TRANSACTIONS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 174, December 1972
REAL LENGTH FUNCTIONSIN GROUPS
BY
NANCY HARRISON( !)
ABSTRACT.
a 'length'
erties
that we
on groups with
which assigns
lowing axioms
This paper is a study of the structure
function
from
G to the nonnegative
G equipped
The proprequire
this function
to satisfy
are derived
from Lyndon's
work
integer-valued
functions.
A real length function
is a function
to each g í G a nonnegative
real number
|g| such that the folare satisfied:
with
A0.
< \xx\ if x 4 L
Ar
= 0 if and only if x = 1.
of a group
real numbers.
A,.
x
= \x\
A,.
c(x, y) > 0 where c(x, y) = '/2(|x| + |y| - \xy~ |).
A,.
c(x, y) > m and c(y, z) > m imply c(x, z) > m.
In this paper structure
theorems
are obtained
for the cases
when G is abelian
and when G can be generated
by two elements.
We first prove that if G is
abelian,
then G is isomorphic
to a subgroup
of the additive
group of the real
numbers.
Then we introduce
a reduction
process
based on a generalized
notion
of Nielsen
elements
is either
transformation.
of G. We prove
free or abelian.
1. Introduction.
arguments
this
used
Lyndon
in Nielsen's
work he developed
to the nonnegative
such a function
a set of properties
groups
This paper discusses
with
length
into the nonnegative
when
.
amount
G is abelian
G be an arbitrary
nonnegative
y
real
real
Intuitively
number
speaking
of agreement
Received
functions
whose
ranges
theorems
and when
G can be generated
group with a function
this
Define
which
c from
ends
with
of what can be said
than
the
with a length
for G are obtained
inte-
function
for the
by two elements.
assigns
to each
c(x, y) = Y2(\x\ + \y\ - |xy |) where
function
on the right-hand
from a group
are more general
Structure
From
for any group
then arises
numbers.
|x|.
function
G which are equipped
cancellation
for free groups.
and sufficient
The question
groups
certain
theorem
for a 'length'
that are necessary
to be a free group.
about
Let
[l] did some work on isolating
proof of the subgroup
integers
gers.
cases
We apply this reduction
process
to finite sets of
that if G can be generated
by two elements,
then G
G x G to the reals
of x and y.
We require
x £ G a
y means
measures
the
that the
by the editors November 12, 1970.
AUS (MOS)subject classifications (1970). Primary 20E99, 20F15.
Key words and phrases.
Nielsen
transformation,
groups with functions
to reals.
(1) This paper is a portion of a doctoral
dissertation
written at the University
of
Michigan.
The author expresses
her gratitude
to Professor
Roger C. Lyndon for his
guidance
and encouragement.
Also she expresses
appreciation
to the University
of Missouri
for financial
support
for the preparation
of this manuscript.
Copyright © 1973, American Mathematical Society
77
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7S
NANCY HARRISON
following
axioms
iDecember
be satisfied:
A0. |x| < |x2| if x4 1.
A j. |x| = 0 if and only if x = 1.
A2.
|x | = |x|.
Ay
c(x, y) > 0.
A/.
c(x, y) > m and
Before
proceeding
c(x, z) > m imply
we will discuss
cAy, z) > m.
briefly
some
examples
of groups
with
real
length functions.
Proposition
1.1.
of real numbers
Proof.
and
Absolute
value
is a length
junction
on the additive
group
R.
The only property
z be real numbers
such
that
requires
any work to verify
is AA.
Let
x, y,
that
(1) <Ax, y) > m and
(2) cAx, z) > m.
If the signs
of x and
y or the signs
of x and
z are opposite
follows
by A,.
have
the same sign.
Since
both cases
are very similar,
we only present
tails
for the case
their
are all positive.
Note that when
ate both positive
minimum
in which
real numbers
\y, z\ < x.
Then
(1) implies
x > m.
Proposition
a real
length
• • • y,
function
is in normal
• roof.
the normal
or else
1.2.
they
For
form.
For
is a length
x in the form
are from the same
A
that
either
factor
and
holds.
function
ax'ä
c(y
Axioms
z > m.
= Z.. __ \y.\
a and
b
that
If y < z then
Sy, z\.
G. is a group
where
with
x = y,y2
for G.
a, x'
£ G and
y
are from different
J m
, y ,) = 0.
A.
the de-
x < minimum
each
where
y , and
J l
first
z all
ly, z\ > x > 772.
where
||x||
then
x, y, and
Suppose
suppose
z) = minimum
x in G define
This
x £ G write
to see that
c(y,
Secondly
G = G, * G2 * . . . * G
\ \.
that
y > m and (2) implies
c(y, z) = z.
Therefore
Let
assume
b) = minimum ¡a, b\.
(1) implies
form of x' , is such
form it is easy
signs
c(a,
c(y, z) = y and if z < y then
Then
We may then
one another,
772= 0 and the result
and
A,
From writing
are obvious.
yiy2
■• • y
factors
x in this
If b and
r both lie in Gk, then |èc| < |¿>| + |c| by A, in G . From this it follows that
A^ is true
the element
in G.
r from
It remains
to verify
A4.
For this
proof
G, .
Let
y = (sv
j1)(s2,
j2) ...
(sß,
Z = (tv kx)(t2, k2) ••• (/
jß),
ky).
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
(r, k) is used
to denote
1972]
Let
REAL LENGTH FUNCTIONS IN GROUPS
p j be such
4 ipv
that
r^ = s^
and
i
=7
Without
loss
c(y, z) > minimum lc(x,
of generality
17< p , and r? = 7
c(y, z)>2:Pl,_1|S.|
'
—
;= i
Secondly
'
11
suppose
=SM"1|r.|
z=l
pl<p2,
ip2 = kp2 in which
case
Proposition
A
\r(\ +c(spi,
in Gp..
7pi).
verification
for
sp ,).
z) = S.2,
Finally
|r.| + c(rp2> tp2).
Axioms
of groups
p.
A
b) < minimum 11«|, \b\\.
\s ] + c(sp ., t p .) =
7p x) > minimum \c(rpi,
lc(x,
when
\s.\ = c(y,z).
for a, b £ G, c(a,
c(y, z) > minimum
If
|*-| +
We turn now to the case
\r.\ = X/Xi
c(x,
that the free product
|r¿| + c(rM,
It follows
follows that c(y,
c(y, z) = 2
2._,
2.1) that
Now c(spi,
Therefore
Now 7=5
for q < p ^ . It follows
By this and the fact that jp l = kp ,, c(y, 2) > 2._.
2Jxj
that
A
|r.|.
s=T
7p . ) = c(x, y).
= p 2. If ip 2 4 kp ., then c(x, z) = £ ._,
suppose
We show
to establish
c(x, y) ^ S¡=1
pj <p2, then rpi = /pjb, but rpj
rfi , ^
¡¿ spy
S61.
and A , imply (see
p , <pT
rp2/-lp2
'
Then
+ cisp.,
or ip {
= c(x, y).
1 11
zp t = ;Pl.
pP 1 - 11|r.|
c(sp j, í pj) = 2._,
that
suffices
Then c(x, y) = S.Jf
for £7< p 2_ Since
rP} ¿spi
for q < p 2 but either
assume
y), c(x, z)} which
First suppose that ipi4jPl-
that
for q < p { but either
Let p2 be such that r q = 1q and i = k
or Z£,2 ¡¿ ¿p2.
79
spi),
c(rp y TPi)\
This
completes
y), c(x, z)j.
with length
functions
by
the
has a natural
length function.
Let
G be a group
ces of the
axioms.
isomorphic
section
to a subgroup
we introduce
the structure
either
free
2.
with a length
In the third
of G generated
In §2 we present
we prove
of the additive
a reduction
group
process
that
of real
for finite
by two elements
some consequen-
if G is abelian,
numbers.
sets
G is
In the fourth
of elements
is determined
then
in §5;
from
such
G.
And
a G is
or abelian.
Consequences
the axioms
function.
section
of the axioms.
for a length
function.
We want
First
to derive
we consider
some
consequences
an alternative
of
statement
of
A4;
c(x, y) < c(x, z) implies
First
assume
c(x, y) by
the original
A
If c(y,
c(y, z) > c(x, y).
tradiction.
A'
assume
r(x,
lc(x,
A'
y) = c(y, z).
y) < c(x, z).
Suppose
Then
c(y,
on the contrary
z) >
that
z), c(y, z)\ > c(x, y), a con-
the new statement,
y), c(x,
And if c(y, z) < c(x, y), then
which
for now we shall
call
z)\ > 777. If c(y, z) > c(x, y), we are done.
implies
r(y,
z) = c(x, z) > 772. Therefore
the
A . and A . are equivalent.
We now state
using
and that
z) = c(x, y), we are done.
minimum \c(x,
two statements
c(x,
By A , c(x, y) > minimum
Conversely
and that
form of A
that
some
immediate
consequences
of the axioms
throughout.
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that we will be
80
NANCY HARRISON
Proposition
Proof.
2.1.
c(x, y) < minimum ¡|x|,
By symmetry
it suffices
[December
|y||.
to show the result
implies that c(xy, y) > 0. It follows
for |x| > |y|.
Axiom A,
by A that
c(x, y) = V2(\x\ + \y\ - \xy\) < Y2(\x\ + \y\ + \y\ - |x|) = \y\.
Proposition
2.2.
c(xy, y) = |y| - c(x, y) [l,
Proposition
2.3.
c(x, y) + c(y, z) > |y|
[1, Proposition
Proposition
2.4.
2.5.
\xyz\ < |x| - |y| + |z|
c(x, y) + c(y, z) < \y\ implies
that
c(xy,
// |x| = |y|
2.6.
and
c(x, y) + c(x, y) > |x|,
If d = c(x, x), then
nonnegative
It follows
reduced
that
z) = c(y, z) [1,
then
x = y [l,
2.4, /). 211].
Proposition
£ G such
implies
2.3, p. 211].
Proposition
arbitrary
2.1, p. 210].
2.2, p. 211].
Proposition
Proposition
Proposition
Proposition
Corollary
| = |x| + n(\x\ - 2d) where
n is an
2.5, p. 212].
that the lengths
c(x, x) = 0, we say that
x, |x"|
|x"|
x is cyclically
are unbounded.
reduced.
If x
For cyclically
=w|x|.
2.7.
Since
= 0 by A
[l,
from this proposition
that
Proof.
integer
\xn+
Every
x £ G, x 4 1, ¿as
x 4 1> |x| > 0 by A,.
But then
|x"|
< |x|,
infinite
order.
If x has finite
contradicting
order,
Proposition
say
2.6.
72, then
Hence
|x"|
the order
of x is infinite.
Finally
from
we need to establish
G to the nonnegative
requires
a number
is a key factor
generated
reals
of case
distinctions,
in one case
proof
for the remainder
Theorem
These
f(x) = |x | - |x|
The proof is long and
but it is not difficult.
following,
frequently.
of the paper
2.8.
For arbitrary
Since
We remark that this
of the structure
of a group
Consider
A2 and its immediate
constant
referral
we will use them without
G
c(g, x) and
is trivial.
Henceforth
c(g, x).
1. c(g, x) = 0 and c(g, x) = 0.
equations
are equivalent
to
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consequence,
is burdensome,
specific
g, x £ G, \x2\ - \x\ = \gx2g\
If g or x is 1, the equality
x 4 1.
Case
function.
of the determination
and most of those
c(x, y) = c(y, x), are used
1 and
that the function
by a pair of elements.
In this
Proof.
the fact
is a class
mention.
- \gxg\.
assume
that
g 4
81
REAL LENGTH FUNCTIONS IN GROUPS
1972]
If c(gx,
g) 4 0, then
|gx| = |g| - |x|.
c(g, x) < c(g,
But this
gx).
contradicts
Axiom
A . then
(1) since
implies
c(x,
|x| > 0 by Aj.
gx) = 0 so
Therefore
c(gx,
g)
= 0 and
(ii>
l«**l= 2|Ä|+ |*|.
If c(g, x ) 4 0, then
|x |=0.
But this
c(g, x) < c(g, x ).
contradicts
Axiom A
AQ since
(Ü,)
implies
|x| > 0 by A
c(x, x ) = 0, hence
Therefore
c(g, x ) = 0 so
lg*2| = |g| + l*2|.
If c(gx , g) 4 0, then c(g, x) < c(g, gx ); so c(x, gx ) = 0 by A4. Using (iij)
and (1) it follows
fore
c(gx
that
|x | = 0, a contradiction
, g) = 0, and using
(iiil}
|x| > 0 by A
There-
(iij)
|g*2g|=2|g|
Then (i.) and (iii,)
of AQ since
+ |x2|.
imply
lg*2g|- lg*g| = |*2| - |*|.
Case 2. c(g, x) 4 0 and c(g, x) - 0.
Then
(2)
Since
|g*l= |g| + l*|¿|g*|.
c(g, x) < c(g, x),
A
implies
(V
|x2|-|x|
Consider
c(gx,
First
g).
assume
We will treat
c(gx,
= |x|.
the zero and nonzero
separately.
