On the hypergroups with four proper pairs and without scalars

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We remember that a hypergroup H is a non-empty set equipped with a hyperoperation such that the following two conditions are satisfied: (1.1) V(x,y,z) E H3, (xy)z=x(yz) (associativity); (1.2) Vx E H, Hx=xH=H (reproducibility).
Given a hypegroup H, we say that a pair (x,y j ~ H~ is proper, if the hyperproduct x'y is not a singleton. Moreover an element x E H is said to be a left scalar (respect. right scalar), if (x,y) (respect. ( y.x)) is not a proper pair. An element is called scalar if it is at the same time left scalar and right scalar.
In [3], [4], [5], Freni, Gutan C., Gutan M. and Sureau Y. have determined all the hypergroups which have at the most three proper pairs. In the present paper the authors succeed in finding all the hypergroups with exactly four proper pairs and without right scalars. The corresponding left case can be c.btained for symmetry. (We remember that if the set of scalars is non-empty, then the set of left scalars is equal to the set of right scalars (see [3]). This case is handled in other papers from the same authors ).
Obviously we have that 3, 4}. otherwise there would exist at least one scalar.
In the rest H denotes a hypergroup without right scalars, P(H) the set of proper IThis work is produced by support of the Italian M.C R.S.T. (quota 40%). No  (II) 2. SOME PRELIMINARY RESULTS. -In this section, H will denote a hypergroup (finite or infinite). Every element a E S defines a map â:H~H such that ~ x ~ H, â(x) = a.x. This map is clearly surjective, in consequence of reproducibility.
We prove now : Proof.-Certainly there exist an integer n > 1 and an element bE S such that en . b = b and so , for ( Moreover, there exists such that the pair (b, y) is proper and consequently b ~ y ~ a ~ x . y implies that (x, y) is also proper. But H is thin to right and thus b = x. This proves the implication.
(j j ). There exist x and y such that bex.a and (b, y ) is proper. Now and so (x, y) is also proper, and thus b = x . Whence the thesis. We have to find, up to isomorphisms, all the hypergroups H={a,b.c}, having exactly four proper pairs and such that j E H there exists at least an element i ~ H for which the pair (ij) is proper.
We can suppose that the four proper pairs are of the following type: (a,m), (a,n), (b,p) and (c,q). We have, up to isomorphisms, only the following eight types, which are listed in the lexicographic ordering:  (3,) We will study separately the eight possible types. As regards the types (3~), {33~, (3~), one does not obtain any hypergroup. For everyone of the types (3,), (3~), (3~), one finds only one hypergroup. The type (31) gives rise to four hypergroups. The richest type of all is the type ~36), which we will study lastly. (bac) sc=br=bb=t=b then s=a.
(aca) ra=au, that is a=au whence u=c and b=r=ac=au=a, impossible.
There are no solutions.  We begin to observe that: a; a ; / b,c b,c b b,c ) I a / a / i c; b~c ! 3 a ) a t As regards the remaining cases s=b, s=c, one observes that by exchanging band c we obtain an automorphism for the type 36 and, by using the 2. and 4. above, we can limit ourselves to study only the following five cases, where s=b. The table summarizes in advance the results: v t u semi-hypergroups hypergroups The case (c,c,c) is isomorphic to C3. The last two cases C4 and C5 are symmetric one another.    (aba) ba=a. In the same manner, (caa) and (aca) give ca = ac = a . Now, (dac) , (dab) and (bad) give da ~ {b,d} and ad ~ c and so. by (1.2), M ~ {b,c,d}. Finally (daa) and (aad) give da = ad = a. It is easy to see that H' = {b,c:d} is a sub-hypergroup of H and this is absurd because in [3] one has showed that there are no hypergroups on H' of this kind.
Then there don't exist hypergroups.
We have a table of the following symmetric type:    permutations (a,b,c,d): (a,c)(b,d); (a,d,c,b) give tables of the same kind of (-~5). Let   In consequence of (2.9) and (2.6), (2.11) we have respectively the following statements:  ra=bX and therefore r ~ {b.c}. If r=b, then b=Nb and so N=={a,c}. Moreover (bccj implies that b=bQ, and thus Q={a,c}. For (1.2), But, if M={a,c}, then from (bbb) we obtaiñ b} ~ P={b}, which is impossible. Thus M=H. For ~1.?1, b ~ P. (cbc) implies that Pc=P and finally. being a P«c E P, P=H. So in case of r=b, one obtains the following hypergroup: H , a,c If r=c, one finds: (a,b) c Q. From (cbc) we obtain Pc=Q and Q=H. Moreover since |Mc|=1, |bc|=1, ) cc j > 1 , one deduces, for (2. 1 1 ), M= (a,b) . Finally we have, for (1.'i'), C E P ~ N; moreover the relations (bcb) and (bca) imply P=bP, N=bN. Necessarily it must be P=N=H. Therefore we have another hypergroup: In all, in this case there exist two hypergroups. If r=c, then s = a and t = b. This is impossible because (bba) gives a = b.
Therefore it follows r = b, s = c and t = a. %.%"e can suppose u = c (because otherwise, by exchanging b and c, %ve should obtain ba=c, that is the preceding absurdity) and so v = a and w = b. (ii) j S j =2; S={a,b}. For (2.6), we can suppose that a is a left scalar identity. ' Up to isomorphisms, we have the following 6 cases selected so that there are at least two proper pairs (d.x) and (d,y) and listed in lexicographical ordering.   Ultimately, the case S ~ =2, has given rise to eight hypergroups. At last, we study the case (iii) S ~ =1 ; S = ~a?. By (2.6), a is a left scalar identity.
There are, unless isomorphisms, the following seven cases selected so that every type contains the proper pair (b,a) and listed in lexicographical ordering: In the rest. these hypergroups will be called x-enlargements of (K, o ). Now we are ready t) prove the following result: (6.2) PROPOSITION.-In the cases (67), (69), (611), by putting x=d, one obtains all and alone the hypergroups, which are x-enlargements of the hypergroups K of size 3, with =3, ~ = S~(K) =1, j S(K)) =~, whose tables are known (see (3j).
Without loss of generality we can set cd=dc=d, (since otherwise, by putting cd=dc=c, one obtains hypergroups which are isomorphic to those ones in which cd=dc=d). D Ultimately, this case gives rise to the following four hypergroups: with M ~ {H,HB{d}}. Now we come to the case (6~). For ( ~.I~;~, we can limit ourselves to determine the products cd, dc. ( 1. ~) implies a E cb and so (cbd) gives cd=d. Finally (dcd) gives dc= d.
Ifr=c then N=P=H and s = d.
Proof.-  By turning to account again (2.14), the following one can be taken as initial table: