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