%\input verbatim
\parindent = 0 pt
\parskip = 6pt
\outer\def\heading#1\endheading{\vskip0pt plus.2vsize\penalty-100
\vskip0pt plus-.2\vsize\bigskip\vskip\parskip
\message{#1}\leftline{\bf#1}\nobreak\smallskip\noindent}
\outer\def\title#1\endtitle{\centerline{\titling #1}}
\magnification=1200
\font\titling=cmr17\font\proclaiming=cmr12
\def\comment #1{\sevenbf\par\noindent\hrule\smallskip #1
\smallskip\hrule\medskip}
\def\empty {\emptyset}\let\bold=\bf
\def\Z{{\bf Z}}
\def\Q{{\bf Q}}
\def\lr{\longrightarrow}
\def\R{{\bf R}}
\def\N{{\bf N}}
\def\C{{\bf C}}
\def\m{{\bf m}}
\def\a{\beta_1}
\def\b{\beta_2}
\def\h{{\bf h}}
\def\s{{\bf s}}
\def\f{{\bf f}}
\def\g{{\bf g}} 
\def\t{\thinspace}
\def\v{\vskip .5cm} 
\def\vm{\vskip .1cm}
\def\n{\hfil \break}
\def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
         \hbox{\vrule width.#2pt height#1pt \kern#1pt
              \vrule width.#2pt}
         \hrule height.#2pt}}}}
\outer\def\proclaim#1#2\endproclaim
{\medbreak\noindent{\proclaiming#1.}
\enspace{\sl#2}\par\ifdim\lastskip<\medskipamount
\removelastskip\penalty55\medskip\fi}
\outer\def\demo#1{{\proclaiming#1}}
\def\enddemo{}
\def\square{\mathchoice\sqr68\sqr68\sqr{4.2}6\sqr{3.0}6}
\normalbaselineskip=12pt
\normalbaselines
\centerline{FINITE  p-QUOTIENTS OF SOME CYCLICALLY PRESENTED GROUPS}\v
\centerline{HUSEYIN AYDIN AND  GEOFF C. SMITH}\v
\v
\centerline{\it  1  Introduction}\vm

We define the {\it Fibonacci}  groups. These are groups
parameterized by natural numbers $r$ and $n$ and described by the
presentations
$$F(r,n)=<x_1,x_2,x_3,\cdots,x_n: x_1x_2x_3\cdots x_r=x_{r+1},
x_2x_3x_4\cdots x_{r+1}=x_{r+2},$$ $$ \cdots , x_{n-1}x_nx_1\cdots
x_{r-2}=x_{r-1}, x_nx_1x_2\cdots x_{r-1}=x_r>.$$

We shall assume $n,r>1$ to avoid triviality. These groups were introduced
following {\bf [7]}, and have been extensively studied, particularly
by D. L. Johnson and co-workers {\bf [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 {\bf [14]} and {\bf [15]} give an excellent summary of what is known.
The work in this paper is parallel to that in {\bf [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 {\bf [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
$R(2,5).$\v


In {\bf [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 {\it Wall's Number}. 

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

He also showed that the division symbol may be replaced by equality
providing $k(p) \not = k(p^2).$ 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
$\pi: F(2,k(p)) \lr \Z/p\Z$ with $x_1\pi = 0$ and $x_2\pi = 1,$ then there
exists a group homomorphism 
$\overline \pi: F(2,k(p)p^{u-1}) \lr \Z/p^u\Z$ with $x_1\overline \pi = 0$
and $x_2\overline \pi = 1.$

Notice that it is possible to have epimorphisms from $F(2,n)$ to
$\Z/p\Z$ which are not of this form. Consider the map 
$\psi: F(2,4) \lr \Z/5\Z$ such that $x_1\psi = 1$ and $x_2 \psi = 3.$
The point is that $(1,3,4,2)$ is a Fibonacci sequence in $\Z/5\Z$
containing no 0. Such loops must have adjacent terms in a golden ratio.
Such loops are studied elsewhere {\bf [1] [2] [4] [16]} and {\bf [17]}.\vm