\gXg\ = \gx\ + \g\.
c(g,
gx) = 0 < c(g, x) implies
(i2a)
by A , that
!gx2|
Consider
c(gx
, g).
that
|gx g| - |gxg|
sult.
Next assume
First
= |x|.
assume
Since
= |gx|
c(gx
|x
c(g, gx ) 4 0.
c(gx,
, g) = 0.
| - |x| = |x|
It is easy
c(gx, gx) > 0 = c(g, gx), A4 implies
but from (2a),
a contradiction.
of (*?), this
Secondly
From
= |gx| + |g|.
from (¡2
) and (2a)
is our desired
c(gx,
c(g, gx) = 0.
Since
c(gx, gx) = 0, (2), and (2a),
is our desired
assume
It follows
by (*2), this
to see that
the contrary,
|gxg|
x) = 0, so
+ |x|.
= |g| + |gx|,
A,.4
cases
g) = 0, so
(2a)
Now
c(x, x) = 0 and
gx)
= 0.
From this
re-
If, on
|gxg|
|gx| 4 \gx\ by (2), we have
|gx g| - |gxg| = |x|.
In view
result.
c(gx, g) 4 0.
Since
c(g, x) = 0 < c(g, gx), c(x, gx) = 0 by
Hence
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82
NANCY HARRISON
(i2b)
1**1= lg|-1*1-
It is easy
to see that
contradiction.
It follows
A4
that
it follows
[gx\ = |gx|,
c(gx,
that
g) = c(gx, gx).
but this
gx) = c(gx,
gx).
Using
the contrary
A
c(gx,
implies
contradicts
g), hence
(2),
We assume
gx), then
[gx g\ - \gxg\ = - \x\.
g) - c(gx,
desired
c(gx,
If c(gx, g) < c(gx,
implies
c(gx,
LDecember
(2).
and derive
g) = c(g, gx).
If c(gx,
gx) < c(gx,
|gx| - [gx g\ = |g| - [gxg\.
But this
contradicts
[gx g\ - |gxg|
= |x|
a
g), then
Using
(i2fc)
AQ. Therefore
which
in view
of (*2) is our
result.
Case 3. c(g, x) = 0 and c(g, x) ^ 0.
This
follows
immediately
from Case
2.
Case 4. c(g, x) 4 0 and c(g, x) 4 0.
We will treat
separately
the three
possiblilites
that arise
in comparing
c(g, x) and c(g, x).
(4a) c(x, g) = c(x, g).
Axiom
A. implies
\x\ implying
c(x, x) > c(x, g) so Vi |x| > c(x, g).
by Proposition
2.4 that
(*4a}
c(x
|x2|-|x|
Also
c(x, x) + c(x, g) <
Thus
= |x2g|-|xg|.
c(g, x) + c(x, g) < \x\ so Proposition
('Uj
Then
, g) = c(x, g).
2.4 implies
c(gx,
g) = c(x, g).
Thus
\gxg\ = 2\xg\-\x\.
Next we need to examine
c(g, x) <Vi |x|,
c(x, x).
[gx g\; we claim
c(g, x) + c(x, x) < |x|,
It follows
that
c(]>x , ~g) = c(x, g).
c(gx,
that
c(g, x g) = c(g, x).
so Proposition
2.4 implies
x) + c(x, g) < \x\; Proposition
Since
c(gx,
2.4 then
x) =
implies
Thus
(ü4a)
\gx2g\ = \x2g\ - 1*1+ |*«|.
Using
(i4
desired
) we obtain
|gx2g|
- \gxg\ = |x2g|
- \xg\.
In view
of (*4a) this
is our
result.
(4b) c(g, x) < c(g, x).
Axiom A4 implies
c(g, x) = c(x, x); hence
(*4b}
Since
c(g, x) < c(g, x),
we obtain
view
l*2|- 1*1
= 1**1
- 1*1c(gx,
|gx|
g) = c(gx, gx).
of (*,, ) is our desired
4D
(4c)
< \gx\
and
Therefore
c(gx,
g) < c(gx,
g).
Applying
A4 to this
\gx g\ - \gxg\ = \gx\ - \g\, which
result.
c(g, x) < c(g, x).
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
in
83
REAL LENGTH FUNCTIONS IN GROUPS
1972]
This follows immediately
3.
Abelian
groups
Introduction.
We prove
that
numbers
R.
a).
with real
Let
function
Define
length
functions.
G be a nontrivial
G is isomorphic
A new length
d = c(a,
from (4b).
abelian
to a subgroup
on G. Choose
a function
group with a real
of the additive
a nontrivial
|| || from
length
group
element
function.
of real
a of G and let
G to R as follows:
if x 4 1, then ||x|| = |x| - 2d;
if x = 1, then ||x|| = 0.
We claim
that
Lemma
|| || is a length
3.1.
function.
// x and y are nontrivial
elements
of G, then
c(x, y) > d where
d = c(a, a).
Proof.
d and
to show
that
c(a, x) > d for x £ G, x 4 1- For
c(a,
c(a, y) > d imply by A
It suffices
that
c(x, y) > d.
exists
x £ G, x 4 1 such
c(a,
that
c(a, x) < d and derive
a) = d implies
by A
First
the case
consider
that
that
c(ax,
ax = Tix. Then
Secondly
c(a
|a | < \ax\.
, âx).
Proposition
x ) = Vi\x
i
")
|x | > |a
it follows
9
that
2.3 then
|, so that
that
|, Proposition
tion of Corollary
|ax|
|a
= |ax|,
Corollary
< |a
|.
a
that
|x | < |ax|
Since
9
x) <
= \ax\.
|ax| = |ax|,
so Proposition
|ax| = |ax|,
x ) < c(ax,
| = |a x |.
2.5 implies
|ax|
Since
c(a,
some
2.5 implies
2.7.
Since
implies
c(ax,
Now
x) >
- \a \) > V2\a~x\.
xa) > |ax|
a = a, contradicting
assume
Now c(ax,
from this
âx) + c(xa,
that there
a contradiction.
c(a, x) = c(a, x); hence
in which
c(dx, ax) = %(2|ax|
It follows
We assume
9
c(a
a ).
c(ax,
- |a
a ) = V2\a \ =
| + |7ix| < |a
By A4, c(ax,
, a x ) + c(ä
9
= a x . This
yields
x
9
|.
x ) = c(a
, x );
, x ä ) = 2|a
| -
= 1, a contradic-
2.7.
Corollary 3.2. // x £ G, x 4 1. then \x\ > 2d where d = c(a, a).
Proof.
Since
c(x, x ) = Vi \x \ and
c(x, x) = |x| - Vi \x \, Lemma
|x| > d + lA \x2\ > 2d. If |x| = 2d, then |x \ = 2d = \x\, contradicting
1. Therefore
Define
nontrivial
3-1 implies
AQ as x 4
|x| > 2d.
c(x,
then
Lemma 3.3.
y) = '/2(||x||
c(x,
c(x,
+ ||y|| - ||xy||).
Note that
y) = c(x, y) - d.
y) < c(x,
x) for x, y eG.
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if x, y, and
xy are all
84
NANCY HARRISON
Proof.
suppose
that
The assertion
that
x ^ 1 and
|x| < |y|
|x| < \y\.
is trivial
y ¡¿ 1.
since
it implies
It suffices
to show
LDecember
if x or y is the identity,
Furthermore
no generality
c (x, x) < c(y,
that
y).
First
whence
is lost
consider
we may
in assuming
the case
c(x, y) < \x\ - d; for from this
that
it follows
that
c"(x, y) = c(x, y) - d < \x\ - 2d = c (x, x).
We assume
on the contrary
that
c(x, y) > |x| - d which
\y\ - \x\ + 2d and derive
a contradiction.
which
c(xy,
yields
conclude
by A4 that
that
|xy| = 2|y|
< d, a contradiction
c(x, y) > |x| - d, then
2.5 implies
that
x) = c(x, y).
- |xy|
of Lemma
Since
From this
> \y\ + \x\ — 2d.
3-1.
is equivalent
|x| < \y\, c(xy,
c(x,
But this
is equivalent
We turn now to the case
y) = c(x,
x).
|xy| <
and our assumption
when
c(x, y) + c(x, y) > \x\ = \y\ by Corollary
x = y so
to
x) < c(xy,
y)
we
to c(x, y)
|x| = |y|.
3.2.
If
Proposition
If c(x, y) < |x| - d, then
c(x,
y) =
c{x, y) —d < |x| - 2d = c (x, x).
Proposition
3.4.
The function
from
G to R defined
by \\x\\ = \x\ - 2d is
a length function.
Proof.
AQ. If x/
for the original
length
1, then
||x|| < ||x ||. This follows immediately
from AQ
function.
A . ||x|| = 0 if and only if x = 1. If x = 1, then ||x|| = 0 by definition.
if ||x|| = 0, then x = 1 or |x| = 2d.
By Corollary
3-2, |x| > 2d unless
And
x = 1.
Thus x = 1.
A .
||x|| = ||x||.
This
follows
immediately
from A2 for the original
length
function.
A3. c(x, y) > 0. If x = 1, then c(x, y) =H(||1||
The case
in which
y = 1 is handled
similarly.
+ ||y|| - |ly||) = 0 by Ar
We now assume
that
x ^ 1 and
y 4 1. H x 4 y, then c(x, y) = c(x, y) - d > 0 by Lemma 3-1. If x = y, then
c (x, x) = |x| - 2d > 0 by Corollary
A4.
If c'lx, y) > 7?zand c(x, z) > m, then
and the result
that
3-2.
c(y, z)>m.
follows by A . If y = z, the result
If 1 £ \x, y, z\, then m = 0
follows from Lemma 3.3.
1 i \x, y, z\ and y 4 z. If x' and y' are both nontrivial
Í c (x-, y') + d
cU', V') = j ^
( c(x', y') + 2d
Now 'c(x, y)>m
c(x, z) >m + d.
and
c(x,
m + d. Since y 4 z, lt follows
We have
adjusted
that
the original
length
We may assume
of G, then
if x' 4 y',
if x' = y'.
z) > m imply respectively
Axiom A4 for the original
elements
that
function
c(x, y) > m + d and
implies
c(y, z) >
c(y, z) > m.
length
function
on G so that
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c(a,
a) = 0
85
REAL LENGTH FUNCTIONS IN GROUPS
1972]
from which
it follows
by Proposition
2.6 that
||a"||
= ra||a||.
Henceforth
denote
||x|| by |x| and c (x, y) by c(x, y).
Properties
of this
length
3.5.
For all
Proposition
Proof.
derive
We assume
that
a contradiction.
A. implies
But this
c(x, ~a) = 0.
|a| + |x|.
there
Since
some
suppose
And since
|a | = 2|a|,
+ \x\.
it follows
that
that
Proposition
3.6.
c(ax, ax)+
a) = 0 < c(a, x),
yields
x, y £ G, \x\ = |y|
|ax|
xa) = 2|x|.
c(a
If |x|
2.7.
xä), whence
that
that
= |ax| =
to Corollary
by A
we have
c(x, â) = 0.
the case
Thus
c(xa,
c(a , xä) = c(xa,
|x| > 0 by A,,
For
c(x, a) = 0.
ax = äx, contrary
2.4 implies
Since
c(a,
We now examine
that
But c(a, x) = 0 < c(a, a ) = |a|
|a x| = 3|a|
Since
c(x, x) 4 0 and
c(x, a) = 0 < c(x, x), A4 implies
c(a, x) 4 0.
2.5 implies
|a| > |x|, then Proposition
\x \ = 2|x|.
x £ G, x 4 1 with
c(a, x) 4 0.
c(x, x) 4 0, A4 yields
Since
- |x|.
exists
the fact that
> |a|, then Proposition
3|a|
x in G, c(x, x) = 0 or equivalently
First
contradicts
c(a, x) = 0.
function.
If
|a x| =
, x) = 0, whence
a contradiction.
implies
y = x or y = x.
Proof. We will show that if c(x, y) 4 0, then y = x and if c(x, y) = 0, then
y = x.
Suppose
first
that
c(x, y) 4 0.
Since
c(x, y) = 0 by A4.
Using
follows
xy) = Vi |xy| = c(x~y, xy).
that
lA\xy\,
A..
c(xy,
but Proposition
Secondly
our hypothesis
3-5 implies
suppose
c(xy,
c(x, y) = 0.
3.5, c(xy, ley) = 0.
Proposition
3.7.
xy) = 0.
c(x, y) = 0 and
c(x, y) = minimum S|x|,
implies
Suppose
above
c(x, y) = 0.
0 = c(xy,
If c(xy,
xy).
c(x, y) = minimum
Then
xy) 4 0, then
Since
\\x\,
S|x|,
since
If c(xy,
Since
c(x, y) > 0.
Establishment
y) = minimum
+ \x\ - \y\).
c(x, y) = |x| < |y|.
implies
it
c(xy, xy) >
|xy| = 0 so y = x by
and Proposition
3.5 then
so y = x by Aj.
x, y £ G either
(ii)
Suppose
Thus
|xy| =0
c(x,
xy) = lÁ(\xy\
then
3.5,
By A4, c(xy, xy) > XA\xy\ and by
It follows that
c(x, y) = 0 and
Proof.
