Math. Proc. Camb. Phil. Soc. (1988), 104, 21 21
Printed in Great Britain
Pregroups and length functions
BY A. H. M. HOARE
Department of Mathematics, University of Birmingham
(Received 27 January 1987; revised 15 October 1987)
Pregroups were defined by Stallings[7] who showed that the elements of the group
they define have a normal form up to an equivalence called interleaving. Recently
Rimlinger[5] has shown that subject to a discreteness and a boundedness condition
any pregroup P defines a graph of groups. We show here that closer analysis of P
makes the boundedness condition superfluous. In § 1 we give results of Stallings and
Rimlinger and prove some key lemmas. In §2 we show that the discreteness condition
gives an integer-valued length function in the sense of Lyndon [4]. It follows from the
work of Chiswell [2] and Serre [6] that this defines a graph of groups. I would like to
thank the referee for his careful reading and useful comments on this paper.
1. LetP be a set. e a distinguished element of P, i:P->P an involution on P possibly
with fixed points, D a subset of P x P and m a function from D to P. We write aT
1
for i(x); we say xy is defined for (x, y)eD; and we write xy for m(x. y) whenever
(x,y)eD. We suppose that the following axioms, which are due to Stallings [7], hold
for all x, y and z in P:
PI xe = ex = x;
P2 x~
x
x = xx~
l
= e;
P4 if xy and yz are defined then x(yz) is defined if and only if (xy)z is defined, in
which case they are both equal.
PROPOSITION 1-1 [7]. If xy is defined then (xy)y"
1
is defined and is equal to x.
Proof, x = xe = x(yy~
l
) by PI and P2,
= (xy)y’
1
by P4.
The following result was assumed as an axiom by Stallings [7] and first proved by
Squicr.
PROPOSITION 1-2. // xy is defined then y~
l
x~
x
is defined and equal to (xy)~
l
.
Proof. y~
l
= ((xy)-
l
(xy))y-
1
by PI and P2,
= (xy)~
x
x by P4 and Proposition 1-1.
Therefore (xy)’
1
= (xy)’
1
(xx~
1
) by PI and P2.
= ((xy)-
1
x}x~
1
by P4,
The following transitive order relation on P is due to Stallings (see [5]).
Let L(x) = {aeP; ax is defined}. Put x ^ y if L(y) £ L(x) and x < y if L(y) c L(x)
and L(y) # L(x). Write x ~ y if L(x) = L(y).
PROPOSITION 1-3 [5]. If x ^y or y ^x then x~
l
y and y~
l
x are defined.

22 A. H. M. HOABE
Proof. By P2, y~
x
eL(y), so if x ^y then y~
x
eL(x), i.e. y~
x
x is defined. The result
follows by Proposition 1-2 and symmetry.
PROPOSITION 1-4 [7]. If’ xa and a~
l
y are defined then (za)(a~
1
y) is defined if and only
if xy is defined in which case they are equal-
Proof. By Proposition 1-1, (xa)a~
1
is defined and equal to x. The result follows by
P4 applied to xa, a~
x
and y.
Henceforth we will assume PI, P2 and P4 and Propositions 1-1. 12. 1-3, and 1-4
without explicit reference. If P4 applies we will say that xyz is defined.
THEOREM 1-5. The following conditions on P are equivalent:
P5 (i) [7]. for all w, x, y and z in P, ifwx, xy and yz are defined then wxy or xyz is
defined;
P5 (ii) [5]. for all x, y and a in P, if x~
x
a and a~
x
y are defined but x~
x
y is not defined
then a < x and a < y;
P5 (iii) [5]. for all x and y in P, if x~
x
y is defined then x ^ y or y ^ x.
Rimlinger[5] showed that (i) implies (ii) and (iii) which is all that we need here.
However we give the proof of equivalence for completeness.
Proof. Suppose (i) and the hypotheses of (ii) hold. If w. is in L(x) then by (i) applied
to ux, aT
1
, a and a~~
x
y, (ux)x~
x
a is defined, i.e. u&L(a). Thus L(x) £ L(a) but y~
l
is in
L(a)\L(x) so a < x. By symmetry a < y.
Now assume (ii) and suppose x’
x
y is defined and x ^ y. Then there exists an
element w in L(y)\L(x) so wy and y~
x
x are defined but wx is not defined. Therefore by
(ii) y <x.
Finally suppose (iii) holds and wx, xy and yz are all defined. Then by (iii) applied
to a;"
1
and y we have x~
x
^ y or y ^ x~
x
. Now (yz)’
1
is in L(y) and wx is in L(x~
x
). Thus
(yz)~
x
x~
x
is defined if aT
1
^ y, and (ivx)y is defined if y ^ aT
1
.
P is called a pregroup [7] if it satisfies the conditions P5. Henceforth suppose that
P is pregroup.
PROPOSITION 1-6 [5,7]. If x ^ z and y < z then x ^ y or y ^ x.
Proof. Under the hypotheses, x’h and z~
x
y are defined. By P5 (ii) x~
l
y is defined,
otherwise z < x and z < y. The result follows by P5 (iii).
We now prove the following key lemmas.
LEMMA 1-7. If x ^y then y~
l
x ^ y’
1
.
Proof. Suppose u is in Liy’
1
); then uy~
l
is in L(y) and hence in L(x) i.e. (uy~
l
)x is
defined, since x ^ y. Thus, since y~
l
x is defined, u(y *x) is defined, i.e. ueL(y~
l
x).
LEMMA 1-8. If a < x then a"
1
< a~
l
x and x~
l
~ x~
x
a.
Proof. Since a < x there exists an element u in L(a)\L(x). Now P5 (ii) can be
applied to (ua)~
l
, a~
l
x and a~
x
since ux, and hence (ua) (a~
1
x), is not defined. Thus
a~
x
< a~
x
x. Moreover applying Lemma 1*7 to a < x and to a~
l
^ a~
x
x we have
x~
x
a ^ x~
x
and (x~
x
a)a~
l
^ x~
x
a, i.e. x~
x
~ x~
x
a.
LEMMA 1-9. // (x~
x
y)z is defined but yz is not defined then y~
x
x < z and x ~ y.