FINITE p-QUOTIENTS OF SOME CYCLICALLY PRESENTED GROUPS

HUSEYIN AYDIN AND GEOFF C. SMITH

1 Introduction

We define the Fibonacci groups. These are groups parameterized by natural numbers r and n and described by the presentations

F (r, n) =< x_{1}, x_{2}, x_{3}, ⋯, x_{n} : x_{1}x_{2}x_{3} ⋯ x_{r} = x_{r+1}, x_{2}x_{3}x_{4} ⋯ x_{r+1} = x_{r+2},
⋯, x_{n−1}x_{n}x_{1} ⋯ x_{r−2} = x_{r−1}, x_{n}x_{1}x_{2} ⋯ x_{r−1} = x_{r} > .

We shall assume n, r > 1 to avoid triviality. These groups were introduced following [7], and have been extensively studied, particularly by D. L. Johnson and co-workers [9],[10],[11]. The main progress has been in understanding when such groups are finite or infinite, and in analyzing their abelian quotient structure. The survey articles [14] and [15] give an excellent summary of what is known. The work in this paper is parallel to that in [5] where recurrences in non-abelian finite simple groups are studied.

Fibonacci groups can have very large finite p−quotients. For example, it takes just a few minutes running CAYLEY [6] to demonstrate that R(2, 5), the restricted Burnside group of exponent 5 on 2 generators, is a quotient of F (2, 20). The group R(2, 5) has order 5^{34} and nilpotency class 12.

It is clear that every two-generator finite group must arise as a quotient of F (2, n) for some natural number n. It is, at first sight, somewhat curious that n can be as small as 20 for a group as substantial as R(2, 5).

In [17] Wall investigated the length of the fundamental period of the Fibonacci sequence modulo a natural number m. By the Chinese Remainder Theorem this amounts to studying the case where m is a prime power. He used k or k(m) to denote the fundamental period of the sequence, and we retain that notation. We call it Wall’s Number.

THEOREM (D.D. Wall [17]) If p is a prime number then for each natural number u we have k(p^{u}) | k(p)p^{u−1}.

He also showed that the division symbol may be replaced by equality providing k(p) ≠ k(p^{2}). He conjectured that this never happens and verified the result for all p < 10^{4}. We have checked this result using more modern computers, and can confirm that there is no counter-example less than 10^{8}. Utterly unjustifiable probabilistic arguments suggest that Wall’s conjecture is indeed true. Wall’s Theorem enables us to understand 2-step Fibonacci sequences in any finite abelian group. We seek to expand the compass of this theorem to enable us to understand more general sequences in finite nilpotent groups.

Wall’s result can be rephrased in the following way. Suppose there exists a group homomorphism π : F (2, k(p)) ⟶ Z/pZ with x_{1}π = 0 and x_{2}π = 1, then there exists a group homomorphism \overline{π} : F (2, k(p)p^{u−1}) ⟶ Z/p^{u}Z with x_{1}\overline{π} = 0 and x_{2}\overline{π} = 1.

Notice that it is possible to have epimorphisms from F (2, n) to Z/pZ which are not of this form. Consider the map ψ : F (2, 4) ⟶ Z/5Z such that x_{1}ψ = 1 and x_{2}ψ = 3. The point is that (1, 3, 4, 2) is a Fibonacci sequence in Z/5Z containing no 0. Such loops must have adjacent terms in a golden ratio. Such loops are studied elsewhere [1] [2] [4] [16] and [17].

2. Notation

We will be able to strengthen Wall’s result so that it may be applied to finite p−groups. Let G be any finite group, and suppose g_{0}, g_{1} ∈ G. Let k(g_{0}, g_{1}) denote the length of a fundamental period of the Fibonacci sequence in G defined by g_{i}g_{i+1} = g_{i+2}, and put k(G) = lcm{k(g_{0}, g_{1}) |g_{0}, g_{1} ∈ G}.

