm,i . Lemma 3.3 For m, n ∈ N, (m, n) = 1 and i ≥ 2 , πmn,i = πm,i πn,i . i Proof. This proof is similar to that of the previous lemma. Let l = M inLEN (D2(mn) ). As in Lemma 3.1 we look at

zl+k =

2k−2 l(l + (2i + 1)) . . . (l + (k − 3)(2i + 1)) , (k − 2)!(2i + 1)k−2

where 3 ≤ k ≤ 2i. Now πmn,i is found by seeing which primes in the prime decomposition of mn are factored out by the (k − 2)! denominator. Any primes that are factored out by the denominator are multiplied back so that mn|zl+k . Now since (m, n) = 1 and the layers of the Fibonacci orbit are the same length the result follows immediately. Lemma 3.4 If p is an odd prime and α ∈ N then πpα ,i = πp,i . Also for any β ∈ N, π2β ,i = 1. Proof. Here we use induction on i. i When i = 2, letting l = M inLEN (D2p ) = 2p5/(4, p) = 10p and using Proposition 2.5 we obtain zl+3 = 2(10p), zl+4 = 2(10p)(10p + 1). i i Both zl+3 and zl+4 are divisible by p and so l = M inLEN (D2p ) = LEN (D2p ) α and πp,2 = 1. Using an analogous argument we see that πp ,2 = 1.

Now assume that πpα ,i = πp,i for all i ≤ µ. We examine the case i = µ + 1. By Lemma 3.2 there are three possibilities: Case 1: πpα ,µ+1 = πpα ,µ and πp,µ+1 = πp,µ Here there is nothing to prove. Case 2: πpα ,µ+1 > πpα ,µ


12

µ α α α Let l = LEN (D2p α ), 3 ≤ n ≤ 2µ, l = l/(2µ+1) = 2p πpα ,µ /(4, p ) = 2p πpα ,µ so the power of b1 in the Fibonacci orbit is given by

zl+n =

2n−2 2pα πpα ,µ (2pα πpα ,µ + 1) . . . (2pα πpα ,µ + n − 3) . (n − 2)!

We must have pα |zl+n (since |b1 | = pα ). µ+1 Now consider what happens in the D2p α case. Keeping l to be the Fibonacci µ length of D2pα but letting 3 ≤ n ≤ 2µ + 2, as above we obtain zl+n

2n−2 2pα πpα ,µ (2pα πpα ,µ + 1) . . . (2pα πpα ,µ + n − 3) = , (n − 2)!

and since we require πpα ,µ+1 > πpα ,µ we have pα 6 |zl+n for some n, 3 ≤ n ≤ 2µ + 2. For n in the range 3 ≤ n ≤ 2µ we have pα |zl+n (this is just the previous µ case), so pα does not divide one or both of zl+n , n = 2µ + 1 or 2µ + 2. This means that (2µ)! contains a power of p, pγ say. Thus πpα ,µ+1 = pγ πpα ,µ . µ+1 We now examine the powers of b1 in the Fibonacci orbit of D2p and letting µ l = LEN (D2p ), 3 ≤ n ≤ 2µ + 2 gives zl+n =

2n−2 2pπp,µ (2pπp,µ + 1) . . . (2pπp,µ + n − 3) . (n − 2)!

For 3 ≤ n ≤ 2µ, p|zl+n but for either n = 2µ + 1 or 2µ + 2 we introduce a factor µ+1 γ of (2µ − 1)(2µ) in the denominator, as in the D2p α case, which contains p . So πp,µ+1 = pγ πp,µ . By the inductive hypothesis we know that πpα ,µ = πp,µ . Thus πp,µ+1 = γ p πp,µ = pγ πpα ,µ = πpα ,µ+1 . Case 3: πp,µ+1 > πp,µ Using an analogous argument as the above we see that in this case πpα ,µ+1 = πp,µ+1 . The second statement of the lemma holds since if we examine the powers i i of b1 in the Fibonacci orbit of D2(2 β ) we have, for l = M inLEN (D2(2β ) ) = 2(2i + 1)(2β /(4, 2β )), i ≥ 2, 3 ≤ n ≤ 2i, zl+n =

2n−2 2(2β /(4, 2β ))(2(2β /(4, 2β )) + 1) . . . (2(2β /(4, 2β )) + n − 3) . (n − 2)!

Now this is always divisible by 2β .

Definition For i ≥ 2, let m = 2α0 pα1 1 . . . pαk k where pl are distinct odd primes, and let tl = blogpl (2i − 3)c. Define Φm,i = pt11 . . . ptkk , and for all α ≥ 1, Φ2α ,i = 1.


13

We note here that Φm,i is obviously multiplicative in that for (m, n) = 1, Φmn,i = Φm,i Φn,i . Lemma 3.5 For all primes p, for α ∈ N and i ≥ 2, Φpα ,i = Φp,i . Proof. The definition of Φm,i is equivalent to finding for m = 2α0 pα1 1 . . . pαk k , tl where pl are distinct odd primes, Q thetl largest natural number tl such that (pl + 3)/2 ≤ i and letting Φm,i = pl |m pl . Since the power of a prime in the prime decomposition of m is not used in computing Φm,i , the lemma holds. i Proposition 3.6 For D2p , i ≥ 2 and p a prime, πp,i = Φp,i .

