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CLIFFORD

By Lemma 18(3), g=a*b=a*f. This gives the first equation, and the second is dual.

At this point we establish the Remark after Fantham^s Theorem.

LEMMA 26. Assume I, II, III*. Then (for x = ABC, CBA, ВАС, or CAB) equation (31-x) holds for arbitrary e >- f >- g in E and a in G if and only if equation (3 0-x) holds under the hypotheses stated for it in dition IV of Fantham^s Theorem.

Proof . As remarked after Fantham^s Theorem, only the implication (30-x) =» (31-x) requires proof. The proof is trivial in the case x = ВАС (and similarly x = CAB). For if e >- f >- g and a € G , then we apply (3 0-BAC) to e, fe, g to obtain

^fe e^^^ * g = (fe * a" )(^ * g)?

which reduces to (31-BAC) by Lemma 16(1).

The same argument shows that, to prove the tion (30-ABC) =Ф(31-АВС), it suffices to assume that ef = f >- g and a € G . Let us do so, and set

fj^ = a" 5* f and g^ = f^^g = (a" * f)g.

The hypotheses of (30-ABC) hold for e, f, and g-, ,

- 1 since (a * f )gi = g-i ь and we conclude from (30-ABC)

that

^f e^^^ *g^=a*g^=a* (a" * f)g. To establish (31-ABC), it suffices to show that

b * g = b * g^

where b = л^г (a) . г ,e

By Lemma 16(1), f^^f = f ^. Since e >• f, we see

that f^ ~ f, and hence f = ff^f = ff^. From Lemma

16 ( 1 ) and b € G^, we conclude that

b * g-, = b * fg^ = b * ff-jg = b * fg = b * g.

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