One expects the Fibonacci recurrence to be slightly special; we set up a general linear recurrence. Let G be a group and let

x_{i} = x_{i−v}^{a_{v}}x_{i−v+1}^{a_{v−1}} … x_{i−1}^{a_{1}}  (∗)

be a v−step linear recurrence which is invertible in the group; this means that given g_{i−v+1}, …, g_{i} ∈ G there exists a unique g_{i−v} ∈ G satisfying the recurrence. We say that (∗) is a good recurrence for Gi if p |̸ a_{v}. For example. if G is a finite p−group then (1) is a good recurrence if p |̸a_{t}. The r−step Fibonacci recurrence is good in any finite group.

A sequence of (finite) group elements satisfying the recurrence is called a loop. Such a sequence must be ultimately periodic, and the length of its fundamental period is called its Wall number. Notice that given such a sequence (g_{i}) we may select a natural number α and obtain another loop (g_{i+α}). We call this process rotation. Notice that the Wall number is rotation invariant. In the event that the recurrence is good, we may extend the sequence backwards to obtain a bi-infinite sequence, and admit rotation by negative integers.

From now on, for the sake of simplicity, we shall assume that all recurrences are good for the particular group in question. We can get away with less, but this will render our arguments more straightforward.

Fix a recurrence (∗) and a prime p, and let k_{∗} denote the fundamental period of the standard loop (st_{i}) in Z/pZ beginning 0, 0, 0, … 0, 1. Notice that any loop of (∗) in Z/pZ must have Wall number dividing k_{∗}, since any such loop must be a linear combination of rotations of (st_{i}). We may define k_{∗}(G) in the obvious way.

Given a finite p−group G, we say it has p−class u if u is the minimum length of a central series of G in which adjacent factors have exponent p.

3. Results

THEOREM A. If G is a finite p−group, then k(G) |k(p)p^{u−1} where u denotes p−class of G.

One can calculate that k(5) = 20, so Theorem A tells us that k(R(2, 5)) divides 20 · 5^{11}, which still leaves us with the mystery as to why k(R(2, 5)) = 20. We cannot give a complete answer to this question, but we shall demonstrate why, in groups such as R(2, 5), Wall’s number is smaller than one might expect. Theorem A is a clean generalization of Wall’s Theorem, but it is subsumed by Theorem B.

THEOREM B. Suppose G is a finite p−group of p−class u then k_{∗}(G) |k_{∗}(p)p^{u−1}.

We now revert to the two step Fibonacci recurrence, and address the question as to whether these factors of p^{u−1} are strictly necessary.

THEOREM C. Suppose that p is a rational prime and let G be a two generator free group in the variety groups of exponent p and nilpotency class 4, then it follows that k(G) = k(p).

Observe that the relatively free group G is finite, and that the property that k(G) = k(p) is inherited by all groups of exponent p and nilpotency class at most 4, irrespective of the size of a generating set. The theorem cannot be strengthened in a crude manner. Calculations in CAYLEY indicate that the result is false (in general) if class 4 is replaced by class 5.

In a forthcoming paper [9], Johnson and Odoni will characterize exactly when F (r, n) made abelian is cyclic. This result will, in turn, characterize exactly when Theorem C applies – since it may be deployed whenever F (r, n) made abelian is non-cyclic.

There are considerable advantages to working in the relatively free group. Any pair of generators will generate G as free generators, so lowest common multiples do not cause any difficulty.

4. Proofs

Proof of THEOREM B. Fix the good recurrence (∗), a loop (g_{i}) in G, and recall that k_{∗} denotes the Wall number of the standard loop (st_{i}) in Z/pZ. Choose N a central subgroup of G such that G/N has p−class t − 1, and N has exponent p. The result is trivial if u = 1 so we may assume u > 1 and put λ = k_{∗}(p)p^{u−2}. By induction we may assume that k_{∗}(G/N) | λ. Pick a transversal T for N in G. Suppose that for each i, we have g_{i} = t_{i}n_{i} where t_{i} ∈ T and n_{i} ∈ N. Notice that (t_{i}) is a periodic sequence of period dividing λ though it is not in general a loop.