3-5,
and Proposition
Axiom A
(i)
c(xy,
A
For all
|x| = |y|
The hypothesis
yield that c(xy, ~xy) = Vi \xy\ = c(xy, xy).
Proposition
c(x, x) = 0 by Proposition
that
|y|!
|y|!.
|y | = 2|y|
xy) = 0, then
since
|x2| = 2|x|,
or
by Proposition
Jxy|
c(xy,
xy) = 0 by Proposition
it follows
c(y, y) = 0, A4 implies
that
Define
c(x, y) = 0.
cp: G —> R as follows:
|x|
if c(a, x) 4 0,
|x|
if c(a, x) = 0.
cp(x) =
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3-5,
c(x, y) = |y| < |x|.
|y|!.
of the isomorphism.
3.5,
= |y| - |x| > 0, whence
By the
86
NANCY HARRISON
Lemma
3.8.
Proof.
Let
= 0.
Assume
x, y £ G.
If x or y is 1, clearly
x / 1 and
oS(xy) = - |xy|.
for definiteness
implies
cf>: G —> R is a homomorphism.
now that
= 0, whence
say
by A4 that
First
c(a, y) 4 0.
<f>(x) and
by A4 that
shows
c(a, y) and
Then
the case
(p(y) have
Then
signs;
<7j(xy) = - |xy| = - \y\
cp(y) both have positive
c(xy,
x) = 0 so
contradicts
Since
signs.
|xy| = \y\ - \x\.
It follows
that
|x| =
as cp(x) and
Proposition
signs.
c/>(l)
c(a, xy)
c(a, xy) < c(a, x)
Hence
|xy| = |x| - \y\.
negative
that
opposite
c(a, y) both are nonzero
But this
<7j(x) and </>(y) both have
3.7 implies
<p(x) and
<p(y) = - \y\.
suppose
signs.
<p{xy) = <p(x) + <p(y) since
consider
|xy| = \y\ - \x\.
argument
Then
4>\y) both have positive
suppose
and
x) implies
c(a, y) > 0, a similar
so x = y by Aj.
suppose
x) = 0 so
+ |x| = cp(x) + cf>(y). Secondly
|y|
y ¡¿ 1. We first
(p(x) = + |x|
c(xy,
Now 0 = c(a, xy) < c(a,
Since
Lüecember
3.7.
Thirdly
c(a, y) = 0, Proposition
0 = c(a, x) < c(a, y) so c(x, y) = 0 by A4.
Hence
cp(xy) = - |xy| = - |x| - \y\ = <-/j(x) + cp(y).
Next we examine
the case
eG, <£(ï~)= - q>(z). If c(a,
definition
c(cz, z) = minimum {|a|,
\z\\.
c(íz, xy) = 0.
is trivial.
By the first
and by the above
cation
We will first
3.7 implies
c(a,
Axiom Aj implies
If c(a, z) > 0, then by the above
And if z = 1, the assertion
This
0 < c(a, xy).
show that for z
z) = 0.
Thus
by
of <rS,r/>(z) = — \z\ =— |z| = - q>(z). It c(a, z) = 0, then Proposition
3-7 implies
z = 1.
that
z) 4 0, Proposition
3.9.
Proof.
Let
q>(z) = - r/j(z).
c(a, xy) 4 0, Proposition
we discussed
0(xy)
3-7 implies
= </>(xy) = t/>(x) + cf>(y),
</j(xy) = - r/>(x) - <j>(y). Another appli-
cp(z) = - cf>(z) yields
the proof that
Lemma
Since
that
c(a, z) > 0 or
<7j(z) = - 4>{z), whence
4>(xy) = — rf>(xy). Thus
of the fact that
completes
case
that
cp(xy) = <f>(x) + cp(y) as desired.
cf> is a homomorphism.
<p: G —►R is one-to-one.
x £ G such that
cp(x) = 0.
The definition
of cp forces
|x| = 0
so x = 1 by A j.
These
last
Theorem
3.10.
G is isomorphic
4.
two lemmas
Let
A reduction
We will first discuss
process.
sets
tity provided
and he proved
there
we set out to prove.
group with a real
of the additive
In this section
of elements
some results
some cancellation
free groups,
the result
G be an abelian
to a subgroup
that we apply to finite
isolated
yield
ideas
group
length
of real
we introduce
numbers
process
from a group with a real length
from Nielsen's
[l].
proof of the subgroup
that a product
is not too much cancellation
paper
of elements
in forming
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Then
R.
a reduction
that we use from Lyndon's
a theorem
function.
function.
Lyndon
theorem
is not the iden-
the product
of any
for
1972]
REAL LENGTH FUNCTIONS IN GROUPS
three
consecutive
with real
length
Lemma
length
factors.
The
arguments
he uses
for this
87
are valid
for groups
functions.
4.1.
Let
function.
x.,
x2, • • • , x
If c(x._,,
be elements
of a group
x.) + c(x., x~~[) < \x]
for all
G with a real
i such that
1 < i < n,
then
(i) Ix.x,
. . . x n' | = S™
. |x.| - 22"í=l~ c(x,í
1 1 2
i =1 ' î1
x.i + ,)
and
l
(ii) x.x2 •■• x 4 1 for 72> 3 [l, Lemmas 6.1 aW 6.2, pp. 222—223].
Theorem
4.2.
a real length
Let
x,,
¡unction.
x~, • • • , x
be nontrivial
of a group
G with
Assume
(4.2a)
c(x.,
x~~x) < minimum {lA\x .\,lA\x
(4.2b)
c(x
,, x .) = c(x.,
factors
elements
x~~.)
= Vi |x.|
x\\ for
1 < i < n;
a'oes 720/ ioW /or awy two consecutive
x.;
(4.2c)
the equation
of (4.2b)
Then the conclusion
implies
of Lemma
\x._xx.x
4.1 holds
A = |*¿_,|
and x.x,
...
x
- \x{\ + lx¿ + il-
¡¿ 1 ¿/ 72> 1 [1,
Theorem 6.4, p. 224].
Let
G be an arbitrary
formations
of X = (xj,
group and x., x,,
x2, • • • , x ) are called
(1) permuting
the
(2)
replacing
x. by x. for a single
(3)
replacing
x. by one of x.x.,
r
°
If a Nielsen
X generates
l
the same
tion on it.
A\x\
We shall
or equivalently
'
l
subgroup
of this
The following
Nielsen
trans-
transformations:
X if the right
i;
x.x.,
J
}
l
x.x.
!
for fixed
l
to a sequence
i, i 4 /•
' .
'
X, the new sequence
G be a group with a real
half of x is isolated
Similarly
the
|yx| > |y|.
left
half
The right
half of x is isolated
definition
i and
X does.
let
the right
or equivalently
A similar
l
of G as
> |y|.
isolated
isolated
1
section
say that
|xy|
y e X, y 4 x, y 4 x.
elementary
index
x.x.,
is applied
if c(y, x) < A |x|
in a set
£ G.
x.;
transformation
For the remainder
. •. , x
length
func-
from y if c(x, y) <
of x is isolated
from
y
half of x is said to be
from all
y and
can be made for the left
half
y where
of x being
in X. And we shall say that x ends with the right half of y if |yx| =
|x| and begins with the left half of y if |yx| = |x|.
Definition.
X = {x,,L x,,¿
maximum j|x .|, |x .| ! where
Definition.
reduced
• • • , x n ! is Nielsen
reduced
I>
j ' —
e, 8 £ {+ 1, - 1 !.
X = \xx, x , ■■. , x ! is fully Nielsen
and for each
if for i 4 si, |xex
l i
x £ X either
|xyf|
reduced
if it is Nielsen
> |y| for all y £ X, y 4 x, and £ 6
S+ 1, - l! or |xye| > |y| for all y £ X, y / x, and ( £ {+ 1, - 1 !; that is for each
x £ X either its right half or its left half is isolated
Theorem
4.3.
// a set
X of elements
in X.
from a group
G with a real
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length
88
NANCY HARRISON
function
is Nielsen
reduced,
then
mations which we can apply
Proof.
This
proof
there
to X to obtain
is an adaption
[2, Theorem 3.1, pp. 126-128].
set where
Nielsen
reduced.
In this
First
way we obtain
right half of every
is Nielsen
Thus
ing in this
X
X.
If so, then repeat
from X
Finally
ii'
we need
's left or right
Assume
w\2
that
in X'
wt.
the above
} in which
If.ti'i
I
ll,
and Solitar
A is fully
element
in X' from w
A
from w
and
already.
I <-\w, | contradicts
If \w'. w~\ =
the fact that
I
is isolated
the right
to obtain
wt2 and
\w'
w' I = \w'
I,
' p. t l
' I '
Continu-
of every
element
in
half of some
element
in
wt2.
Now the right
wt , that is
\w''.wt2\
halves
> \wt \ and
w'\ €X>-
side
that
at least
one side
in X, then
of wt2
\wt2w'Xl\
of wt
X' U \wtA
is isolated.
Then
= \w'Xl\ = \w,2\
and
is isolated
is fully
there
must
\wt2w\2\
in X.
Nielsen
exist
= \w\2\
the right half of z^ir
If |ia>'x| = \wt2w'x\,
then let w"x = wt2w'x;
otherwise
w'l - w'..
w"
a set
X
which
Theorem
a real
ii^'j,
length
Nielsen
reduced.
If a set
function
., wtA
for w
is fully
4.4.
■ ■■ , iv"
procedure
X = {xy
is fully
is fully
y w
Nielsen
reduced,
then
We
end with
let
reduced.
2, • • • , and
w . Finally
x2, ■■■ , x } of elements
Nielsen
and
= \wt2\.
w'^ that
If
reduced.
w'x.
half of t//(2 from those
the entire
X
in X', k < p.
= \w |}.
half
end with the right
w . If
the right
Then
in it
wx by w^.
tu'. £ {w., w~.\ and the
2, ••• , t — \/\w'w\
from which
from
X' so as to isolate
Repeat
reduced.
in X1.
element
procedure
to assure
suchthat
•• • , w
If there is some w'/-r- £ X such that
might
half is isolated
neither
will modify
w'.t
are isolated
\w'x wt2\ > \wt2\ fora11
Karrass,
half of every
Since the right half of w'
at a wt.
However,
transfor-
Nielsen
in X, then replace
is isolated
À = maximum
way we arrive
of elements
of w'
Next consider
_
to choose
is isolated.
is isolated
half
wf by w" = w'w'.
is fully
X = \w.,
•■• , w'
ÍÍÍA
reduced.
it is best
that
w by w' = w w' . Note that
l
then replace
Assume
half of each
the right
of Nielsen
in Magnus,
X so that the right
of X
the right
number
X' which
of one given
X' = \w'
element
I > \w \, then
\w\, then replace
a set
half of wx is not isolated
a set
Next we isolate
|u^i/}
we modify
If the right
a finite
Let X = iz*,, w0, ■■■ , w \ be a Nielsen reduced
\wy\ < \w2\ < • • ■ < \w \.
is isolated.
exists
[December
we obtain
from a group
H = (xy
G with
x2, • • • , xn ) is a
free group.
Proof.
X, w ¿I,
i ^
Consider
'
and
w.
w - w.w^w^
I
,-¿w.
z+ 1
i
for
z
w..
reduced,
And if w.
c(w.,
...
w, where each
fe
1 < i < k - 1.
—
we will check that the hypotheses
is Nielsen
3
—
w. £ \x, x] for some
It suffices
of Theorem 4.2 hold.
l
to show that
An implies
c(w.,
w ) < Vi\w \.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
so
Let 1 < i < n. Since X
u>. .) < minimum \Vi\w t\, H|w¿+1|l,
= w., then
i*/l
x €
provided
Thus
(4.2a)
w.^
4
is satisfied.
89
REAL LENGTH FUNCTIONS IN GROUPS
1972]
We now show that it is not possible
wi-l=wi
then
c(w._v
c(wf. wi + 1) <A\wj\.
reduced,
the left or right
isolated,
then
j) <lA\w.\.
w
follows
c(w._.,
w
w)
Let
.=w.
in X.
X is fully Nielsen
If w.'s
half
w^j = c(w.,
j) = A\w ] holds
is valid
w.
vacuously.
a reduction
sequence
then
By invoking
for no i.
Theorem
subset
of elements
It
4.2
with free generators
for a finite
x , • • • , x ) be a sequence
left half is
is isolated,
H is a free group
If
implies
Since
and if its right
w 4 1- Hence
X = (x,,
w.
y 4 w..
c(w._v
hypothesis
that
w.
) = %|w>.|.
<Y2\w.\
Next we want to define
from G.
4 w. and
half of w. must be isolated
Thus
that the third
we can conclude
w.) = c(w ., w.
w~) < % \ w. \ by AQ. Similarly,
Assume
c(w.,
for c(w.
X - Si!.
of elements
from G such that
l*ll 5 l*2l - " ' " S \x„\- We define the length of X, denoted |X|, to be 2? j|x.|.
A reduction
of X is an elementary
a new sequence
X" such
by an elementary
Nielsen
(x'j,
x'2, ...
X^
= (x[l\
x[l\
process
...
sequence.
stops,
transformation
applied
of X" is shorter
transformation
which permutes
,x'n ) so that
a reduction
Nielsen
that the length
|xj | < |x2 | < ...