\centerline{\it 2.   Notation }\vm 

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 $g_0,g_1 \in G.$ Let $k(g_0,g_1)$ denote
the length of a fundamental period of the Fibonacci sequence in $G$
defined by $g_ig_{i+1} = g_{i+2},$ and put $k(G) = lcm\{ k(g_0,g_1)\ |
g_0,g_1 \in G\}.$


One expects the Fibonacci recurrence to be slightly special; we set
up a general linear recurrence.
Let $G$ be a group and let
$$x_i = x_{i-v}^{a_v}x_{i-v+1}^{a_{v-1}}\ldots x_{i-1}^{a_1} \eqno (*)$$
be a $v-$step linear recurrence which is invertible in the group;
this means that given $g_{i-v+1}, \ldots , g_i \in G$ there exists
a unique $g_{i-v} \in G$ satisfying the recurrence. We say that
$(*)$ is a {\it good} recurrence for $G$i if $p \not |\ a_v.$
For example. if $G$ is a finite
$p-$group then (1) is a good recurrence if $p \not | a_t.$ The
$r-$step Fibonacci recurrence is good in any finite group.

A sequence of (finite) group elements satisfying the recurrence is
called a {\it 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
$\alpha$ and obtain another loop $(g_{i+\alpha}).$ We call this process
{\it 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 $\Z/p\Z$ beginning
$0,0,0,\ldots 0,1.$
Notice that any loop of $(*)$ in  $\Z/p\Z$
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 $p.$ \v

\centerline{\it 3. Results }\vm
THEOREM A.  {\it If $G$ is a finite $p-$group, then $k(G) \ | k(p)p^{u-1}$
where $u$ denotes  $p-$class of $G.$}\v

One can calculate that $k(5) = 20,$  so Theorem A tells us that
$k(R(2,5))$ divides $20\cdot 5^{11},$ 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.\vm

THEOREM  B.  {\it Suppose $G$ is a finite $p-$group of $p-$class 
$u$ then $k_*(G)\ | k_*(p)p^{u-1}.$}\vm
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.\vm

THEOREM C.  {\it 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 $k(G) = k(p).$} \v

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.

\centerline {\it 4.  Proofs}\vm
{\it 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 $\Z/p\Z.$ 
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 $\lambda = k_*(p)p^{u-2}.$
By induction we may assume that
$k_*(G/N)\ | \ \lambda.$
Pick a transversal $T$ for $N$ in $G.$
Suppose that for each $i$, we have $g_i = t_i n_i$ where 
$t_i \in T$ and $n_i \in N.$
Notice that $(t_i)$ is a periodic sequence of period dividing $\lambda $
though it is not in general a loop.

Consider the loop $(m_i)$ such that 
$m_i = n_{\lambda + i}^{-1}n_i$ for
$0 \le i \le v-1.$ 
The loop $(m_i)$ consists of central elements of $G.$ 

Let $(g^+_i) = (g_{i+\lambda})$ be the rotation of $(g_i)$ through
$\lambda. $ Now form the co-ordinatewise product loop
$(g^+_i)\cdot (m_i).$ This is still a loop (because $(m_i)$ is central).
Notice that $(g^+_i)\cdot (m_i)$ and $(g_i)$ co-incide in the first
$v$ places, and so $(g^+_i)\cdot (m_i) = (g_i).$
Thus we may equate entries. Notice that $(m_i)$ has fundamental
period dividing $k_*,$ and hence dividing $\lambda.$

For all $j$ and for all $i$ in the range $0\le i \le v-1$ we have
$t_in_{j\lambda+i} = t_in_{(j+1)\lambda + i}m_i.$
Thus $n_{j\lambda+i}n_{(j+1)\lambda + i}^{-1} = m_i$
is independent of $j,$ and so $n_{p\lambda + i} = m_i^{-p}n_i
= n_i$ for all $i$ in $0\le i \le v-1.$ 
We conclude that $(g_i)$ has period dividing
$p\lambda$ and we are done.\vm 
{\it 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 $x^ay^bz^ct^du^e$ where
$a,b,c,d,e \in \Z.$ 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 $(\Z/p\Z)^5,$
and obtain a group denoted $H^{\Z/p\Z}$ 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 $x^ay^bz^ct^du^e.$\v We have
$$(a,b,c,d,e) \cdot (a',b',c',d',e') = (a'',b'',c'',d'',e'')$$
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 $\Z/p\Z.$ 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 $n \in \N.$
We summarize the results. In what follows, we use an unadorned
$\sum$ 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.