Consider the loop (m_{i}) such that m_{i} = n_{λ+i}^{−1}n_{i} for 0 ≤ i ≤ v − 1. The loop (m_{i}) consists of central elements of G.

Let (g_{i}^{+}) = (g_{i+λ}) be the rotation of (g_{i}) through λ. Now form the co-ordinatewise product loop (g_{i}^{+}) · (m_{i}). This is still a loop (because (m_{i}) is central). Notice that (g_{i}^{+}) · (m_{i}) and (g_{i}) co-incide in the first v places, and so (g_{i}^{+}) · (m_{i}) = (g_{i}). Thus we may equate entries. Notice that (m_{i}) has fundamental period dividing k_{∗}, and hence dividing λ.

For all j and for all i in the range 0 ≤ i ≤ v − 1 we have t_{i}n_{jλ+i} = t_{i}n_{(j+1)λ+i}m_{i}. Thus n_{jλ+i}n_{(j+1)λ+i}^{−1} = m_{i} is independent of j, and so n_{pλ+i} = m_{i}^{−p}n_{i} = n_{i} for all i in 0 ≤ i ≤ v − 1. We conclude that (g_{i}) has period dividing pλ and we are done.

Proof of THEOREM C. Consider a group H = H^{Z} with the following presentation:

H^{Z} =< x, y, z, t, u : (y, x) = z, (z, x) = t, (t, x) = u >

where those pairs of generators with unspecified commutators are implicitly deemed to commute. This is a torsion-free nilpotent group of nilpotency class 4. In fact this group is generated by just x and y. Each element of the group will have a unique representation as x^{a}y^{b}z^{c}t^{d}u^{e} where a, b, c, d, e ∈ Z. In fact, we may as well think of this group as being a rather non-standard group structure on Z^{5}. The group multiplication law will be given by rational polynomials in 10 variables, and inversion by rational polynomials in 5 variables. These polynomials must, of course, have the property that, regarded as maps, they assume integral values when supplied with integral arguments.

The only primes arising in the denominators of these polynomials are 2 and 3. We can use these polynomials to define a group structure on (Z/pZ)^{5}, and obtain a group denoted H^{Z/pZ} or just \overline{H}. A straightforward calculation shows that \overline{H} has exponent p if p > 3.

The group law, in H or \overline{H}, is given as follows. We use (a, b, c, d, e) to denote x^{a}y^{b}z^{c}t^{d}u^{e}.

We have

(a, b, c, d, e) · (a′, b′, c′, d′, e′) = (a′′, b′′, c′′, d′′, e′′)

represents the group law. A straightforward induction argument yields the following:

a′′ = a + a′
b′′ = b + b′
c′′ = c + c′ + a′b
d′′ = d + d′ + a′c + (_{2}^{a′})b
e′′ = e + e′ + a′d + (_{2}^{a′})c + (_{3}^{a′})b.

These are tractable formulas. In \overline{H} the loop beginning x, y is the object of our attention. We seek to show that it has period k. Let (s_{i}) denote the standard loop for the 2-step Fibonacci recurrence in Z/pZ. This is the loop with s_{0} = 0 and s_{1} = 1. In the vector notation , we suppose that our loop has i^{th} entry (s_{i−1}, s_{i}, z_{i}, t_{i}, u_{i}). The first two entries are correct, for x = (1, 0, 0, 0, 0) and y = (0, 1, 0, 0, 0). We need to show that z_{k} = z_{k+1} = t_{k} = t_{k+1} = u_{k} = u_{k+1} = 0.

Using the formula for multiplication, induction arguments enable us to write down formulas for z_{n}, t_{n} and u_{n} for each n ∈ N. We summarize the results. In what follows, we use an unadorned ∑ to denote a sum taken over a fundamental period, so the dummy variable of summation may be taken to range from 0 to k − 1, or over any set of k consecutive integers.

z_{n} = ∑_{i=0}^{n−1} s_{n−i−1}s_{i}^{2}.

So

z_{k} = ∑_{i=0}^{k} s_{k−i−1}s_{i}^{2} = ∑_{i=0}^{k−1} s_{k−i−1}s_{i}^{2} ∑ s_{k−i−1}s_{i}^{2}.