, x]'])
, ...
For some
that is we reach
further
by applying
elementary
process
continues
indefinitely.
numbers
with absolute
value
where
< \x'n |.
xt!+1]
sequences
a term
from
length
transformations.
as a length
let
•••
of X^
G this
,xj,
is called
reduction
cannot
be reduced
For other
sequences
R be the additive
function.
X" to X' =
X^=(xj,
is a reduction
of elements
For example
of X, followed
the sequence
Then
XL J whose
Nielsen
to X to obtain
than that
The reduction
any
the
group of real
process
for
Í3, 27! is infinite.
As a corollary
to the previous
theorem
Corollary
// the reduction
sequence
elements
4.5.
from a group
. • . , x ) is a free
■• • , x ) be a sequence
sequence
weakly
limit numbers
\x^\
> |x[5 + 1l| for all
sequence
has a limit.
If the reduction
limit number
to be
|xL J|.
ith limit number to be limit
A reduction
from G such
■• ■ be a reduction
bA ^|, Ix^. J|, |x^.
decreasing
function,
sequence
for a reduction
of elements
X^0-" = X, X*- , X^ \
< 2 < 72, consider
of X = (Xj, x2, ■• • , x ), a set of
is finite,
then
H = (xx, x2,
group.
We want to define
and let
G with a real length
we get
sequence.
that
|xj|
sequence
Let
< |x2|
for X.
1, • • ■ • By our construction
s.
Thus
for each
of real numbers
bounded
sequence
at the
stops
And if the reduction
X = (xj,
x2,
< • ■• < |x |,
For each
i, 1
of a reduction
1, \x\ % \x\ J|, ...
is a
below by zero and hence
Nth step,
sequence
define
is infinite,
the ¿th
define
the
|x^ ■'I.
for a set of elements
we claim that the limit numbers of a finite
from G is by no means unique.
reduction
sequence
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
for a sequence
However,
X of
90
NANCY HARRISON
elements
proof
from G depend
is adapted
Proposition
Karrass,
4.6.
S and
Suppose
x2, ■. . , x ), a subset
S and
T stop
Proof.
only on X itself
from Magnus,
after
of elements
a finite
and not on the reduction
and Solitar
of steps,
their
limit
Let H denote the subgroup of G generated
are fully Nielsen
and Zk=
Hk.
(z
£ Z\\z\
there
exists
(y v y 2' ' ' ' ' ?
x
We claim that
and
in H,
such
wy
w. , 4 w
for
w2, ■■• , w
2m .lu/.I -22m~lc(w.,
!=!'!'
l-l
I
derive
that
) = W, we can write
; < 77}, w. 4 1, and
for X - (x.,
length
numbers
function.
If
are equal.
by X. And let (y!, y',
sequences
• • • , y' )
S and T respectively.
H. Let H, = (x £ H\\x\ < k), Y, = (y £ Y\\y\ < k),
Hk <t Y,
some
we know that
< k).
that
sequences
Y - \yy y2, • ■• , yn\ and Z = [zy z , ■■■ , z } which
reduced and generate
We assume
This
3.1, p. 128].
G with a real
and (jZj, z^ , ■■■ , z' ) be the last points in the reduction
By Theorem 4.3 we can obtain sets
process.
[2, Corollary
T are two reduction
from a group
number
LDecember
rYfc= Vfc = Z k.
a contradiction.
|x|
< k and
x = w w2
1 < i < m.
satisfy
...
x i
w
Clearly
Since
Y,.
, where
Since
£ Y
x £
w. £ \y ., y . | 1 <
In view of the proof
the hypotheses
Yfe C
H
of Theorem
of Theorem
4.2.
4.4
Therefore
ÜTT). For Ki<m,
! +1
7-1
7.| + ¿
[(«7.1 - 2ciw., uz.+1)] +
X
¿=1
Since
c(w ., w.
\w \.
Now at least
But then
Li«',-!- 2c^l-
v wi)]'
¿=7+1
,) < minimum \V2\w \, l/2\w. ,|}, the above
one h, 1 < h < m, is such that
|x| > \w, \ > /e, contradicting
our choice
expression
\w,\
yields
|x| >
> k; for if not, x e V .
of x.
Thus
Y, = H,.
Similarly
Z, = H,. By Corollary 4.5, Y - ll) and Z - 11}freely generate the free group H,. Since the
rank of H,
is unique,
Y and
Z have the same
length < k and hence of length
We want to discuss
It is easy
to see that
for a finite
set
without
G = A. *A?
absolute
value
*A,
as a length
of nontrivial
elements
with
k.
the limit
further
X of elements
Suppose
1.2.
briefly
number
from
where
numbers
of infinite
qualifications
G need
two infinite
not have
A. is the additive
function,
and give
reduction
reduction
the same
group
G the length
sequences.
sequences
limit numbers.
of real
numbers
function
with
of Proposition
Take X = 1(2, l), (77, 1), (e, 2), (3, 2)} where (r, k) is the real number r
from A, . We can formulate
(2, 1) and (77, l) but leaves
0, 0, e, 3.
an infinite
reduction
sequence
(e, 2) and (3. 2) alone;
We can also formulate
an infinite
reduction
for X which
its limit
numbers
sequence
reduces
would be
which reduces
both 1(2, 1), (77,1)} and l(e, 2), (3, 2)} and has limit numbers 0, 0, 0, 0. Thus
in contrast
to a finite
reduction
quence do not depend on X alone.
sequence,
the limit numbers
In this paper
we are mainly
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
of an infinite
concerned
sewith
REAL LENGTH FUNCTIONS IN GROUPS
1972]
reduction
sequences
difficulty.
of pairs
We do feel that
defined
in such
problems
a way that
involved
We close
of elements,
its limit
are major,
this
so this
a more restrictive
section
situation
type
numbers
a result
does
of reduction
would depend
so we did not pursue
with
about
91
this
not present
sequence
only on X.
question
reduction
any
can be
But the
any further.
sequences
for a pair of
elements.
Proposition
a group
equal
coordinates,
Proof.
all pairs
|a|
Let
S be a reduction
G with a real length
first
quence
4.7.
Let
takes
for all
ß where
(a, y) have
Then
a=
|a|
beyond
sequence
some fixed
first
coordinate
and there
equal
...
exists
from
pair have
is finite.
a< ß be the limit numbers
their
the form (a, y), (a, y ), (a, y2),
i.
for a pair of elements
If all pairs
then the reduction
a and
beyond
junction.
sequence
of 5.
to a, that
with
Assume
|y . , | < |y .| but
an integer
N such
that
is the se-
that,
|y | >
for n > N,
\y | < /3 + <x- 2d where d = c(a, a"). Let z = yN and z. = yN . fot i > 1. Then
the sequence
from (a, yN) is as follows:
(a, z), (a, Zj),
We will show that
assume
aza.
If z_ = aaz,
From this
this
then
it follows
are similar
case
- azaa,
|aza|
is
c(a,
a) < c(az,
is impossible;
or the sequence
the sequence
Hence
c(z,
|za| < |z|
tradiction.
process
a),
then
|z,|
implies
A4 implies
And if c(az,
+ |z| = 2|a|.
stops
and
AJmplies
It is easy
|z,|
This
< |z2|
c(a,
a) = c(a,
the pair
z
aza,
az).
2d.
z, = aaza.
c(az,
a) = c(az,
a) < c(z, a),
then
Since
|aza|
< |z|.
so |aza|
that
< |za|,
= |za|
|aza|
< ß.
(a, z ) must be Nielsen
and
= |aza|
c(az,
|aaza|
We
\za\ + \az\ =
|zJ
A4 implies
< \az\,
Since
But then
a) = c(z, a).
z) and
|z,|
groups
with real length
functions.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
>
c(az,
> |a|
a)
or
a) > Vi\a\.
< |aza|,
c(a,
a)
- |a| + 2d". Since
But this
reduced
is a con-
so the reduction
stops.
5. Two-generator
In
And if
a).
to see that
But
= aza.
4 za.
c(a, a) = c(aza,
suppose
or
the arguments
for z
< \za\,
is a contradiction;
aza)
Since
the details
|az|
|za|
a) = c(a,
|z| < ß + a-
so that
and |z| < ß + \a\ - 2d, it follows
Therefore
c(a,
of generality
is aaz,
Finally
c(z,
< |z|.
or equivalently
loss
z
or aza.
< |az|
z2 = az was chosen
stops.
Without
Then
implies
aza
only present
Since
so A,
< |z|.
|aza|
a)>A\a\
< c(a,a~za~);
we will
a),
|za|
2|a|.
z) so
is either
+ 2d < ß, a contradiction.
For if c(z,
= c(az,
z
za, or azaa.
to show that
This
of steps.
= |z| - \a\ + 2d < ß since
of these,
c(a, a) < c(aza,
= - |a| + |az|
need first
|a|
|az|
....
|azj, |az|!.
a) < c(a, az), so A,
Therefore
aaza,
number
< minimum S|az|,
that
for both
z,
in a finite
|az|
is a contradiction.
this
z
it stops
z. = az and
(a, z2),
In this section
we
NANCY HARRISON
92
determine
the structure
a real
length
function.
a real
length
function
lx, y} where
prove
of a group
which
Throughout
this
| |.
is generated
section
that if the reduction
then
process
by two elements
let
In view of Corollary
x, y £ G is finite,
[December
G be a group
4-5 if the reduction
(x, y ) is a free group.
for \x, yj continues
and has
equipped
with
process
In this
indefinitely
for
section
then
we
( x, y)
is abelian.
Proposition
numbers
a and
Proof.
sequence
for
// a reduction
ß, then
Let
limit _|x
< ß.
5.1.
a=
for x, y £ G is infinite
with limit
ß.
x, y £ G with
(x, y)
sequence
with
|x| < |y|.
limit
| - a< ß, there
Let
numbers
exists
(x, y), (xr
a and
a positive
ß
y2), ...
where
integer
be a reduction
a < ß.
N such that,
Since
for n > N, \x \
We claim that x^ = xN . for all k > 0. The new element p in the (N + l)th
pair has length
and x..
less
j = x...
sequence
than that
Similarly
of yN.
Since
\xN\ < ß,
x,, = x», , for k > 1.
\p\ > ß.
Hence
By Proposition
yNi=P
4.7 the reduction
is finite.
Proposition
5.2.
Let
(x ., y .) be an infinite
reduction
sequence
for x, y £ G
where limit.¡—.oc' Ix.lit = limit.7 —.oo'^r1ly.l = 0. Then (v x, ■'y )' is abelian.
Proof.
For an arbitrary
[g. tí], [g, h]\
a Nielsen
where
transformation
respondence
elements
reduction
between
the sets
For each
sequence
T
f is a class
of elements
by Theorem
lx | = 0 = limit
is a one-to-one
such
that corresponding
to verify
f(g) = \g | - |g|.
2.8, the above
limit
teger
N such that, for all n> N, \y \ <%(.
n'
h')
and known,
And for each
that
so we omit
pair in our
K. = Kl for all
any <r> 0 there
°
cor-
a £ Y(x n , y
)}. Since
' n
implies
ly I, given
n — "ounl
When we apply
there
K n = minimum \f(a)
'
Since
ri —.00 i
(g1, h'),
is straightforward
g £ G define
Y(g, tí) = í[g, h], [g, h],
of g and h.
Y(g, h) and Y(g',
This
(x n , y
) define
J n
function
g, h £ G let
the commutator
to (g, h) to obtain
are conjugates.
the details.
pair
[g, h] = ghgh,
'
exists
z.
an in-
Now
Kn —
< f(x
y x y ) < lx y x y I < 2|x
I + 2|y
I < (.
'
n}n n' n - ' nJ n nJ n1 ' n'
u n' —
Hence
K. = 0 so x and y commute.
We turn now to the case
nite with limit numbers
(x, y).
There
a. = ß > 0.
are several
point on in our sequence
we verify
and
the
limit
some simple
lengths
c(y ., y)
when every reduction
parts
no element
lemmas
of elements
= limit
.
Let
(x., y)
to this case.
occuring
concerning
x.) ='/Sa.
for (x, y) is infi-
be a reduction
First
sequence
for
we show that from some
is cyclically
the amounts
in our sequence.
c(x.,
sequence
of cyclic
Thirdly
Finally
commute.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
reduced.
reduction
we show
we prove
Secondly
that
that
x and
y
1972]
REAL LENGTH FUNCTIONS IN GROUPS
If a and
and
b ate the elements
y but it is not determined
use the notation
(a/b)
Lemma
Let
limit ■_^
that
5.3.
element
Proof.
Let
and yM = v.
whether
|y .| = a > 0.
0 < e < (l/3)a.
For definiteness
\u, v\ is not Nielsen
reduced,
that
c(u, uv)
implies
2. If \uvu\
Since
implies
that
reduced
exists
n>N
|y„ | < a + c. Let
such
xf,=u
< \v\.
in the
(M + l)th
so the following
then
and
pair
of our
holds:
u is not cyclically
reduced.
Then
\uv\ = \v\ - \u\.