Proof. Let p be a fixed odd prime. We use induction on i, so πp,k = Φp,k for k < i. • i = 2. It follows by Lemma 3.4 that πp,2 = 1 and by definition that Φp,2 = 1. We now prove the result for i. First we note that πp,i forms a monotonic increasing sequence by Lemma 3.2. There are two cases to consider: 1. πp,i−1 = πp,i ; 2. πp,i−1 < πp,i . • Case 1, πp,i−1 = πp,i . There are two possibilities for the value of Φp,i ; either Φp,i−1 = Φp,i or Φp,i−1 < Φp,i . We will assume the latter case and reach a contradiction and so obtain Φp,i = Φp,i−1 = πp,i−1 = πp,i , the desired result. Assume that Φp,i > Φp,i−1 . If Φp,i−1 = pc , where c = blogp (2(i − 1) − 3)c, then Φp,i = pc+1 , where c + 1 = blogp (2i − 3)c. We know that πp,i−1 = πp,i so, using Proposition 2.5 and Lemma 3.1, the prime p divides

zl+(2i−1) =

22i−3 l(l + 1) . . . (l + (2i − 1) − 3) , ((2i − 1) − 2)!

i where l = LEN (D2p )/(2i + 1) = 2pπp,i /(4, p). But pc = πp,i−1 , so l = 2p.pc = 2pc+1 giving

22i−3 (2pc+1 )(2pc+1 + 1) . . . (2pc+1 + 2i − 4) (2i − 3)! c+1 c+1 c+1 c+1 2p 2p + 1 2p + 2 2p + 2i − 4 2i−3 = 2 ... . 2i − 3 1 2 2i − 4

zl+(2i−1) =


14

Note that the bth bracketed term, 2 ≤ b ≤ 2i − 3, in the above product may c+1 2p +pi mb be written as , where (mb , p) = 1 and 0 ≤ i ≤ c + 1 (this last inpi mb equality follows from the out the obvious simplification c+1−i definition of c). Carrying Q 2i−3 we obtain 2p mb +mb . Now notice that ( b=2 mb ) is coprime to p. Recall that p|zl+(2i−1) and, by assumption, blogp (2i − 3)c = c + 1 or 2i − 3 = pc+1 . From this last equation it follows that p 6 |zl+(2i−1) , a contradiction. Hence Φp,i = pc = Φp,i−1 = πp,i−1 = πp,i .

• Case 2, πp,i−1 < πp,i . Let l = 2pπp,i−1 . So by Lemma 3.1 we have l = 2pk for some integer k. Using Proposition 2.5 for 3 ≤ n ≤ 2i gives 2n−2 2pk (2pk + 1) . . . (2pk + n − 3) (n − 2)! k k k 2pk 2p + 1 2p + 2 2p + n − 3 n−2 ... . = 2 n−2 1 2 n−3

zl+n =

Since πp,i−1 < πp,i we must have p 6 |zl+2i−1 or p 6 |zl+2i . Using an argument similar to that given above we see that pk = (2i − 1) − 2 or pk = (2i) − 2. Hence k = blogp (2i − 3)c. Finally by Lemma 3.1 we have πp,i = pk−1 .p = pk , and so pk−1 = πp,i−1 = Φp,i−1 and pk = Φp,i = πp,i . i i Theorem 3.7 For i ≥ 2, LEN (D2m ) = Φm,i M inLEN (D2m ).

Proof. The theorem follows from Lemmas 3.3, 3.4, 3.5 and Proposition 3.6. Corollary 3.8 2 ) = 10m/(4, m) LEN (D2m

Proof. In this case, for m = 2α0 pα1 1 . . . pαk k where pl are distinct odd primes, we have tl = blogpl (1)c = 0 giving Φm,2 = 1. So by Theorem 3.7 we have 2 2 LEN (D2m ) = M inLEN (D2m ) = 10m/(4, m). Corollary 3.9 3 LEN (D2m ) = 14mΦm,3 /(4, m)

where Φm,3 = 3 if m ≡ 0 mod 3, and Φm,3 = 1 otherwise. Proof. Using the same notation as in the proof of the previous corollary we obtain tl = blogpl (3)c which is equal to 1 if pl = 3 and 0 otherwise.