Similarly

z_{k+1} = ∑_{i=0}^{k} s_{k−i}s_{i}^{2} = ∑ s_{k−i}s_{i}^{2} = ∑ s_{i}^{3}(−1)^{i+1}.

Next, we find that the formula for t_{n} is

t_{n} = ∑_{j=0}^{n−1} s_{n−1−j}s_{j}z_{j} + ∑_{i=0}^{n−1}(_{2}^{s_{i}})s_{i}s_{n−i−1}.

In particular,

t_{k+1} = ∑_{j=0}^{k} s_{k−j}s_{j} ∑_{i=0}^{j−1} s_{i}^{2}s_{j−i−1} + ∑_{i=0}^{k}(_{2}^{s_{i}})s_{i}s_{k−i}

which, after cosmetic treatment, becomes

t_{k+1} = ∑_{j=0}^{k−1} ∑_{i=0}^{j−1} s_{j}^{2}s_{i}^{2}s_{j−i−1}(−1)^{j+1} + ∑_{i=0}^{k}(_{2}^{s_{i}})s_{i}^{2}(−1)^{i+1}.

Similarly we find that

t_{k} = ∑_{j=0}^{k−1} s_{k−1−j}s_{j} ∑_{i=0}^{j−1} s_{i}^{2}s_{j−i−1} + ∑_{i=0}^{k}(_{2}^{s_{i}})s_{i}s_{i+1}(−1)^{i}.

Finally, the expression for u_{n} is

u_{n} = ∑_{r=0}^{n−1} s_{n−r−1}s_{r}t_{r} + ∑_{j=0}^{n−1} s_{n−j−1}(_{2}^{s_{j}})z_{j} + ∑_{i=0}^{n−1} s_{n−i−1}(_{3}^{s_{i}})s_{i},

which expands to

u_{n} = ∑_{r=0}^{n−1} ∑_{j=0}^{r−1} ∑_{i=0}^{j−1} s_{n−r−1}s_{r}s_{r−j−1}s_{j}s_{j−i−1}s_{i}^{2} + ∑_{j=0}^{n−1} ∑_{i=0}^{j−1} s_{n−j−1}s_{j}(_{2}^{s_{i}})s_{i}s_{j−i−1}

+ ∑_{j=0}^{n−1} ∑_{i=0}^{j−1} s_{n−j−1}(_{2}^{s_{j}})s_{j−1−i}s_{i}^{2} + ∑_{i=0}^{n−1} s_{n−i−1}(_{3}^{s_{i}})s_{i}.

In particular we have

u_{k+1} = ∑_{r=0}^{k} ∑_{j=0}^{r−1} ∑_{i=0}^{j−1} s_{r}^{2}s_{j}s_{i}^{2}s_{r−j−1}s_{j−i−1}(−1)^{r+1} + ∑_{j=0}^{k} ∑_{i=0}^{j−1} s_{k−j}s_{j}(_{2}^{s_{i}})s_{i}s_{j−i−1}

+ ∑_{j=0}^{k} ∑_{i=0}^{j−1} s_{j}(_{2}^{s_{j}})s_{j−i−1}s_{i}^{2}(−1)^{j+1} + ∑_{i=0}^{k} s_{i}^{2}(_{3}^{s_{i}})(−1)^{i+1}

and

u_{k} = ∑_{r=0}^{k} ∑_{j=0}^{r−1} ∑_{i=0}^{j−1} s_{k−1−r}s_{r}s_{r−j−1}s_{j}s_{j−i−1}s_{i}^{2} + ∑_{j=0}^{k−1} ∑_{i=0}^{j−1} s_{k−j−1}s_{j}(_{2}^{s_{i}})s_{i}s_{j−i−1}

+ ∑_{j=0}^{k} ∑_{i=0}^{j−1} s_{k−j−1}(_{2}^{s_{j}})s_{j−i−1}s_{i}^{2} + ∑_{i=0}^{k} s_{k−i−1}(_{3}^{s_{i}})s_{i}.

