Proceedings of the Edinburgh Mathematical Society (1987) 30, 57-67 '
LENGTH FUNCTIONS AND PREGROUPS
by I. M. CHISWELL*
(Received 10th August 1985)
The idea of a pregroup was introduced by Stallings and provides an axiomatic setting
for a well-known argument, due to van der Waerden, used to prove normal form
theorems. Details are provided in [7], Section 3.
The normal form theorem for a pregroup ([7], 3.A.4.5) gives a corresponding notion
of length on its universal group, which will be described later. In [8], Question Bl, p.
372, it was asked whether or not this length satisfied the axioms of Lyndon [4] for a
length function, the main point being whether or not Lyndon’s axiom A4 is satisfied (we
shall list all axioms used later). We show that A4 holds if and only if the pregroup
satisfies an extra axiom, called P6. This result was obtained independently and at about
the same time by Nesayef ([5], Theorem (2.10)). In Section 2 we given an example of a
pregroup not satisfying P6, so that Axiom A4 need not be satisfied by the length
associated to a pregroup.
Originally, our argument involved showing that if P6 was satisfied, then one obtained
a "normal form structure" in the sense of Hurley [3] and one could then appeal to his
results to see that A4 was satisfied. Indeed, this was how we were led to Axiom P6.
However, motivated by some of the lemmas in a recent paper by Promislow [6], we
have investigated exactly what axioms are satisfied by the length associated to a
pregroup, using arguments which are essentially contained in [1]. In the course of this
investigation, we give a fairly simple argument, closely related to some of the results in
[6], that P6 implies A4. These results are presented in Section 3. Finally, some examples
are given to clarify the logical relationships between the new axioms which are
introduced.
We should like to thank A. H. M. Hoare for several helpful conversations concerning
this paper.
2.
We consider sets with a partial multiplication, that is sets P, together with a subset D
of PxP and a mapping D->P,(x,y)r-*xy. Instead of saying that (x,y)eD, we shall say
that xy is defined.
A pregroup is a set P with a partial multiplication, together with a distinguished
element denoted by 1 (and called the identity element of P) and a map P->P,x-+x~
l
,
*This paper forms part of the Proceedings of the conference Groups-St Andrews 1985.
57

58 I. M. CHISWELL
satisfying five axioms:
(PI) for all xeP, xl and lx are defined and equal to x;
(P2) for all xeP, xx"
1
and x~*x are defined and equal to 1;
(P3) if xy is defined, then so is y~
1
x~
1
and y~
l
x~* =(xy)~
1
;
(P4) suppose xy and yz are defined; then (xy)z is defined if and only if x(yz) is defined,
in which case they are equal;
(P5) if wx, xy and yz are all defined, then either w(xy) or (xy)z is defined.
If G is a group and P is a subset of G closed under taking inverses and containing 1,
let D = {(x,y)ePx P\xyeP} the partial multiplication being obtained by restricting
multiplication on G to D. Then Axioms (PI) to (P4) are automatically satisfied.
We introduce some additional conditions on a pregroup P:
(P6) if (x,y)$D, but xa and a~
i
y are both defined, then au and ua are defined for all
ueP.
(P7) ax is defined for all xeP if and only if xa is defined for all xeP.
Proposition 1. In a pregroup P, Axiom P7 is equivalent to:
(P7’) if ax is defined for all xeP, then xa is defined for all xeP
and to
(P7") if xa is defined for all xeP, then ax is defined for all xeP.
Proof. It follows easily from PI, P2 and P4 that (x"
1
)~
1
=x for all xeP. It then
follows from P3 that (P7’) and (P7") are equivalent, so both are equivalent to P7.
Proposition 2. In a pregroup P, Axiom P6 is equivalent to:
(P6’) if(x,y)$D but xa and a~
l
y are both defined, then au is defined for all ueP
and also to
(P6") if (x,y)$D but xa and a~
1
y are both defined, then ua is defined for all ueP.
Moreover, P6 implies P7.
Proof. It suffices to show (P6’) implies (P7’) and (P6") implies (P7"). Assume (P6’),
and suppose ax is defined for all xeP. Assume va is not defined for some veP. Since
av~
l
is defined, va’
1
is defined by P3. Also, aa is defined, so by (P6’), a~
l
u is defined
for all ueP, in particular, a~
l
v~
l
is defined. By P3, va is defined, a contradiction. Hence
xa is defined for all xeP, and (P7’) holds. The proof that (P6") implies (P7") is similar.
The following equivalent statement of P6 was given by Nesayef ([5], Theorem (2.7)).
In a pregroup, P6 holds if and only if: if (x, y) $ D and (ax)y is defined, then (ax)z
and z(ax) are defined for all zeP.
The reason for this is that x(ax)"
1
is always defined, while if xa and a~
l
y are defined,
then b = a~
l
x~
l
is defined by P3 and (bx)y is defined. In view of Proposition 2, we can