$$ z_n = \sum_{i= 0}^{n-1} s_{n-i-1}s_i^2. $$

So

$$ z_{k} = \sum_{i=0}^k s_{k-i-1}s_i^2 = \sum_{i=0}^{k-1} s_{k-i-1}s_i^2
 \sum  s_{k-i-1}s_i^2.$$
Similarly
$$z_{k+1} = \sum_{i=0}^k s_{k-i}s_i^2 = \sum s_{k-i}s_i^2
= \sum s_i^3(-1)^{i+1}.$$
Next, we find that the formula for $t_n$ is
$$ t_n = \sum_{j= 0}^{n-1} s_{n-1-j}s_jz_j
 + \sum_{i=0}^{n-1} (_2^{s_i})s_is_{n-i-1}.$$

In particular,
$$ t_{k+1} = \sum_{j= 0}^k s_{k-j}s_j \sum_{i=0}^{j-1} s_i^2 s_{j-i-1}
+ \sum_{i=0}^k (_2^{s_i})s_is_{k-i}$$
which, after cosmetic treatment, becomes
$$ t_{k+1} = \sum_{j=0}^{k-1}\sum_{i=0}^{j-1} s_j^2s_i^2s_{j-i-1}(-1)^{j+1}
+ \sum_{i=0}^k (_2^{s_i})s_i^2(-1)^{i+1}.$$
Similarly we find that 
$$t_k = \sum_{j=0}^{k-1}s_{k-1-j}s_j\sum_{i=0}^{j-1} s_i^2s_{j-i-1}
+ \sum_{i=0}^k (_2^{s_i})s_is_{i+1}(-1)^i.$$
Finally, the expression for $u_n$ is 
$$u_n = \sum_{r = 0}^{n-1} s_{n-r-1}s_rt_r
       + \sum_{j=0}^{n-1} s_{n-j-1}(_2^{s_j})z_j
       + \sum_{i = 0}^{n-1} s_{n-i-1} (_3^{s_i}) s_i,$$

which expands to

$$u_n = \sum_{r=0}^{n-1}\sum_{j=0}^{r-1}\sum_{i=0}^{j-1}
        s_{n-r-1}s_rs_{r-j-1}s_js_{j-i-1}s_i^2
+ \sum_{j=0}^{n-1}\sum_{i=0}^{j-1}
        s_{n-j-1}s_j(_2^{s_i})s_is_{j-i-1}$$
$$+\sum_{j=0}^{n-1}\sum_{i=0}^{j-1}s_{n-j-1}(_2^{s_j})
        s_{j-1-i}s_i^2
       +\sum_{i=0}^{n-1} s_{n-i-1}(_3^{s_i})s_i.$$
In particular we have
$$u_{k+1} = \sum_{r=0}^k\sum_{j=0}^{r-1}\sum_{i=0}^{j-1}
            s_r^2s_js_i^2s_{r-j-1}s_{j-i-1}(-1)^{r+1}
+\sum_{j=0}^{k}\sum_{i=0}^{j-1}
        s_{k-j}s_j(_2^{s_i})s_is_{j-i-1} $$
$$+ \sum_{j=0}^k\sum_{i=0}^{j-1}s_j(_2^{s_j})s_{j-i-1}s_i^2(-1)^{j+1}
+ \sum_{i=0}^k s_i^2(_3^{s_i})(-1)^{i+1}$$
and
$$u_k = \sum_{r=0}^k\sum_{j=0}^{r-1}\sum_{i=0}^{j-1} 
s_{k-1-r}s_rs_{r-j-1}s_js_{j-i-1}s_i^2
+ \sum_{j=0}^{k-1}\sum_{i=0}^{j-1}
        s_{k-j-1}s_j(_2^{s_i})s_is_{j-i-1}$$
$$+ \sum_{j=0}^k\sum_{i=0}^{j-1} s_{k-j-1}(_2^{s_j})s_{j-i-1}s_i^2
+ \sum_{i=0}^k s_{k-i-1}(_3^{s_i})s_i.$$

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 ${\bf [4]}$.
 