2.1,
But
reduced.
c(u, u) <
\v\ - \u\ <(<a
implies
|f| > maximum \\u\,
given
any
'
Lemma
5.4.
limit.
Let
'
(x., y)
|x.|
= limit.
where
JÎ+1
cyclically
Z
any
N there
c(v, û~)> 0, for A4
c(v, ü) = 0, so
But this
\vu\ -
contradicts
our
\uv\\.
\uv\ \, then
to show that
N there
reduced.
uv is not cyclically
c(u, uv) > 0 and
c(uv,
reduced.
u) > 0.
And c(uv,
If
u) = 0
of the hypothesis.
exists
n > N such
that
some
element
of
reduced.
,, y . ,) = (x./z)
Then given
that
\uv\ = \v\ - \u\ <e < ex, a contradiction.
(x , y ) is not cyclically
Jn
to show that
on the contrary
\uvu\ = \uv\ + \u\, a contradiction
Therefore,
v is not cyclically
\uvu\ > \vu\ - \u\ = |v|.
3- If \uvu | < maximum S|«|,
c(u, uv) = 0, then
then
It suffices
Assume
In view of A4 it suffices
!+l
for x, y £ G where
N there
that
\uv\,\uv\\
< maximum S| u |, | i/f|},
c(v, J7) > 0.
since
Case
(x.
any
u is cyclically
0 = c(u, uv)
By Proposition
hypothesis
where
we will
|, \üvu\ , \üvü\ \< maximum S\u\, \üv\ \.
| uv \ < \v\ , c(u, v) > 0.
\v\ + \u\.
n
sequence
uv is the new element
\u, uv\ is not Nielsen
on the contrary
so A
length,
is a contradiction.
Case
then
\ûv\,
1. If \vu \ < maximum i | u \, \uv\\,
Suppose
for x
reduced.
M such
minimumi|î<22
so this
Then given
Choose
assume
Again
Case
reduction
of (x , y ) is not cyclically
Since
sequence
a or b has the shorter
(x., y .) be an infinite
minimum S\ûv\,
sequence.
in a reduction
fot the pair.
\x \ = limit.__
some
of some pair
93
be an infinite
reduction
ly.l = a > 0.
z has
exists
Furthermore
the shortest
n > N such
sequence
length
Oí
for
assume
x, y £ G
that for all
of all the possible
that neither
element
i,
choices.
of (x^, y)
is
reduced.
Proof. Let 0 < c < (l/3)a-
Choose Q such that |y„| < a+ c. Let xQ =
u and y„ = v.
Let
uv be the new element
hypothesis
pair
is (ul uv) where
this
in the
(Q + l)th
|hi>| < minimum ||zvv|,
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
pair.
\uv\.
By our
\uv \\.
We
94
NANCY HARRISON
consider
three
Case
cases
1.
vuu is a coordinate
By Lemma
neither
is v.
since
|aii|
as in the previous
v) > 0.
pair following
u is not cyclically
c(u, v) > 0.
< \v\, c(u,
lemma.
of the second
5.3 we know that
Suppose
[December
Since
Axiom
(u, v).
reduced.
We show that
c(u, u) 4 0, c(u, v) > 0 by A4.
A4 then
implies
c(v,
v) > 0.
And
Therefore
if
c(u, v) > 0, we are done.
We show that
which
it follows
2d < a+
(.
c(u, v) - 0 leads
to a contradiction.
by A4 that
c(u,
v) - 0 also.
If d < V2CLthen
c(u,
u) < c(vu,
\vu\ = |f| - \u\ + 2d < ( + 2d. Since
Let
u).
Assume
d = c(u,
Axiom A
\vu\ > a, a-
c(zz, v) = 0, from
u).
Note
implies
that
a-
e<
d = c(u, vu), so
e < 2d < a . And if d > l/2a,
then a<2¿<|zz|<a+t.
By Proposition
integer
2.6, the lengths
p such that
It follows
that
\up\ > 2\v\;
Ivu^l > |v|.
< \vu\ < \v\ and
\vun+
want to determine
> \vum\ + \u\ - 2d.
72. Therefore
for
77 be the integer
|.
Since
contradicts
c(zz, u) < c(vum,
Note that
2.1, c(v, u q) < \v\ for all
that
|t>«" + \ < \vu"\
\vu\ < \v\, we know that
the fact
a), So A4 implies
if vu" and
so we can find a positive
such
1 < ttz < 77. If c(vum,
But this
\v\ - m\u\ + 2md.
are unbounded
and by Proposition
Let
| > \vun+
\vum\
|zr|
that
\vum+
occur
t < a.
Also
if vun
and
vun+
|v|; for \un +2\ -\v\ =(n+2)|«|
Any pair
integer
\s, t\ in \\u,
will be called
are only a finite
(x
such that the
r + 1 terms
of u.
occuring
Let
M—
< N, then
Then
(xM, yM) = (u/vum)
the
= (vum/vum-1)
\x M' yM)
'N.
there
\,{vum,
pairs.
exists
vuv\\
\t\\ < \v\.
but the
then
\vun\ + |u| <
\vun+
\ 4 \un+
where
Since
\-
Suppose
(N + l)th
then
pair
an integer
_
ttz is a positive
\vw\
that
> \v\, there
every
r _between
beyond
i|iTZ7r|, \vur+
(vur/vur+
(M + l)th
|S < |zz|,the
) is less
terms
pair
is not normal.
than
If
1 and _ M + 1
(u, v) are (u, vu), (u, vu ),...,
maximum
Mth and
+ ï), (u/vum~
(u, vur'
length
the length
are both normal.
(xM, yM) = (vum/vuv)
Uu .„ y„ .,) € ÍW-
X), (vum/vum-
£\(vuv/vum),
Now
l), (vum/vum^)\;
(vuv/vum'l)\;
implies
V^z'), (v«m+1Mv),(w'n-
Vv«™), (í«m+1/í«m)¡.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
),
of any
implies
(xM + 1, yM+1)
yM + l
- |«| + 2d =
\v\ =-
then
777<
implies
(xM + 1, yM+1) £\(u/Vum
cM+l'
vum~
normal
following
And since
in the sequence
M < N.
\vum,
if maximum \\s\,
of distinct
where
element
vum\,
, y „) is normal
u ir lx„,Ni Jy.,}
M
(t7z/, u), (v~ur/vur + 1).
in the sequence,
\
- 2(« + Dd - \v\ = |«| - |t7zz"| + |a| -2d<2(<a.
normal
number
from (u, v) through
occur
since
in the sequence,
\v u , + 1| 4 |z7n+ 1| - \v\; for |¿n + 1| - |v| = (77 + 1) \u\ -2nd-
We
\vum+
\ < \vum\
\vum\ = |7z7m"1|
vun+
72> 0.
u) < c(u, u) = d, then
q.
< ■■■
1972]
REAL LENGTH FUNCTIONS IN GROUPS
From this
fot all
it is easy
m < s occurs
Suppose
vumvum-1
).
c(vum,
yN\
Then
\vum\
v)<c(vum,
c(vum~l,
but
Suppose
earlier
m - I <n
Thus
0 = c(vum~
> 72 + 2.
Then
vu"+
R be the term in which
(u/vu"+3),
vu"+
first
term where
S < R.
replace
Since
Proposition
vu"+
the element
that this
does
Let
c(fan
contrary
the
to our choice
Secondly
assume
pair
\v\.
umv) > 0.
Since
T = c(vun+
to what
vum.
If
a
\vu
-
Now assume
part
v
that
m
of the sequence.
Let
Rth term is necessarily
vun + l must occur
in the St h
in the
Rth term, the fact that we
implies
that
\vun+
\ < \vun+
\.
If T < \u\, then
\vun+2\ > \Vun + 1\.
|^"+2|
< \vun + l\ - \u\ + 2d < \vun + 1\,
of 72.
that
(x,,,
is either
Since
\vu\ < \vu\,
y..) = (vum¡vuv).
vuvumv
Then the new element
or vuvumv.
Suppose
c(t>, vum) = 0 as shown
c(v, u) < c(v, u).
above,
It equality
if the inequality
\v\ - 2d, in which
|yz<i>| = |a| - 2d < ( , also
Then
0 =
- \u\ < c, a contra-
, u) + c(u, u).
|tv| < 0, a contradiction;
vuvu~mv.
case
0 = c(v, vum) < c(vuv,
is strict,
umv),
z = vuvumv
A4 implies
holds,
then
then
\vuv\ = \vu\ \vuv\ = |vw| -
A4 implies
\vu\ = \u\ +
a contradiction.
so A4 implies
z in
so
Suppose
z =
\vuv\ = \vu\ - \v\ <
0, a contradiction.
Thirdly
assume
pair is either
latter
Assume
the
, yN) = (u/vum).
or uvum■
0 = c(~u~,v) < c(u, vum).
\um\ -\v\.
Since
(x
uvum
now that
(R + l)th
m > n + 2.
Let
we have our desired
But
contradiction.
contradicts
in both cases
what we established
R be the term in which
Rth term is necessarily
\vun+2\
Hence
z of the
(N + l)th
0 = c(u, v) < c(u, vum),
By A4, 0 = c(v, vum)
u e \xN, yN\, the
is (u/vun+2).
The new element
If the former
If m < 72+ 2, then this
R < N and
)
Thus
+ |«| -2d>
c(vun + \ Ü) > \u\ - d so
(N + l)th
c(vuv,
length
+ , u) = c(u, u).
|7Ja" + 3| = |^"+2|
If T > \u\, then
Now
not occur
of the larger
impossible.
2.4 implies
z = vum~
Then the
vuv).
y
v) > c(u, v) = 0,
= \vum~2\
provided
appears.
(vu"+i,
(x^,
- \v\, contrary
Suppose
\vum~l\
that
Therefore
in some term of the normal
(vun + 2/vun + i), or
assume
, vum); A 4 implies
m < n + 2.
earlier
vum
pair is either
If c(vum,
= \um~l\
+ 2.
, v) < c(vum~
occurs
l.
\vum-l\
A4 implies
to what we established
We claim
z = vumvum~
provided
contrary
First
(N + l)th
- \u\ < c, a contradiction.
v) > c(u, v) = 0, then
diction.
always
z in the
A4 implies
S < N, then
5th pair.
vum + 1 4 \xN, yN\.
= \vum~ï\
wtm~l);
we established
in the 5th pair where
the new element
or vu™-1™™.
A4 implies
occurs
in some pair prior to the
vum £¡xN,
= (vu™, vum~
then
to see if vus
95
vu"+
and if the
so
earlier.
first
(u, vun+
> \vun + 1\ by our choice
\vum\ =
occurs.
) and
of 72. Thus
c(u, v) > 0 and v is not cyclically
reduced.
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9<>
NANCY HARRISON
Case
2.
uvu
By Lemma
assume
A.
is a coordinate
5.3, v is not cyclically
the contrary
implies
Let
lAa.
and derive
0 = c(u,
d = c(v, v).
Since
v).
Since
\uvu\
implies
c(uv,
v) = d.
tradiction.
Therefore
Let
72 be such
reduced.
We claim
c(v,
that
after
Since
v) > 0, A.
d > l/2a.
(u, v).
We claim
a contradiction.
And since
u\.
We want
< maximum
that
c(u,
\uv\\,
c(uv,
that
is u.
v) > 0 as
implies
Assume
neither
0 = c(u,
We
\uv\ < \v\,
v).
on the contrary
that
d <
Axiom A4 implies
u) > Ma>
c(u, v).
Axiom
d A4
\uv\ = \u\ - \v\ + 2d < 2d < a, a con-
/2a<d.
that
\v"+
u\ < \vnu\
to determine
|fmzz|
-2d>
c(v,
it follows
vmu),
\\u\,
But it follows
\vm + 1u\ > \v\ + \vmu\
\vm~
pair
|zzî7| < |tz|, c(u, v) > Vi\u\ > V2a. > d = c(v, v).
c(u, v).
\vn+
of the second
[December
from which
< • • • < \vu\ < \v\ and
for
1 < ttz < 72. If c(v,
\~vmu\, a contradiction.
by A4 that
vmu) < c(v,
Therefore
d = c(v,
u\ - \v\ + 2d = |zt.| - ra|i>| + 27T2Í7'.Also
\v "+ u\ >
vmu).
note that
v), then
c(~v, v) <
Therefore
\v mu\ =
if vn + lu an¿
~y~nu occur
in the reduction
sequence,
then \vn+ u\ 4 |^"+ | - |a|; for \vn+ u\ > a. and
"
+
1!
I„l
_
l.,l
I77"„l
<Q;m;iorl„ if
\( ,,"
v
u\ = \v\ — \vnu\ < e, s< na. Similarly
v"+ + 2u and v "u occur in the sequence,
then
\vn + 2u\ 4 |t7" + 2| - |u|;
for
|v" + 2zz| > a but
\vn + 2\ - \u\ = \v\ -
2d + \v\ - \v"u\ < 2e < a.
