30

CLIFFORD

( 31 - ABC ) T)^^ (a) л g = a * (a"-^ * f)g , ex 5 e

( 31~CBA ) g * n^g g(a) = g(f * a"^) * a , (31-BAC) n^g g(a) * g = (f * a"^)(a * g), (31-CAB) g * n^xr ^ = (g * a)(a""-^ * f ) .

Since each of the hypotheses of IV(a,b,c,d) requires e >• f >• g, it is evident that (31-x) implies (30-x) for X = ABC, CBA, ВАС, and CAB. The converse will be proved in the next section (Lemma 26).

Proof of converse half of Fantham^s Theorem

Let S = и S^ be an orthogroup, and let us adopt aeY ^

the notation of the second paragraph of the present

section . Then (I) holds with E = E^ = U E^, and

^ aeY with G the maximal subgroup of S containing the

element e of E.

Define the operation * on S by (19). To show

( II ) , let e >- f (e,f in E) . Equations (28) are obvious.

Let e 6 E , f e Eo (a,3 in Y), so that a ^ 3. We are a p

to show (21)-(25) for all a,b in G^ and all f,g in

The elements af, a ^ f, and g all belong to the rectangular group S j , and af ^ a * f, so (af)gK^*f)g5 and hence

( a * f)g = afg = a * (fg).

Thus (21) holds, and (22) is its left-right dual. By the

same argument, f(a * g)^f(ag) and (fa)gPC(f * a)g.

Hence the idempotents f(a * g) and (f * a)g are both

equal to fag, and (23) follows. By Lemma 11,

a ( b * f)Ka(bf) = (ab)f, which implies (24), and (25) is

its dual.

Now define the mappings rixr^*G-^G^(e>-f inE)

X , e e X

by (20). Let S be represented as in Preston's Theorem, so that the product in S is given by (6) and (7). Let

with a > ß in Y. By (20) and (7),

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