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Isi >lthen5(') is a direct product cf groups of order 3. This is an immediate consequence of the fo11owing two theorems

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- 4 -

b is a fixed e1ement of S, then 5(+,') is a (Z,p)··sernifie1d and -l -l -l

b+b = b .

And now we want to prove that if S(+,·) i, a (2,p)-semifie1d and

Isi > l then 5(') is a direct product cf groups of order 3. This is an immediate consequence of the fo11owing two theorems

= -

1

.

1S•

heN exists such that m h

that the coset m+(n)

THEOREM 5. S(+) is a group and a is its zero-e1ement.

k+1 k+1-1

PROOF. In fact for all beS one has b+b=b ·(l+l)=b 'a ; then,

. -1 2 p k+1 p k+ 1 p k+l k+'i

S1nce a =a =a , 4b=(b+b)+(b+b) = b ·a +b a =(b +b )·a =

=(b k+1)k+1. a-1. a = b[(k+l)2]. Now then, since the coset k+1+(n) is invertib1e in (~)(.), the e1ement m = (k+l)2 is such

invertible too. As a consequence an e1ement

h (m h)

(mod n), then 4 b = b = b. The conclusion now fo11ows in the same way as in the proof of theorem 2.

Q.LO.

THEOREM 6. The subset M coincides with S.

PROOF. In fact for al1 xeS one has:

2 2 2

l+x=a 'a+a ·a·x=a(a+a·x) = a·a·x = a ·x,

1+x=a.a2+x.a.a2=a.aP+x.a.aP=(a+x.a).a=x.a.a=x.a2

Then a~ is a centra1 element in sto) and hence a = (a2)2 is central too.

Q.E.O.

R E F E R E N C E

[1J A. LENZl Su cU

uruI

l>.tJu..Lttu;l.Q. .intJwdo:U.a. da. J. Szép

to be patilished.

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