The second
the former.
or u
let
vu.
pair after
Then
Since
\vu\ > \uv\,
that
follows
\w\ < \vu\ — \u\.
that
(vu/uvu).
the sequence
and the next pair,
Rth
pair
v).
is
Let
u) < c(u, v),
Therefore
w = uvu
it is
it z = u vu, and
Axiom A4 and
\uv\ < \vu\
\vu\ < \v\ + \u\ — 2d.
the second
ate unbounded.
then
v u), (üvu,
say the Rth pair,
is not
uvuvTu,uvruvTu,ot
Since
but we showed
c(uvu,
c(v,
\vpu\
vu), (üvu,
vr+
c(u, v) = 0, c(uvu,
\u\ < 0, a contradiction.
uvuvru,
to vu.
Suppose
(u, v) is vu, u vu,
Let
pair
after
It
(u, v) is
r be the integer
such
that
from (u, v) on is as follows:
(u, v), (vu, u), (uvu,
c(uvu,
or (vu/uvu).
0 = c(u, u) < c(u, w).
Since
\w\ < e, a contradiction.
Now the lengths
(u/uvu)
z of the third pair after
z is not equal
w = uvu if z = u 2vu, then
imply
the
(u, v) is either
the new element
above
And
that
v) < c(uvu,
vTu).
u.
c(uvu,
vTu) ; and
The
new term
c(vru,
v) = 0 = c(vr+
u, v).
Since
is either
that
\vru\
= \vr\ - \u\ and
c(vT+
u, v) > 0 implies
c(u,
that
c(uvu,
that
v) = 0 =
\uvu\ = \uv\ - \u\;
\vTu\ = \vT~
First
v) > 0 implies
by A4 that
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
u\ - \v\ < 0,
(vru/vr+
u or vT+ uvru.
v) - 0, c(vru,
If z =
then
the Rth term is
vTuvr+
z in
\uvu\ = \uv\ —
a contradiction.
z = uvuvru
A4 implies
u), (uvu/vru)
The new element
by A4 that
is also
that
T/~
by A, that
if
z = vrAr u, and hence
pair
note
v) > 0 implies
In both cases
(R + l)th
u).
First
\vu\ — \u\ < e so this
Therefore
in the
(uvu/vr+
v) > 0 implies
0 = c(û~vu, v) < c(uvu,
a contradiction.
v u), ■ ■ ■ , (üvu,
\vr+
u\ =
u).
note
by A4
REAL LENGTH FUNCTIONS IN GROUPS
1972]
|z/+
| - \u\.
72+1.
Since
Thus
|«|.
Hence
c(vru,
\vTu\ < \vT~
by what
c(vru,
u\ < • • • < \v u\ < \vu\
we established
earlier
v) = 0 = c(vT*
vr+ u) implying
by A
if z = vr+ uvTu, then
u, v).
that
Case
3.
Case
reduced
2, we will present
omit most details.
either.
Let
72 be such
\(üv)n+2ü\
and
u and
c(u, v) = 0.
by A4 that
\vru\ =
(u, v).
let d = c(uv,
that
v) <
And
u) 4 0.
to those
and any difficult
c(u, ~u) = 0.
And also
uv). We claim
are similar
of the development
For
0 < 722< 72, \(Uv)mü\
in the reduction
(uv)"u
0 = c(vru,
a-
Then
that
in
parts
but
c(u, v) > 0,
e < 2d < a+
c.
|(íh/)" + 1a| < \(üv)n~ü\ < \(üv)n~ 1w[ < • • • < \iïvû\ < \v\ and
> \(üv)" + 1ü\.
(üv)n~ü occur
(uv)"*
that
implying
the arguments
on the contrary
c(u, v) = 0, c(u, v) = 0, and
u, then
pair following
reduced;
Since
an outline
Assume
c(u,
of thé second
By Lemma 5.3, uv is not cyclically
u is not cyclically
u, vru),
Therefore
uvu is a coordinate
of r, r <
|7/+1zz| 4 |t7r+ | -
u\ = \vru\ - \v\ < 0, a contradiction.
0 = c(vT+ u, v) < c(7/+
\vT~ u\ - \v\ < 0, a contradiction.
by our definition
\vru\ 4 \vT\ - \u\ and
Now if z = vruvTJr
\vr+
97
occur
sequence,
= |«| - 722|â72| + 2772^. If (üv)n + lÜ
then
in the reduction
¡(üv)"*
sequence
ü\ 4 \(uv)"*
then
\(uv)"+
\ - \u\; if
u\ 4 \(uv)n*
|
- \u\.
The second pair after
the new element
case
(u, v) is either
0 = c(u, u) < cAll, uvu).
to c(û, vu) = |a|.
It follows
u2vu.
that
Thus
Let
Then
\uvu\
(uvu/u)
or (vu/uvu).
_y _
z in the following
Axiom A implies that
by A
= \u\ - \uv\ < 0, a contradiction.
pair
r be the integer
following
such
(u, v) must
that the sequence
(u, v), (üv, u), (üv, uvu), (üv, (üv)
in the Rth term is ~üvu(vu)r, jjuu(vu)r,
ä or a contradiction
new element
of these
in the
lead
This
the proof
Let
(x., y.)
|x.| = limit.
the reduction
It is easily
We assume
be an infinite
u).
The new element
u or (uv)r*
N there
be shown
exists
ü(üvYü.
z
to be
((uvYuf (uvY*
u is not cyclically
any
z 4
ü), (uv/(û~v)ru~)
u; it can easily
(üv)r~ü'(üv)r+
that given
that
from (u, v) on has the form
the Rth term is
Therefore
seen
u).
The
But both
reduced.
n > N such
that
reduced.
reduction
sequence
for
x, y e G where
ly J = a> 0. // the elements of (x* y*), the Nth pair of
sequence,
/¿e?2 for n > N neither
Proof.
term is
\vli\ = 2\u\ - \uv\.
u), ■■• , (ûv, (üv)r~
or (uvY*
Thus
of (x , y ) is not cyclically
Lemma 5.5.
limit.
(R + l)th
to contradictions.
completes
some element
results.
that
be (vu/uvu).
and the next term, say the Rth term, is not (ûv/(ûv)T+
(üvY*
_2 _
z = u vu, in which
\ûvu\ = \vü\ - \u\, which is equivalent
c(u, v) < c(û, vu) implying
the second
If it is the former, then
2_
pair is u vu or u vu. Suppose
are not cyclically
x
7?or y
the existence
reduced
is cyclically
and have
length
less
than
reduced.
of 72> N such
that some element
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of
2 a,
NANCY HARRISON
98
U » y ) is cyclically
reduced
and derive
[December
a contradiction.
Choose
imal and let z be the element
of the 72th pair that is cyclically
z = x'y'
(72 - l)th pair in the reduction
where
(x'( y')
is the
easy
to see that neither
c(y',
c(x',
y')
this and
>1/2|x'|.
Using
x')
nor c(y',
x')
is zero.
the fact that
77 to be min-
reduced.
Write
sequence.
Since
It is
\x'y'\
< \y>\,
y' is not cyclically
reduced,
c(y>, x') > 0 by A4. And from c(x', x>) > 0 and c(y', x') > 0 it follows by A
that
c(y', x') > 0.
A4 that
0 = c(x'y',
is cyclically
c(x'y',
Also
y')
= 0 so
a contradiction.
The next step
Lemmas
Assume
to |x'y'|
< c(x'y',
= |x'|
x').
implies by
- |y'| < 0.
Since
z
Axiom A4 then implies
\y'\ < 2a and
|x'|
z in the reduction
Let
some trivial
reduced.
numbers
Let
pair
lemmas
(x ., y .) be an infinite
for this
> a, \x'y'\
sequence
< a,
after
that we need
reduction
sequence
a and that no group element
that the following
x>) < c(y', x')
reduced.
that the limit
cyclically
0 = dx'y',
Since
no element
is to establish
5.6—5.8.
zero real number
x'y')
= \y'\ - \x'\.
Therefore
For
is equivalent
0 = c(x'y',
\x'y'\
(x* y*) is cyclically
G.
which
reduced,
x')
c(x'y '. x') 4 0.
occurs
(a, b) be a pair in the reduction
is (a/ab)
where
sequence
are both equal
which
in the sequel.
for x, y £
to some
non-
in this sequence
sequence
\ab\ < minimum l|ßr?|,
is
and assume
\ab\,
\ab\\.
Let
c = c(a, ~a) and d = d(b, b).
Lemma 5.6. The following hold:
(5.6a) If c = d, then c = c(a, b) = c(a, b) < c(a, b) < c(a, b).
(5.6b)
// c < d, then one of the following
occurs:
(i)
c(a, a) = c(a, b) = c(a,
b) < c(b, b) = c(a,
b) < c(a,
(ii)
c(a, ~a) - c(a, tí) = c(a,
b) < c(a, tí) = c(a,
b) < c(b, tí);
b);
(iii)
c(a, ö) = c(a, ~tí) = c(a,
b) < c(a, T) = c(a,
b) = c(b, T>).
(5.6c) If d < c, then c(b, tí) = c(a, b) = c(a, b) < c(a, Ö) = c(a, J) < c(a, V).
Lemma
second
5.7.
// c < d < l/2a or d < c < V2CL,then
pair after
Lemma 5.8.
after
(a, b), \ab\ = |£>| - |a| + 2c, 072^ c(ab,
If d < c and
(a, tí), c(ab,
Proof of 5.6.
Since
ab occurs
c(a, tí) so A
implies
b), c(a, tí), c(a,
c = d.
Then
Compare
c < d.
ah ) = d
a b occurs
in the pair after
b)\. These
c(a,
in the second
pair
(a, b), c(a, ~a) < lÁ\a\ <
Also by our choice
results
of ab, c(a, V) >
hold for all three
c(b, tí) < c(a, tí) so A4 implies
b) we conclude
Then
of the
\ab\ < \a\ or \ab\ = \b\ - \a\ + 2c.
c(a, o) = c(a, tí).
And from c(a, a) = c(a,
Assume
2d < a< 2c, then
a~tí)= d, and either
maximum \c(a,
Assume
a b is a component
by A4 that
c(a,
b) < c(b, tí) so A4 implies
cases.
c(b, tí) = c(a, b).
b) > c(a,
c(a,
a).
b) = c(a,
b).
c(b, tí) and c(a, tí). If c(b, tí) < c(a, ~b~),then c(b, b) = c(a, b) by A4.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
1972]
If c(a, V) < c(b, V), then c(a, è) = c(a, b) by A4 . Finally
c(b, b).
Axiom
of the proof
d < c.
c(a,
b).
Then
And since
Proof
when
A4 implies
c(a, T>)> c(a,
Assume
c(b, a).
b) > c(a, b).
Therefore
But as we noted
c(a,
The arguments
for both the case
when
We will only present
c(b, V) = d < Aa,
applicable
here.
for the new element
(a, b) ate
aba,
The possibilities
aba,
and
|aè|!.
< c(a, ba).
c(b, ba)
l/2a < c(a,
|aè|
first
ba).
that
that
c = c(a,
b) by (5.6b(i))
Then
Ao. < c(a,
c(a, a) = c(a, ~aS) which
yields
in (5.6b(i))
in the second
that
ba).
Axiom
are
pair after
is
\aba~\ <
b) = d by (5.6b(i))
b) < c(a,
But then
and
A4 implies
|a¿>| < a, a contradiction.
suppose
aèaoccurs.
and c <Act,
But then
ab)
for c < d < Ace.
the results
Now c(a,
c(a,
= |a| - \b\ + 2d.
|aè| = |a| - \b\ + 2c.
a b must occur.
c(b, è) =
c < d < A.a and the case
aba occurs,
> maximum S|a| , |aè| !. Secondly
Since
so
Suppose
so it follows
b) = c(b, ba ) so
\aba\
a b.
Then
by hypothesis,
Therefore
in the beginning
the details
c(a, V) > A\a\ > l/2a and
d <Aa
c(a, V) =
b) = c(a, V).
d = c(b, T) < c(a, b) = c, so A4 implies
d < c < A<x are similar.
maximum{|a|,
suppose
c(a, V) > c(a, ~a)> c(b, V), c(b, è) = c(a, b) by A4.
of 5.7.
Since
c(a,
99
REAL LENGTH FUNCTIONS IN GROUPS
A,
implies
Then Aa
c(a, b) =
\ab\ < a, a contradiction.
so c(a, a) < c(a,
\ab\ = \b\ - \a\ + 2c.
aè).
Axiom
Note that
|aè|
Therefore
A4 implies
> a but 2c
< a so |a| < \b\.
It remains
to prove
that
c(ab,
a~b) = d.
We first
Compare c(ab, a) with c(b, a) which by (5.6b(i))
then A4 implies
which
is impossible.
earlier
If c(a,
in the proof,
yields
that
c(ab,
a) = d.
c(a, b) = c(b, ab) so \ab\ = \a\ - \b\ + 2d. But then \ab\ < a
so \aba\ = 2|a| - \b\.
maximum S|a|,
establish
is d. If c(a, b) < c(a, ab),
ab) < c(a,
But since
that
\ab\ \, c(a,
b), then
A4 implies
|a| < \b\ this contradicts
\aba\
> maximum l|a|,
a~b) > minimum i/4|a|,
c(a,
ab) = c(ab,
the fact,
\ab\\.
Since
b)
established
|a b\ <
A\ab\ \> Act> d.