One is faced with the task of showing that the relevant terms vanish. Each expression is made up of a number of separate sums. We can show that, for all but a finite number of bad primes, these individual sums vanish. The bad primes are not artifacts of the algebra; sometimes the individual sums do not vanish. Happily, however, they cancel one another in such circumstances. The easy way to handle the bad primes is simply to make a list of them, and then allow CAYLEY to verify that Theorem C is true for those primes. The bad primes which arise in the course of the analysis are 2, 3, 5 and 11.

In the case of good primes the proofs take some 40 A4 pages, and we do not wish to presume too much on the hospitality of this journal. Some of the techniques deployed are illustrated in Section 5. A full treatment can be found in [4].

We move on to show how the proof may be completed for good primes, given that it has been shown that z_{k} = z_{k+1} = t_{k} = t_{k+1} = u_{k} = u_{k+1} = 0. The group \overline{H} is not the ‘correct’ group for our purposes.

Let T be the two generator relatively free class 4 group, and suppose that T is relatively free on x and y. We assume the reader has a passing familiarity with the theory of torsionfree finitely generated nilpotent groups, and refer him or her to [8] or [13] in the event that our presumption is incorrect. The group T has Hirsch length 8, and a Mal’cev basis may be easily described. It is

x, y, (y, x), ((y, x), x), ((y, x), y), (((y, x), x), x),
(((y, x), y), x) = (((y, x), x), y), (((y, x), y), y).

Thanks to [12] page 349 exercise 9, and to Eamonn O’Brien for pointing out this reference, we know that \overline{T} = T/T^{p} has order p^{8}. It follows that the Mal’cev basis we have described induces a central series in \overline{T} where adjacent terms of the series have quotients which are cyclic of order p. We abusively denote the images of x and y in \overline{T} by the same letters. The group \overline{T} is the relatively free group which we seek. We have a Fibonacci automorphism ϕ : \overline{T} ⟶ \overline{T} defined by xϕ = y and yϕ = xy. Let M =< (y, x, y) >\overline{T} , a normal subgroup of order p^{3}. A routine calculation shows that Mϕ ∩ M ∩ Mϕ^{−1} = 1. Also, one can verify that \overline{T} /M is naturally isomorphic to \overline{H}. We are in the happy position of understanding a particular loop in \overline{H}.

We have natural maps \overline{T} ⟶ \overline{T} /Mϕ^{i} for each i, and these induce an embedding

θ : \overline{T} ⟶ \overline{T} /Mϕ × \overline{T} /M × \overline{T} /Mϕ^{−1}.

Now Im(\overline{T}) is a subdirect product in \overline{T} /Mϕ × \overline{T} /M × \overline{T} /Mϕ^{−1}. Each group \overline{T} /Mϕ^{i} is a copy of the group H studied in this chapter.

When we compute the Fibonacci sequence in \overline{T} the images in \overline{T} /Mϕ × \overline{T} /M × \overline{T} /Mϕ^{−1} of the first four terms are as follows.

(x(Mϕ), xM, x(Mϕ^{−1}))  (a)
(y(Mϕ), yM, y(Mϕ^{−1}))  (b)
(xy(Mϕ), xyM, xy(Mϕ^{−1}))  (c)
(yxy(Mϕ), yxyM, yxy(Mϕ^{−1}))  (d)

In the second co-ordinate, we are simply enumerating the Fibonacci loop in \overline{H} which we studied earlier. This is not quite true of the first and third co-ordinates. The groups are the same (up to isomorphism), but the starting values are different. However, in these co-ordinates we are actually dealing with rotations of the series in the second co-ordinate. This is because (xM)ϕ = y(Mϕ) and (yM)ϕ = xy(Mϕ) yielding a rotation by 1 place in the loop of the first co-ordinate. Similarly ϕ^{−1} induces a rotation through 1 place in the opposite direction in the third co-ordinate. Thus the loop beginning (x, y, …) in \overline{T} has length k = k(p) as required. The fact that \overline{T} is relatively free on x and y shows that the theorem must hold in any group in the variety of groups of p−class 4.

