REAL HILBERTIANITY-TOTALLY REAL NUMBERS
11
without loss of generality assume that the map
(3:
K
~
K' is an inclusion of
fields. Hence
(3*
= resp. In the commutative diagram (2) we may replace
Aut(G) by
H,
so that h1 is an isomorphism. The assertions follow from the
commutativity of that diagram.
3. Group-theoretic lemmas
For a group
G
and r
2:
0 put
G
= G"-{1}. Let .C(G) be the nontrivial
conjugacy classes of G, and put l = I.C(G)I. For an r-tuple
lT
= (IT1, ... , O"r)
E
(;r
and
C E
.C(G), let
nc(tr)
be the number of indices, 1
~
i
~
r,
with
O"i
E C.
Then
L:ca(c) nc(tr)
=
r.
LEMMA
3.1.
Let G be a finite group. Every su/Hciently large multiple
r
of
4l
satisfies the following. Let E
E
Aut(
G)
be of order
2
and let
I
be involutions
in (GXJ(E))"-G withEE I. Let e = 8 ·IGI! if III
2:
2, and e = 0 if III=
1.
Put
m
=
r- e. Then there are sequences
tT
E
(;e,
T
E
(;m with these properties:
( 1)
E
-1 -1 -1
t"
h
1
.
a
O"i
=
0"1· ·
·O"i-llTi
!Ti-l· ··0"1 ,
10r eac _
t _
e;
(a2)
rJ
=
T~~l-j
for each 1
~ j ~
m;
(b)
I=
{E, EITI
1
EIT!0"2
1
••
1
EITIIT2 · ·
·O"e};
(c)
(tr,r)
E
£(r)(G); or (ITI,···
,O"e,Tl,···
,Tm)
=
G,
lTI···O"eTl···Tm
=
1,
and
nc(tr,r)
=
r/l
for each C
E
.C(G).
PROOF. We may consider E
E
Aut( G), lEI= 2, and one set I of involutions.
A. Separation of
tT
from r. Define an equivalence relation on .C(G): the
class [C) is
{c,c-1,C,c-}.
Part B constructs
tT
E
(;e with (a1), (b),
(cl) ITI · ·
·O"e
= 1; and
( d1) for each [C) there is
J.L[C]
2:
0 such that
nc (
lT)
=
4J.L[C]·
Observe that
e
=
L:cEC(G)
nc(tr)
=
4
L:ca(G)
J.L[C]·
Also, for each [C) let
V[c]
be a positive integer and m =
4 l::cEC(G) V[C]·
Part C shows there is
T
E
(;m
satisfying (a2),
(c2) (r1, ... , Tm) =
G
and TI · · ·
Tm
= 1, and
(d2)
nc(r)
=
4v[c],
for each
C.
Let
n
=
ft,
and assume
n
J.L[c]
for each [C). In the last step with
V[c]
=
n-
J.L[c],
e
+
m =
4
L
J.L[c]
+
4
L
V[c]
=
4
L
n
=
4ln
=
r.
CEC(G) CEC(G) CEC(G)
Substituting (c1), (c2), (d1), and (d2) in the expressions for (c) shows (c) holds.
In fact,
nc(tr,r)
=
4n
=
f
for each
C
E
.C(G).
B. Construction of
tT.
If I
= { E}, let
e
= 0 and
tT
= (). Otherwise put
lT
= (E1E2, E2E3, ... , EeEI), where E1 = E, E2, ... , Ee
E
I,
not necessarily distinct,
but E
1
-1-
E
2
-1- · · · -1-
Ee
-1-
E1. Then
tT
satisfies (a1) and (cl). Furthermore,
if
I
= { E 1, . . . , Ee}, then
lT
also satisfies (b). To construct such E 1, . . . , Ee, let
n' =
211
~{}!·
Note: n' is an integer divisible by 4, because II"-{E}I IGI. Let
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