THE 3-TORUS IS KERVAIRE
5
of G1, using Lemma 2 to verify that the associated subgroups are each free of rank
2. It is clear that
G
is a subgroup of
G
0 ,
hence of G
1
and hence of
H,
and that the
three required equations hold in
H.
Proof of Lemma 2.
Sets=
lx- 1i
E
Go,
and
t
=
eyf
E
G
1
.
Then we may regard
G
1
as the HNN extension
with stable letter
t.
An easy argument, involving the normal form for words in an
HNN extension, shows that s,
t
generate a free subgroup of G
1
of rank 2 provided
the subgroup
s
has trivial intersection with each of the associated subgroups
as-
1{3
,
"(S8
,
and from the free product structure of
G
0
as
G
*
s it
is clear that this holds provided at least one of a,
{3
is nontrivial in G, and at least
one of"(,
8
is nontrivial in
G.
By hypothesis at least two of a,
{3, "(, 8
are nontrivial,
so
we must consider the case where either a = 1 =
{3, 'Y =1-
1
=1- 8,
or a
=1-
1
=1- {3,
'Y
= 1 =
8.
But then it again follows that
s, t
generate a free subgroup of rank 2 by
a similar argument, this time regarding G
1
as an HNN extension
The argument to show that
jxk,gy-
1
h
generate a free rank 2 subgroup of G
1
is
entirely analogous, and we omit it.
Theorem 3.
If
some face of the cube in Figure 1 has all four corner labels trivial
in G, then the system of equations has a solution in an overgroup of G.
Proof.
Without loss of generality we assume that the hypothesis of the Theorem
applies to the top face of the cube, in other words that a' =
{3'
= "(1 =
8'
= 1 in
G.
Define
G2 =
G,x,y
I
[jxk,gy-
1
h]
=
[lx- 1i,eyf]
= 1
.
Then G2 contains
G
as a subgroup, because the natural map
G ___.
G2 is a split
injection. A splitting can be defined, for example, by
x
~----
il, y
~----
hg.
If N
is the normal closure of
Gin
G
2
,
then
G
2
/N
~
Z
2
,
with basis
{xN, yN}.
It
follows easily that the subgroups of G2 generated by
{jxk, gy-
1
h}
and
{lx- 1i, eyf}
respectively are each free abelian on the given sets of generators. We can thus form
the HNN extension
which contains G2
,
and hence
G,
as a subgroup.
Note that we have the equation
axbycx- 1dy-
1
=
(hj)x(kh-
1
)y(g-
1
k-
1
)x-
1(j-
1g)y-
1
=1
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