Observations. The final ruse in the proof of Theorem B was used because we did not know that all loops in H had length dividing k, we only knew it for a specific loop of \overline{H}. Now, however, we are suddenly liberated. In particular, understanding one loop in \overline{H} has enabled us to understand all loops in \overline{H}.

5. Methods

We need to develop a calculus to deploy in order to demonstrate that various sums mentioned in the proof of Theorem C actually vanish.

Let (s_{i}) denote the ordinary Fibonacci sequence. The following identities hold for each integer j. Some are well-known. The proof of each each equation is trivial.

s_{−j} = s_{j}(−1)^{i}
s_{j}^{2} = s_{j+1}^{2} + s_{j−1}^{2} − 2s_{j}^{2} + 2(−1)^{j+1}
3s_{j}^{2} = s_{j+1}^{2} + s_{j−1}^{2} + 2(−1)^{j+1}
6s_{j}s_{j−1}^{2} = s_{j+1}^{3} + s_{j−2}^{3} − 2s_{j}^{3}
6s_{j}^{2}s_{j−1} = s_{j+1}^{3} − s_{j−2}^{3} − 2s_{j−1}^{3}

Now one can develop ploys to deal with sums. From now on we will be working modulo a prime p. The following are samples – and are easy to see.

∑_{i=0}^{k−1} s_{j} = ∑_{i=0}^{k−1} s_{k−j}

∑_{i=0}^{k−1} s_{j+a} = ∑_{i=0}^{k−1} s_{j}

We will do some example calculations in order to demonstrate the techniques involved. We choose these results almost arbitrarily.

DEMONSTRATION RESULT 1 The following equations concerning sums of powers of Fibonacci numbers over a fundamental range all hold good.

∑_{i=0}^{k−1} s_{i} = 0.  (1)

∑_{i=0}^{k−1} s_{i}^{2} = 0.  (2)

∑_{i=0}^{k−1} s_{i}^{3} = 0.  (3)

Proof. The results are true by inspection when p = 2 so we may assume p > 2.

∑_{i=0}^{k−1} s_{i} = ∑_{i=0}^{k−1} s_{i−1} + ∑_{i=0}^{k−1} s_{i−2} = ∑_{i=0}^{k−1} s_{i} + ∑_{i=0}^{k−1} s_{i}.

Equation (1) now follows. Note how we have exploited the periodicity of the loop (s_{i}).

∑_{i=0}^{k−1} s_{i}^{2} = ∑_{i=0}^{k−1}(s_{i−1} + s_{i−2})^{2}

so

∑_{i=0}^{k−1} s_{i}^{2} = 2 ∑_{i=0}^{k−1} s_{i}^{2} + 2 ∑_{i=0}^{k−1} s_{i}s_{i−1}.  (4)

Similarly, we express s_{i−2} as s_{i} − s_{i−1} to obtain

∑_{i=0}^{k−1} s_{i}^{2} = ∑_{i=0}^{k−1} s_{i−2}^{2} = 2 ∑_{i=0}^{k−1} s_{i}^{2} − 2 ∑_{i=0}^{k−1} s_{i}s_{i−1}.  (5)

Adding (4) and (5) we obtain

2 ∑_{i=0}^{k−1} s_{i}^{2} = 4 ∑_{i=0}^{k−1} s_{i}^{2}

from which the equation (2) follows unless p is 2.

∑_{i=0}^{k−1} s_{i}^{3} = ∑_{i=0}^{k−1} s_{i+1}^{3} = ∑_{i=0}^{k−1}(s_{i} + s_{i−1})^{3}

so

∑_{i=0}^{k−1} s_{i}^{3} = 2 ∑_{i=0}^{k−1} s_{i}^{3} + 3 ∑_{i=0}^{k−1} s_{i}^{2}s_{i−1} + 3 ∑_{i=0}^{k−1} s_{i}s_{i−1}^{2}.  (6)