4

15

Other dihedral group generators

2 We now examine the case of other presentations of D2m when m is odd. 2 Theorem 4.1 For m odd, D2m is defined by the presentation

h x, y | x2m = 1, (xm y)2 = 1, y 2m = (xy)2 i. 2 In this case LEN{x,y} (D2m ) = 6. 2 Proof. That the presentation defines D2m is shown in [4]. We first show that the 2 following relations hold in the presentation for D2m , m odd:

xy 2 xy 2 xyxy 2 xy = x and yxyxy 2 xyxy 2 xy 2 xyxy 2 xy = y. In [4] it is shown that the following relations also hold in the presentation for 2 D2m :

(i) y 2m = 1, (ii) (xy)2 = 1, (iii) xy m−1 x−1 = y −m+1 , (iv) y −m xy m = x−1 . Now 1 = xy m−1 x−1 xy m+1 x−1 by (i) and 1 = xy m−1 x−1 xy m+1 x−1 = y −m+1 xy m+1 x−1 by (iii). We now have 1 = yy −m xy m yx−1 = yx−1 yx−1 by (iv). Now using (ii) we see that 1 = yx−1 yx−1 = yx−1 y 2 xy = yx−1 yxyxy 2 xy = y 2 xy 2 xyxy 2 xy. 2 Thus xy 2 xy 2 xyxy 2 xy = x in D2m , m odd. The second result follows since

yxyxy 2 xyxy 2 xy 2 xyxy 2 xy = yxyxyyxyx = yyxyx = y. We now recall the well-known result that every finite 2-generator group G = ha, bi is a homomorphic image of some Fibonacci group F (2, n), where n =


16

2 LEN{a,b} (G). In the preceding paragraph we have shown that LEN (D2m )≤6 2 for all m , where D2m is defined by the presentation

h x, y

|

x2m = 1, (xm y)2 = 1, y 2m = (xy)2 i,

m odd; the proof is complete on noting that 6 = min{n : |F (2, n)| is infinite}. Relating this result to that of Corollary 3.8 we see that the generating set for m odd, used in this section always gives a constant Fibonacci length of 6 while the generating set used in the previous section gives a Fibonacci length of 10m/(4, m), which is always greater than 6. 2 D2m ,

2 Note The presentation given in Theorem 4.1 for D2m , m odd, is efficient. For 3 odd values of m, D2m has an efficient presentation

h a, x, z

|

x2m = z 2m = (xz m )2m = a2 = [x, a] = [a, z m ] = [x, z m−1 ] = 1, (xm−1 z m )2 = (az m−1 )2 = 1 i.

In this case our computations, using the GAP computer algebra package [8], imply that the Fibonacci length is equal to 8m. This remains as a conjecture.

Acknowledgments The third author wishes to thank the University for Teacher Education of Iran and the University of St Andrews for their support while this research was being undertaken.

References [1] H. Aydin and G.C. Smith, Finite p-quotients of some cyclically presented groups, J. London Math. Soc. 49 (1994), 83-92. [2] H. Ayik, C.M. Campbell, J.J. O’Connor and N. Ruˇskuc, The semigroup efficiency of direct powers of groups, in: Proc. Internat. Conf. on Semigroups eds. P. Smith, E. Giraldes and P. Martins, World Scientific, Singapore, 2000, 19-25. [3] C.M. Campbell, H. Doostie and E.F. Robertson, Fibonacci length of generating pairs in groups, in: Applications of Fibonacci numbers 3, eds. G.A. Bergum et al., Kluwer Academic Press, Dordrecht, 1990, 27-35. [4] C.M. Campbell, E.F. Robertson and P.D. Williams, On the efficiency of some direct powers of groups, in: Groups-Canberra 1989, ed. L.G. Kov´acs, Lecture Notes in Math. 1456 Springer, Berlin, 1990, 106-113.


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[5] H. Doostie, Fibonacci-type sequences and classes of groups (Ph.D. Thesis), University of St Andrews, Scotland, (1988). [6] H. Doostie and C.M. Campbell, Fibonacci length of automorphism groups involving Tribonacci numbers, Vietnam J. Math. 28 (2000), 57-65. [7] H. Doostie and R. Golamie, Computing on the Fibonacci lengths of finite groups, Int. J. Appl. Math. 4 (2000). [8] GAP, GAP-Groups, Algorithms, and Programming, Version 4.3, Aachen, St Andrews, 2002. (http://www-gap.dcs.st-andrews.ac.uk/~gap) [9] D.L. Johnson, Presentations of groups, 2nd edition, London Math. Soc. Student Texts 15, Cambridge University Press, Cambridge 1997. [10] J.C. Lennox and J. Wiegold, Generators and killers for direct and free products, Arch. Math. (Basel) 34 (1980), 296-300. [11] C.C. Sims, Computation with finitely presented groups, Encyclopedia of Mathematics and its Applications 48, Cambridge University Press, Cambridge 1994. [12] A.G.R. Stewart and J. Wiegold, Growth sequences of finitely generated groups II, Bull. Austral. Math. Soc 40 (1989), 323-329. [13] R.M. Thomas, The Fibonacci groups revisited, in: Groups - St Andrews 1989 eds. C. M. Campbell and E. F. Robertson, Cambridge University Press, Cambridge, 1991, 445-454. [14] D.D. Wall, Fibonacci series modulo m, Amer. Math. Monthly 67 (1960), 525-532. [15] J. Wiegold, Th Schur multiplier: an elementary approach, in: Groups St Andrews 1981 eds. C. M. Campbell and E. F. Robertson, Cambridge University Press, Cambridge, 1982, 137-154. [16] H.J. Wilcox, Fibonacci sequences of period n in groups, Fibonacci Quart. 24 (1986), 356-361.