Axiom
A4 then
c(ab, alb) = d.
Proof
of 5.8.
a b must occur
Since
d < c, the results
in the term following
of (5.6c)
(a/ab).
c(a, ba) > minimum \V2\a\,y2\ab\\>A.CL'>d.
implies
Secondly
d = c(b,
assume
A . yields
ba)
so
aba
|aè|
Suppose
Since
Then
which
d < c(a, ba).
as seen
above
One of aba
aba
occurs.
aba,
A4
\ab\ < a, a contradiction.
Since
implies
and
Then
c(a, b) = d by (5.6c),
= |a| - \b\ + 2a". But then
occurs.
d = c(b, ba)
hold.
c(a,
b) = d by (5.6c),
a contradiction.
Therefore
a b must occur.
Now consider
And
c(ab,
aè).
c(a, b) = d by (5.6c).
in the sequence
Since
|a| < \b\ and
Therefore
c(a, ab) > minimum\A\a\,
A4 implies
A\ab\\
|«¿>| > a>
c(ab,
2d, c(ab,
a) = d.
> Y2a> d.
Since
b) > d.
a b occurs
An application
of
A4 yields c(ab, a~B)= d.
Finally
consider
\ab\.
If c > A\ab\, then
\ab\ < \a\.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
And if c <l/2\ab\,
then
100
NANCY HARRISON
c = c(a, a) < c(a, aV) so A4 implies
Proposition
5.9.
Let
c = c(a, «¿).
(x ., y .) be an infinite
Assume
that the limit numbers
for this sequence
number
a. and that no element
which
Furthermore
length
assume
of all
the
that for all
possible
[December
occurs
i, (x.
Thus
\ab\ = \b\ - \a\ + 2c.
reduction
in this
sequence
., y . ,) = (x./z)
choices.
Then
sequence
are both equal
is cyclically
where
limit.
for x, y £ G.
to some nonzero
z has
reduced.
the shortest
c(y , y .) = % a =
limit. I — oo c(x.,i' x.).
i'
Proof.
that
Fix
\b \ — a<
of generality
where
0 < e < %cl.
e. Let
assume
(a1, b') be a pair
(a, b) as the
c = c(a, ä) and d = c(b, b).
\ab |, |flt?|}.
By our assumption
It suffices
thus
that
Without
is the second
that
such
loss
this pair
we will refer
(a/ab)
to prove
assume
sequence
(a1, b').
For convenience
(72 + l)th pair;
2( < 2d < a+ e. For the time being
in our reduction
pair beyond
ab is in the next pair.
\ab\ < minimum [|aè|,
pair following
Let
(a, b) be an arbitrary
is (a/ab)
to the 72th
pair.
a-2e<2c<a+(
Let
and
a —
d < V2a.
Case 1. c < d.
By Lemma 5.7, 2c = \ab\ + \a\ - \b\. Since
+ e, 2c > a-
e. Hence
a> 2d > 2c > a-
|<2¿>|> a, \a\ > a, and \b\ < a
e.
Case 2. d < c < V2a,.
By Lemma
with
c(ab,
5-7, c(ab,
~ab) = d.
If \ab\ < \a\, then the second
ab) < c(a, a) < V2<x. Thus
we are in Case
\ab\, then consider the third pair (a/bS).
= d.
Since
If \aab\
< \a\, we are in Case
Proposition
4.7 implies
that 2c and 2d are within
'a'
1.
1 and hence
is (ab, a)
done.
If \a\ <
By Lemma 5.7, b- = aab and c(b-, bA
Otherwise
becomes
pair
consider
a 'b.',
we can
the fourth
conclude
pair
(a/b.).
by Case
1
so that
2c =
2e of a.
Case 3. c = d.
(3.1)
a b occurs
Then
c(cz, a) < c(a,
\a\ + \ab\ - \b\.
in the third
ab).
Since
pair.
Axiom
\a\ > a,
A4 implies
\ab\ > a, and
c(a, ä) = c(a,
ab),
\b\ < a + e, it follows
that
2c > a
- t.
(3.2)
aba
occurs
in the third
Then
c(a, a) < c(a,
ba).
A4 so 2c = \b\ + \ab\ - \a\.
pair.
Since
This
(5.6a)
implies
c(a,
is a contradiction;
b) = c, c(a,
2c < a but
b) = c(b, ba) by
\b\ + \ab\ - \a\
> \ab\ > a.
(3.3)
aba
Then
c(ab,
occurs
third pair is (a/aba),
aba
\aba\
, then
in the third pair.
a) > c(a,
a), so
c(ab,
so a coordinate
c(a, ä) < c(aba,
First
of the fourth
"a). Axiom
+ \a\ - \ab\ > <x- e. Suppose
a) = c.
A
implies
a ba occurs
assume
pair
is aba
c = c(aba,
in the fourth
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
\a\ < \ab\.
Then
or a ba.
the
If it is
a), so that
2c =
pair.
c(c7, a)
Then
1972]
< c(a, aba),
need
101
REAL LENGTH FUNCTIONS IN GROUPS
so A4 implies
to consider
\ba\.
by (5.6a),
M > \b\ implies
assumption.
Therefore
|aèa|
< \ab\, c(ab,
c = c(a, aba).
Let
M = c(a,
c(a,
Hence
b) >A\b\.
Thus
Since
But then
M < \b\ so c(ab,
a) > A\a\.
2c = |a| + \aba\ - \ba\.
b) + c(b, a).
c(b,
|aè|
b) < c(a,
< |a|,
contrary
a) = c(b, a) by Proposition
c(b, a) > A\a\,
so
|èa|
We
b) < c(a,
to our
2.4.
< |è|.
b)
Then
Since
2c > |a| +
\aba\ — \b\ > a - e.
Secondly
assume
pair is (aba/ab).
|(arV)"|
Let
\ab\ < |a|.
Since
are unbounded
Then the second
\(ab)"a\
> |(ai))"|
by Proposition
m be the positive
integer
(ab, a), (ab, (ab)la),
- |a|
pair is (ab, a) and the third
by Proposition
2.6, the lengths
such
|(af3)"a|
that the sequence
(ab, (ab)2a),
(ab, (ab^a),
...
, (ab, (ab)m~la),
say the Rth term, is not (ab/(ab)m+
in the
(ab)maba,
which
case
that
c((ab)ma,
c((abYa,
ab) > c(a,
ab), A4 implies
- |a6| < 2c.
Since
|(aè)'a|
Suppose
c(a, a) < c(ba,
that
c(ba,
c.
Let
(ab)l~
Finally
0 < t < m.
a) < c(a,
Since
are similar
of the former.
c((a,
Assume
Since
c(a,
By A4,
a).
c((ab)m
Then
let
z = (ab)maba
t be the first
integer
c((ab)la,
Hence
c((ab)ma,
c((abY~
Then
for both of these,
a occurs
that
c((abYa,
is
in the
it is less
in the sequence,
- |aè|
> a-
(.
Secondly
integer
((ab)ma/(ab)m
(R + l)th
than
+ la).
the details
term.
or equal
c((ab)ma,
suppose
result,
or (ab)m + 1a(ab)ma.
Con-
to
(ab)m + 1a) >
+ 1a, (ab)m + 1a) = c((ab)m + la, (ab)ma)<2c.
/ be the smallest
ab)
ab), AA im-
we will only present
a occurs
first
+ ia
If
c. Assume then
our desired
the Rth pair
e as
c(a, a).
< c((ab)'a,
yields
pair is (ab)ma(ab)m
Suppose
such
\(ab)la\ + |(a¿)'-1íj|
such that
a, ab)
this
z
in
ab) > c(a, a)
ab) >Acl>
integer
As above
that_(a¿>)ma(a¿)CT+
(ab)ma(ab)m+
|(aè)m + 1a| + \(ab)ma\
c(a, a).
Since
(R + l)th
b)m+ a, (ab)m+a).
c(a, a).
case
z = (ab)m + la.
of the
The new element
in Case 3 (3-1) yields 2c > a-
a) = c.
(ab/(ab)ma)
a| > a, and |aè| < a+ e, 2c > a-
in which
t be the first positive
suppose
the arguments
Let
a).
Suppose
(abY~1a)<c.
> a, \(abr~
The new coordinate
sider
a).
a.
0 < t < m. Since
c((abYa,
a), then the argument
a) < c(a, a).
c((ab)'a,
2c > a-
or (ab)m+
a) = c(ab,
z = (ab)maab
> c(a, a) and note that
plies
(ab)maab,
ab) > c(a, a) and note that
>c((ab)l~la,
desired.
is
are unbounded.
from (a, b) is as follows:
and the next term,
Rth pair
2.1 and the lengths
Thus
2c>
c((ab)m + 1_a,_(ab)m + 1a) >
such that
c((ab)la,
(abYa)
> c(a, a).
Note that 1 < t < m + 1. Let T = c((ab)'a,
c((ab)'~la,
c((abY~
(abY^)
a, (abYa).
(abY~rJa). If T > c, then since
< c < T, A4 implies that c((abY~la, (abY~la) =
Hence
2c > \(ab)'~la\
If T <c, then since c < c((ab)'a,
Hence
2c > \(ab)'a\
have our result.
c(ä, ab).
+ \(abY~
- |aè| > a-c
as desired.
(abYa), A4 implies that T = c((abYa, (abY~la).
a\ - \ab\.
We claim that this
+ |(af3)'a|
If |èa| < a+ 2f, then 2c > a-
is always
the case.
If c(â, b) < c(ä, ab), then A4 implies
Compare
2e, and we
c(a, b) and
c(ä, b) = c(b, ab). Since c(b, ab)
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
102
NANCY HARRISON
> Yi\ab\, it follows that \ab\ <\a\
[December
+ \b\ - \ab\ < a. + 2e. If c(«, ab) < c(â, b), then
\ab\ < \b\ + \aba\ — \ab\ < a + 2e.
Case 4. d < V2a< c.
By Proposition
(a, b ), (a1, a).
we see that
(a'/a'b').
4.7 our sequence
b.) = d = c(a',
Compare
c(a',
c(a',
a') = c(a',
A4 implies
If c(a',
a1).
Assume
aT) = d and c(a\
a'b')
the pair
'a'b').
a'b').
Therefore
< c(a',
a'), then
following
It c(a',
a'b')
Since cÇF, aTbt) < d, 2d > \a'\ + \a'b'\ - \b'\ > aa'b').
a'b'a'.
\a'a'b'\
(a'/a'b1).
If the latter,
(a'/a'b')
> maximum !|cz'|,
Hence
c(a'b',
is a pair such that
Case
3.
Assume
a'b'a'
|a'¿>'|},
the new element
a') > d.
TPT'), then
- \b | > a-
so
t and
\a'a'b'\
=
e. Suppose c(a', aT) =
so a'a'b
is not a term in the
in this pair
Axiom A4 implies
d = c(íz', fl7) = c(a'b',
occurs
ô7) < c(a',
= c(äT, a'b')
\b'\.
Then
(a1, b ) is
2d = \a'\ + \a'b'\
c(a',
c(a',
pair following
••• ,
Let a = b . By Lemma 5.7 if 2c = a and by Lemma 5.8 if 2c > a,
c(b.,
we are done.
has the form (a, b), (a, bA, (a, bA,
is either
a'b'a'
or
c(a'b', a'b') = d. Then
IFtí') < Via. and we are done by
in the pair following
(a'/a'b1).
Consider
c(a', ~P) + c(b\ aT). Since c(a', tí7) > xA<xand c(b', b7) > Via, A4 implies
c(a',
b') > xAa. This together
if c(a',
b')
< \b'\ - d, then
Y2a, contrary
Thus
if d < Y2a, then
d > V2a. Then
becomes
In the proof
2.4 implies
c(a'b',
< maximum {|a'è'|,
> \a'b'\
c and
c(a',
c(a,
c(y ., ~)
a') = c(b',
|ß'|}.
a')
Now
= d <
And if c(a',
b')>
> a, a contradiction.
a) = c <V2oi.
= limit.
(x, y)
b') = d.
d ate within a couple of i's of V2a.
a— 2( < 2c by the above
that
of a product
Proposition
both
a') = d < V2a yields
a< 2d < \b\ < a + e. If c > Ma, then
Assume
a 'b',
proof that limit.
length
\a'b'a'\
2z¡?> \b'\ - \a'\ + \a'b'\
and we are done.
'a'
c(a',
Proposition
to the fact that
\b'\ - d, then
assume
with
Since
Proposition
arguments.
This
Now
a< 2c < \a\ < a+
4.7 implies
completes
e
every
the
c(x ., ~x1)= Via..
is abelian
we need
a proposition
concerning
the
of elements.
5.10.
// a. £ G such that
\a \ > 2d and c(a., a. A > ¿ iOT \ <
z < 72, then
\a,a' l
Proof.
1. Assume
. < • a n |<2<i
' —
¿
We will prove
this
1Í1Í77
by induction
it is true for T2- 1 elements
> d,' then \a,a• ■• a «1\ —
< \a,
a- ...