Similarly we may write

∑_{i=0}^{k−1} s_{i}^{3} = ∑_{i=0}^{k−1} s_{i−2}^{3} = ∑_{i=0}^{k−1}(s_{i} − s_{i−1})^{3}

and so

∑_{i=0}^{k−1} s_{i}^{3} = −3 ∑_{i=0}^{k−1} s_{i}^{2}s_{i−1} + 3 ∑_{i=0}^{k−1} s_{i}s_{i−1}^{2}.  (7)

Add (6) and (7) to obtain

2 ∑_{i=0}^{k−1} s_{i}^{3} = 2 ∑_{i=0}^{k−1} s_{i}^{3} + 6 ∑_{i=0}^{k−1} s_{i}s_{i−1}^{2}

from which it follows that

3 ∑_{i=0}^{k−1} s_{i}s_{i−1}^{2} = 0.  (8)

Now we adjust (8) using s_{i} = s_{i−1} + s_{i−2} to obtain

3 ∑_{i=0}^{k−1} s_{i}^{3} + 3 ∑_{i=0}^{k−1} s_{i}^{2}s_{i−1} = 0.  (9)

From (6) and (8) we know

∑_{i=0}^{k−1} s_{i}^{3} + 3 ∑_{i=0}^{k−1} s_{i}^{2}s_{i−1} = 0.  (10)

Subtract (10) from (9) to obtain the result, since p > 2. Corollary. If p > 3 then

∑_{i=0}^{k−1} s_{i}s_{i−1} = 0,  (11)
∑_{i=0}^{k−1} s_{i}^{2}s_{i−1} = ∑_{i=0}^{k−1} s_{i}s_{i−1}^{2} = 0  (12)

Note, by the way, that ∑_{i=0}^{k−1} s_{i}^{4} is not 0 in general, but it does vanish when p is 2 or 3. DEMONSTRATION RESULT 2.The following double sum vanishes for all primes p > 3.

ν = ∑_{j=0}^{k−1} ∑_{i=0}^{j−1} s_{j}^{2}s_{j−i}

Proof. We have

ν = ∑_{j=0}^{k−1} ∑_{i=0}^{j−1} s_{j}^{2}s_{j−i+1} − ∑_{j=0}^{k−1} ∑_{i=0}^{j−1} s_{j}^{2}s_{j−i−1}

= ∑_{j=0}^{k−1} ∑_{i=−1}^{j−2} s_{j}^{2}s_{j−i} − ∑_{j=0}^{k−1} ∑_{i=1}^{j} s_{j}^{2}s_{j−i}.

We can tinker with the ranges of summation providing we only add or subtract terms which vanish, or compensate by inserting new sums. Thus

ν = (∑_{j=0}^{k−1} ∑_{i=0}^{j−1} s_{j}^{2}s_{j−i} + ∑_{j=0}^{k−1} s_{j}^{2}s_{j+1} − ∑_{j=0}^{k−1} s_{j}^{2}s_{1}) − (∑_{j=0}^{k−1} ∑_{i=0}^{j−1} s_{j}^{2}s_{j−i} − ∑_{j=0}^{k−1} s_{j}^{3} + ∑_{j=0}^{k−1} s_{j}^{2}s_{0}).

In this expression for ν, the first and fourth terms cancel, the sixth vanishes since s_{0} = 0, and the remaining terms – the second, third and fifth, vanish by equations (12), (2) and (3) respectively. Thus ν = 0 and we are done.

The general technique should now be clear. We build up an armoury of sums which vanish, gradually assembling enough material to complete the proof of Theorem C.

Parts of this work are based on the first authors doctoral thesis. He acknowledges with gratitude the financial support of Ataturk University and the Government of Turkey which made this work possible. Both authors also acknowledge helpful conversations with D.L.Johnson and many others.

[1] H. AYDIN ‘Recurrence Rations in Finite Nilpotent Groups’, Ph.D. thesis. University of Bath 1991.

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Huseyin Aydin and Geoff C. Smith School of Mathematical Sciences University of Bath Bath BA2 7AY England

Present address of the first author:

Ataturk University Fen-Ed Faculty Department of Mathematics Erzurum Turkey