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 torsion-free finitely generated nilpotent groups, and refer him
or her to {\bf[8]} or {\bf [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 {\bf [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 {\it Fibonacci
automorphism } $\phi : \overline T \longrightarrow \overline T$ defined
by $x\phi = y$ and $y\phi = xy.$ Let $M = <(y,x,y)>^{\overline T},$ a normal
subgroup of order $p^3.$ A routine calculation shows that
$M\phi  \cap M \cap M\phi^{-1} = 1.$ 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 $\overline T
\longrightarrow {\overline T}/M\phi^i$ for each $i,$ and these induce
an embedding
$$\theta : \overline T \longrightarrow {\overline T}/M\phi \times {\overline T}/
M
\times {\overline T}/M\phi^{-1}.$$
Now $Im(\overline T)$  is a subdirect product in
${\overline T}/M\phi \times {\overline T}/M
\times {\overline T}/M\phi^{-1}.$
Each group $\overline T/M\phi^i$ is a copy
of the group $H$ studied in this chapter.

When we compute the Fibonacci sequence in
$\overline T$ the images in ${\overline T}/M\phi \times {\overline T}/M
\times {\overline T}/M\phi^{-1}$
of the first four terms are as follows.

$$ (x(M\phi) , xM, x(M\phi^{-1})) \eqno (a)$$
$$ (y(M\phi)
 , yM, y(M\phi^{-1})) \eqno (b)$$
$$ (xy(M\phi), xyM, xy(M\phi^{-1})) \eqno (c)$$
$$ (yxy(M\phi),yxyM,yxy(M\phi^{-1})) \eqno (d)$$

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
$(xM)\phi = y(M\phi)$ and $(yM)\phi = xy(M\phi)$
yielding a rotation by 1 place in the loop of the first co-ordinate.
Similarly $\phi^{-1}$
induces a rotation through 1 place
in the opposite direction in the third co-ordinate.
Thus the loop beginning $(x,y,\ldots)$
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 $4.$ 


{\it Observations.}  The final ruse in the proof of Theorem B was used 
because we did not
know that {\it  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 $\overline H.$\v

\centerline{\it 5.  Methods}\vm
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.
$$s_{-j} = s_j(-1)^i$$
$$s_j^2 = s_{j+1}^2 + s_{j-1}^2 - 2s_j^2 + 2(-1)^{j+1}$$
$$3s_j^2 = s_{j+1}^2 + s_{j-1}^2 + 2(-1)^{j+1}$$
$$6s_js_{j-1}^2 = s_{j+1}^3 + s_{j-2}^3 - 2s_j^3$$
$$6s_j^2s_{j-1} = s_{j+1}^3 - s_{j-2}^3 - 2s_{j-1}^3$$

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. 

$$\sum_{i=0}^{k-1}  s_j = \sum_{i=0}^{k-1} s_{k-j}$$
$$  \sum_{i=0}^{k-1} s_{j+a} =  \sum_{i=0}^{k-1} s_j$$

We will do some example calculations in order to 
demonstrate the techniques
involved. We choose these results almost arbitrarily.\v
DEMONSTRATION RESULT 1 {\it The following equations concerning sums of powers 
of
Fibonacci numbers
over a fundamental range all hold good. 

$$ \sum_{i=0}^{k-1} s_i = 0.\eqno (1)$$

$$ \sum_{i=0}^{k-1} s_i^2 = 0.\eqno (2)$$

$$  \sum_{i=0}^{k-1} s_i^3 = 0. \eqno (3)$$}

{\it Proof.} The results are true by inspection when
$p = 2$ so we may assume $p>2.$
$$\sum_{i=0}^{k-1} s_i = \sum_{i=0}^{k-1}s_{i-1} + 
\sum_{i=0}^{k-1}  s_{i-2} = \sum_{i=0}^{k-1}s_i + \sum_{i=0}^{k-1} s_i.$$
Equation (1) now follows. Note how we have exploited the periodicity
of the loop $(s_i).$
$$ \sum_{i=0}^{k-1} s_i^2 =  \sum_{i=0}^{k-1} (s_{i-1} + s_{i-2})^2$$
so

$$ \sum_{i=0}^{k-1} s_i^2 = 2 \sum_{i=0}^{k-1}s_i^2 + 2 
\sum_{i=0}^{k-1} s_is_{i-1}. \eqno (4)$$