' l z
1 1 l
applied to \a,a-...a
on 72. The result
where
is obvious
for 72=
72> 1. If c(a,a• • • a n- ,,
an )
i l
I
« n— ,l1I + \a
I - 2d. ByJ the induction hypo' n'
Jr
—
thesis
+ 72maximum
\\a
. - 2d\.
, .
' z'
.1,
\a, a-, ••• a I < 2d + (n - 1) maximum lla.l - 2d\ + \a I - 2d
1 < 1 sn—1
< 2d + n maximum! |a .) - 2(i}.
l<!<n
'
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1972]
103
REAL LENGTH FUNCTIONS IN GROUPS
If c(a,a-,
12
...
a n—i ,, a n ) < d, then A.4 implies
c(a.a-,
• • • a n—I,, a~)
=
r
1 2
n
c(a,a~.
••• a n-l ,, a 7Z-1 ,) since by' hypothesis
c(a n-l ,, a-)7! —
> d. Thus |a,
an
1 2
/r
112
a 72—1,a| n< = la.a.,...a
-I
+
la
|
la
,|.
By
the
induction
hypothesis
applied
112
n— 21
i n1
' n—1'
'
'r
rr
to |aja2
...
an_2|,
la,a^ ... a I < 2d + (n - 2) maximum! la .| - 2a"!+ (la I - 2d) + (2d - \a
112
„l -
l<,<„-2
l
< 2d + 72maximumS|a
Proposition
5.11.
Proof.
Let
reduction
n
length
Assume
for all
of all the possible
then
choices.
for x, y £ G with
*
J
v y ■ ¡) = (x./z)
where
In view of Lemmas
is lost in assuming
that no element
cyclically
reduced.
n, let
which
with
is abelian.
sequence
i, (x.
generality
For each
for x, y £ G is infinite
(x, y)
reduction
J n
a= ß > 0.
sequence
but not zero,
(x , y ) be an infinite
limit numbers
shortest
'
// every
that are equal
, |)
"
| - 2a"!.
Isis*
limit numbers
"
occurs
z has the
5-4 and 5-5 no
in this
sequence
is
T n = S[x
, y ], [xn , yn'
y ], [x
, y ], rx
, y 1!
"- rz' ' n"
L n' yn"
L n' yn1'
where
[g, h] = ghgh.
else
Fix
for an infinite
Define
c
infinite
number
n
Assume
= c(x , x ) and
n
\bab\ - \b\.
n
of 72. Let
(a/ab)
assume
c > 0.
number
First
d
Since
n
J n
then
commute.
\baba\
ba).
Then
i^|a|;
(x, y)
that
implies
are
If c(ab,
Proposition
only finitely
these
such
ba) < c(ab,
a), then
£. Now assume
of these
ba).
From the former
\b\. And from the latter
Combining
(5.6c)
these
implies
c(ab,
occur,
foi an
\b\ < a+
e.
e. Now
Since
c(a, b)
Now if c(bab, a) <
then
lèaèal
> a.
Suppose
bab = a so a and
b
pairs
(x , y ) with
c.
Assume
> \ab\ + \aba\
ba) < c(ab,
so c(ab,
a) = c(a, ba) by A4, so
= d .
the next
- \a\ > a-
a), then
a) < c(ab,
c
that
\abab\
e
>
ba) and c(ab,
a)
|a¿>7i| = \ab\ - \ba\ +
c(ab, a) = c(a, ba) by A4 so \aba\ = \ab\ - \ab\ + \b\.
we get that
|aè|
= |at?| - \ab\ + \ba\.
|ar>| = |a| + \b\ + 2c - 4d.
\ab\ > \b\, a contradiction.
n
commute.
\abab\
a-
= d
|¿>aea| > \bâ\ +
b) = c(b, ba).
\b\ < a+
|a| < a+ e. If c(ab,
< c(ab,
V), then
a) < i^|a|,
|aè| > a, |at?ä| > a, and
n
or
e.
n'
\b\ < a+ f, |i>ac7a| > a-
many
that
'
c
c(a, a) = c(b, b) and
2.5 implies
as
neither
n
IK I > a-
for now that
c(ba,
= (a, b), x and y also
there
n
x and y commute
such that
It follows that |i?aè| = |a|.
then
(a, b) be a pair beyond
pair is (a/ab).
A.
either
£ Y
ba) < c(ba,
> \bab\ > a; and if c(bab,
than
Since
Now assume
Let
J n
If c(ba,
that
Assume
(a, b) be a pair with
= c(a, b) by (5.6a), \ab\ = \ab\.
A\a\,
K
|è5| > a, \bab\ > a, and a<
b) < c(ba,
both are greater
exists
= c(y , y ).
is the next pair.
c(ba,
we prove
of 72 there
Therefore
\ab\ = \a b\ by (5.6b), |aè| = \ab\.
But since
If c(a, ä) > c(b, b), then
d < c < A\a\,
c(a, a) < c(b, b) in this
it follows
situation.
that
Since
It follows that (5.6b(ii)) or (5.6b(iii)) must occur.
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104
NANCY HARRISON
In both of these
c(a,
is always
(a, b) beyond
c(ab,
a pair
ba) < c(ab,
exists
b) < d, so
a) where
\ab\ > |fl|.
(x„,
(a/ab)
But this
contradicts
number
for which
to see that
either
(x , y ) with
given
c(ab,
ba) < c(ab,
(a, b).
72> M, \y
M there
4.7.
Therefore
of rpairs
either
a) or
For if not, there
, | > |x |.
The sequence
(a, yM), (a, yM+1), {a, yM + 2), • • • , where
Proposition
are an infinite
It is easy
is the pair following
M such that for all pairs
from (xM, yM) on has the form
there
y„)
[December
a = x^.
x and y commute
(x n , Jn
y ) in our reduction
sequence
'
or else
such
that
\K
I > a- £ for some Kn e Yn .
1 «'
Suppose
r r
> a-
f.
for an infinite
(g, tí) to obtain
and Y(g',
select
number
As we previously
(g\
noted,
h ) there
is a one-to-one
h ) such that corresponding
a subsequence
Consider
such
of these
a sequence
is some
K n £ Y n such
transformation
correspondence
elements
L , . ...
where
that
\K
I
' n '
is applied
between
are conjugate.
K. such that all elements
L,,
\L2\
< 2c'.I + 8(\b.\
- 2c')Z
1 Z ' —
' z1
> a-
of 72 there
when a Nielsen
to
r(g,
Therefore
tí)
we can
in it are conjugate.
Applying Proposition
c'. 1 = minimum \c(a., till a),
5.10 to \L. |, we get
c(b.,
b.)}.
Since
IL.I
'
z1
c,
\L2\
- 2c') - a+(.
1,11,1 - IL.I < 2c' ■ + 8(\b.\
i,i
By Proposition
5.9 as
que nee are
ß, \b.\ —» a as
a-
i —» 00, 2c' —> a.
as we want by choosing
number
case
A.
a=
implies
ß > 0.
number
indefinitely
limit
limit
Thus
There
Let
can be stated
L = 0.
the proof
for any reduction
can happen
pair
of the
sequence
'
x, y £ G.
It stops
in which
indefinitely
case
'
with
x and
limit
c(x , x ) = c(y , y~) = Via for all
n
sequence
in a
of rank 0, 1, or 2; it continues
|y 1=0,
case
with a reduction
of elements
n—001 •'n1
n
J n
pair of elements
reduced
G be a group
J n
from G generates
with a real length
free or abelian.
in a finite
are some results
Proposition
Therefore
completes
y are
J
rz—001
lx rz I' =
72, and x
either
a
subgroup.
5.12.
Nielsen
then
or it continues
an arbitrary
x, y £ G is either
be fully
that
n1
ly I = a> 0, in which
where
things
lx I = limit
n—ooi-'rz1
This
is abelian,
and we get a free group
rz— 001
for our se
c(x 1., x 1) = c(y ii., y.) = c(x, x).
and commute;
free or an abelian
L . are conjugate.
y commute.
with an arbitrary
reduced
Theorem
all the
the limit numbers
rr
we can make
\L . I - \L .1 as small
2
But |L.| - |L.| is some fixed
e small.
(x, y)
are three
of steps
and y commute.
that
if
2.5 forces
there
with
'
cyclically
11
2.8 since
Note that
(x , y ) associated
finite
and
L. - 1 so x and
(x 1., J1
y .) Proposition
r
In summary
i —»00. Thus
i large
L by Theorem
Axiom
And since
3.7, without
of this
And if (x, y)
number
of steps.
theorem
that
adjusting
function.
the length
is free,
Then
then
(x, y)
\x, y\ can
are worth
mentioning.
First
function
for cyclic
reduction,
as follows:
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
note
1972]
105
REAL LENGTH FUNCTIONS IN GROUPS
Proposition
nontrivial
3.7.
elements
Let
G be a group with a real length
of a commutative
maximum
\c(a,
subgroup
b), c(a,
function.
If a, b, c are
H of G, then
b)\ = minimum
j|a|
- d, \b\ - d\
and
minimumSc(a, b), c(a, b)\ = d
where
d = c(c, c).
Proposition
5.13.
are nontrivial
then
y'
and z'
Proof.
By Theorem
G be a group with a real
of G such
that
y'
5.12 we can fully
y'
y and
and
z.
z'
Let
and
and
z')
M , O AL where
72 is any integer.
that
|xn| > |y|
Similarly
and
|x"|
y)\
that
Sc(x",
But
This
element
proposition
5.14.
since
abelian
Proof.
do not commute.
contradicts
y and
x commutes
of Proposition
with
x and
y, and
x £ AL Pi AL, xn £
2.6 we can choose
Proposition
72
3.7 implies
say
c(x",
y) = \y\ - d.
say
c(xn,
z) =
so c(y, z) < minimum \A\y\,
%|z|!.
It
Now
c(z,
by Lemma
x") > A\z\
3.2 applied
by Lemma
to M ..
3.2 applied
of a nontrivial
y would
abelian
subgroups
of a
then AL O AI = {1¡.
Then there
There
of AL.
maximal
exists
exists
some
with every
element
be an abelian
Now x and
By Proposition
x £ Al o AL such that
z £ AL such that y and z
of AL, then
the subgroup
containing
AL.
group
y commute,
x and
5.13 we know that this
But this
z commute,
but
is impossible.
A4xn A12= {lj.
Actually
maximal
pose
by AL and
Since
to the fact that the centralizer
For if y commuted
the maximality
containing
z.
d - c(x, x).
Since
x must commute
subgroup.
AI. n AL ¡^ {lj.
z do not commute.
Therefore
commute,
we have our contradiction.
y£Al,-(A!,nAl2).
of G generated
If x, y', z'
a contradiction.
to jy, z\.
that
subgroup
x and
// AI. a72a' AL are distinct
Suppose
Let
z'\
\y\ - d > A\y\
c(y, z) = c(z, x").
Thus
and derive
z)\ = \z\ - d; fot definiteness
reduced,
group G with a real length function,
x 4 L
abelian
Let
is equivalent
is a maximal
Corollary
function.
and x and z'
|y| - d; fot definiteness
z), c(xn,
c(y, z) < A\z\.
\y',
it follows
In view
> |z|.
c(y, z) < c(y, x")
Axiom A then implies
to AL.
reduce
containing
equals
Now Sy, z\ is Nielsen
follows
do not commute
= (y, z),
subgroup
y), c(xn,
maximum
\z\ — d.
z'
M , be the maximal
abelian
such
length
x and y' commute
Nielsen
(y;,
AL the maximal
maximum \c(xn,
that
commute.
We assume
with both
both
Let
elements
this
abelian
x, y, and
corollary
subgroups
is equivalent
of G are equal
z are nontrivial
elements
to Proposition
or have trivial
of G such that
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5.13.
Assume
intersection.
xy = yx and
that two
And sup-
xz = zx.
106
Let
NANCY HARRISON
M. be the maximal
containing
x and
z.
abelian
Since
subgroup
containing
x and
1 / x £ M. n /VL, M, = AL.
y and
AL that
Therefore
y and
z com-
mute.
There
are some more immediate
Corollary
disjoint
that
5.15.
union
Corollary
5.16.
trivial
Proof.
M
// G is a group
of its maximal abelian
no pair of subgroups
or has
consequences
//
with
of Proposition
a real
length
subgroups.
bas nontrivial
function,
(Disjoint
elements
5.13.
then
is used
G is the
here
to mean
in common.)
G is a group with a real length
function,
then
G is abelian
center.
Suppose
for z £G
G is not abelian.
to be the maximal
C is the center.
abelian
Let
x, y £ G such
subgroup
that
containing
xy j¿ yx.
C and
Then Mx í My but Mx n My DC. Therefore
\z\
Define
where
Corollary J 5.14 im-
plies C = 11}.
REFERENCES
1. Roger C. Lyndon, Length functions
in groups, Math. Scand. 12 (1963), 209—234.
MR 29 #1246.
2.
of groups
W. Magnus,
in terms
New York, 1966.
A. Karrass
and D. Solitar,
of generators
and relations,
Combinatorial
Pure
and Appl.
group
theory:
Math.,
vol.
Presentations
13, Interscience,
MR 34 #7617.
DEPARTMENT OF MATHEMATICS, BROOKLYN COLLEGE (CUNY), BROOKLYN, NEW YORK
11210
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