Similarly, we express $s_{i-2}$ as $s_i - s_{i-1}$ to obtain

$$ \sum_{i=0}^{k-1}  s_i^2 = 
 \sum_{i=0}^{k-1} s_{i-2}^2 = 2 \sum_{i=0}^{k-1}s_i^2 - 
2 \sum_{i=0}^{k-1} s_is_{i-1}. \eqno (5)$$

Adding (4) and (5) we obtain
$$\ 2 \sum_{i=0}^{k-1} s_i^2 = 4 \sum_{i=0}^{k-1} s_i^2$$
from which the equation (2) follows unless $p$ is 2.\v
$$  \sum_{i=0}^{k-1} s_i^3 =  \sum_{i=0}^{k-1}s_{i+1}^3 =  
 \sum_{i=0}^{k-1}(s_i + s_{i-1})^3 $$
so
$$  \sum_{i=0}^{k-1} s_i^3 = 2 \sum_{i=0}^{k-1} s_i^3 + 
3 \sum_{i=0}^{k-1} s_i^2s_{i-1} + 3 \sum_{i=0}^{k-1} s_is_{i-1}^2. \eqno
(6)$$
Similarly we may write
$$  \sum_{i=0}^{k-1} s_i^3 =  \sum_{i=0}^{k-1} s_{i-2}^3 =  
\sum_{i=0}^{k-1} (s_i - s_{i-1})^3$$
and so
$$  \sum_{i=0}^{k-1} s_i^3 = -3 \sum_{i=0}^{k-1} s_i^2s_{i-1} + 
3 \sum_{i=0}^{k-1} s_is_{i-1}^2. \eqno (7)$$
Add $(6)$ and $(7)$ to obtain

$$ 2 \sum_{i=0}^{k-1} s_i^3 = 2 \sum_{i=0}^{k-1}s_i^3 + 
6 \sum_{i=0}^{k-1} s_is_{i-1}^2$$

from which it follows that

$$ 3 \sum_{i=0}^{k-1} s_is_{i-1}^2 = 0.\eqno (8)$$

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

$$ 3 \sum_{i=0}^{k-1} s_i^3 + 3 \sum_{i=0}^{k-1} 
s_i^2 s_{i-1} = 0. \eqno (9)$$

From $(6)$ and $(8)$ we know

$$  \sum_{i=0}^{k-1} s_i^3 + 3 \sum_{i=0}^{k-1}s_i^2 s_{i-1} = 0. \eqno (10)$$
Subtract $(10)$ from $(9)$ to obtain the result, since $p>2$.\v

{\it Corollary.}  If $p>3$ then\vm
$$ \sum_{i=0}^{k-1}s_is_{i-1} = 0, \eqno(11)$$
$$ \sum_{i=0}^{k-1} s_i^2s_{i-1} =  \sum_{i=0}^{k-1}
s_is_{i-1}^2 = 0 \eqno(12)$$\vm
Note, by the way, that $\sum_{i=0}^{k-1}  s_i^4$ is not 0 in general, but it 
does vanish
when $p$ is $2$ or $3.$


DEMONSTRATION RESULT 2.{\it The following double sum vanishes for all
primes $p > 3.$
$$\nu = \sum_{j=0}^{k-1}\sum_{i=0}^{j-1}s_j^2s_{j-i}
$$}
{\it Proof.} We have
$$ \nu= \sum_{j=0}^{k-1}\sum_{i=0}^{j-1} s_j^2s_{j-i+1}
   - \sum_{j=0}^{k-1}\sum_{i=0}^{j-1} s_j^2s_{j-i-1}$$
$$ = \sum_{j=0}^{k-1}\sum_{i= -1}^{j-2} s_j^2s_{j-i}
   - \sum_{j=0}^{k-1}\sum_{i= 1}^j s_j^2s_{j-i}.$$
We can tinker with the ranges of summation providing we only add or
subtract terms which vanish, or compensate by inserting new sums. Thus
$$ \nu = (\sum_{j=0}^{k-1}\sum_{i=0}^{j-1} s_j^2s_{j-i}
         +\sum_{j=0}^{k-1} s_j^2 s_{j+1}
         -\sum_{j=0}^{k-1} s_j^2 s_1)
         - (\sum_{j=0}^{k-1}\sum_{i=0}^{j-1}s_j^2s_{j-i}
           -\sum_{j=0}^{k-1}s_j^3
           +\sum_{j=0}^{k-1} s_j^2s_0).$$
In this expression for $\nu,$ 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 $\nu = 0$
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.\vm

{\bf [1]} H. AYDIN  `Recurrence Rations in Finite Nilpotent Groups', 
Ph.D. thesis.
University of Bath 1991.\n
{\bf [2]} H. AYDIN, R. DIKICI and G. C. SMITH
`Wall and Vinson revisited' , Proceedings of the
Fifth International Conference on Fibonacci
Numbers and Their Applications (1992){\it (to appear).}\n 
{\bf [3]} H. AYDIN and G.C. SMITH 
`Fourier Analysis in Finite Nilpotent Groups',
Proceedings of the Fifth International Conference on Fibonacci
Numbers and Their Applications 1992 {\it (to appear).}\n
{\bf [4]} H. AYDIN and G.C. SMITH, `Remarks on Fibonacci Sequences in Groups 1',
Bath Mathematics and Computer Science Technical Report 91-50, 
University of Bath 1991.\n
{\bf [5]} C. M. CAMPBELL, H. DOOSTIE and E. F. ROBERTSON,
`Fibonacci Length of Generating Pairs in Groups', Applications of
Fibonacci Numbers, ed. G.E. Bergum, A.N. Philippou and A.F. Horada
Kluwer vol.3 (1990), 27-35.\n
{\bf [6]} J. CANNON, `A Language for Group Theory',
 Department of Pure Mathematics University of Sydney (1982), Australia.\n
{\bf [7]} J. H. CONWAY, `Advanced Problem 5327', {\it Amer. Math.
Monthly} 72 (1967), 915.\n
{\bf [8]} P. HALL, `The Edmonton Notes on Nilpotent Groups',
Queen Mary College Mathematics Notes (1979).\n
{\bf [9]} D. L. JOHNSON and  R.W.K. ODONI  `Some 
results on symmetrically presented groups' {\it (in preparation)}.\n
{\bf [10]} D. L. JOHNSON, `A Note  on the Fibonacci Groups',
{\it Israel J. Math.} 17 (1974), 277-282.\n
{\bf [11]} D. L. JOHNSON, J.W. WAMSLEY and D. WRIGHT,
`The Fibonacci Groups', {\it Proc. Lond. Math. Soc.}
(3) 29, (1974), 577 - 592.\n
{\bf [12]} W. MAGNUS, A. KARRASS and D. SOLITAR,
`Combinatorial Group Theory', Dover (1966).\n
{\bf [13]} D. SEGAL, `Polycyclic Groups ', C.U.P. (1983)\n
{\bf [14]} R. M. THOMAS, `The Fibonacci Groups - a Survey',
Technical Report 23, Department of Computing Studies, University of
Leicester (May 1989).\n
{\bf [15]} R. M. THOMAS, `The Fibonacci Groups Revisited', 
in C. M. CAMPBELL and E. F. ROBERTSON
(eds.), Proceedings of Groups St. Andrews 1989 vol. 2, LMS Lecture
Note Series 160, CUP, (1991) 445 -454.\n
{\bf [16]} J. VINSON,` The Relations of the Period
Modulo m to the Rank of Apparition of m in the Fibonacci Sequence',
{\it The Fibonacci Quarterly} 1 (1963), 37-45.\n
{\bf [17]} D. D. WALL, `Fibonacci Series Modulo m',
{\it Amer. Math. Monthly} 67 (1960), 525-532.\vm

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

{\it Present address of the first author:}\vm
Ataturk University\n
Fen-Ed Faculty\n
Department of Mathematics\n
Erzurum\n
Turkey\vm

\end
