Comment.Math.Univ.Carolin. 34,2 (1993)199–201 199
On a class of commutative groupoids determined by their associativity triples
Aleˇs Dr´apal
Abstract. Let G =G(·) be a commutative groupoid such that {(a, b, c)∈ G3;a·bc 6=
ab·c}={(a, b, c)∈G3;a=b6=cora6=b=c}. Then Gis determined uniquely up to isomorphism and if it is finite, then card(G) = 2ifor an integeri≥0.
Keywords: commutative groupoid, associative triples Classification: 20N02, 05E99
For a groupoidG=G(·) denote by Ns(G) the set of its non-associative triples, i.e. Ns(G) ={(a, b, c)∈ G3; a·bc6=ab·c}. If V is a variety of groupoids and S a non-empty set, then it can be a non-trivial problem to determine all suchN ⊆S3 that N =Ns(G) for a groupoidG =S(·) ∈ V. For example, it is known [1], [2]
thatNs(G)6={(a, a, a);a∈G}for any non-empty groupoidG.
In the present short note we investigate the case when V is the variety of the commutative groupoids andNs(G) ={(a, b, c)∈G3;a=b6=c ora6=b=c}. We shall show that all such non-trivial groupoids can be obtained by a slight modifica- tion of a 2-elementary Abelian group and that these groupoids are determined up to isomorphism by card(G). Moreover, whenever G is finite and non-trivial, then card(G) = 2i for an integeri≥1.
Note thata·ba=ab·afor anya,b∈GwheneverGis a commutative groupoid.
The set{(a, b, c)∈G3; a=b6=c or a6=b =c} thus covers all (a, b, c)∈G3 such that card{a, b, c} ≤2 anda·bc6=ab·c can occur.
Theorem 1. For an Abelian groupG(+)and each06=e∈Gdefine on the setG a commutative groupoidGe by0·0 =e, a·b=a+b anda·0 = 0·a= 0for any a,b∈G\{0}. If G(+) is2-elementary, then Ns(Ge) ={(a, b, c)∈G3;a=b6=cor a6=b=c}. Conversely, ifG(·)is a commutative groupoid wherea·bc6=ab·cif and only if a=b6=c or a6=b =c, and card(G)>1, then there exist a 2-elementary Abelian groupG(+)and an element06=e∈Gsuch thatG(·) =Ge. Moreover,Ge
is isomorphic toGf for any choice ofe,f ∈G,e6= 06=f.
Proof: Only the converse part of the theorem requires a proof. Let us hence as- sume thatG(·) is a commutative groupoid, card(G)>1 andNs(G(·)) ={(a, b, c)∈ G3;a=b6=cor a6=b=c}. AsGis commutative, we have
(1) a·ba=ab·afor anya, b∈G.
200 A. Dr´apal
Let a = bc, where a, b, c ∈ G are pair-wise distinct. If c 6= ab, then aa·b = (bc·a)·b= (b·ca)·b=b(ca·b) =b(c·ab) =bc·ab=a·ab. Hence c=ab and we have
(2) Ifa=bc,b6=a6=candb6=c, thenb=acandc=ab.
Further, we shall prove
(3) Ifa=bc,b6=a6=candb6=c, thena2=b2=c2 anda2∈ {a, b, c}./
To see this, observe thatc2=ab·c=a·bc=a2 by (2) and thata2=aimplies a·bb=a·aa=a=cb=ab.b.
If a∈ Gis such that a, a2, a3 are pair-wise distinct, we obtain from (3) a2 ∈/ {a3, a2, a}, a contradiction. Therefore it holds
(4) a=a2 ora2=a3 ora=a3 for anya∈G.
Leta, b, c∈Gbe again pair-wise distinct and a=bc. Thena6=a2 by (3), and a3=aimpliesa·bb=a·aa=a3=a=c·b=ab·b. Hence we have
(5) Ifa=bc,b6=a6=candb6=c, thena·a2=b·a2=c·a2=a2=b2=c2. We shall now order the setGbya < biffab=banda6=b. Froma < bandb < a it followsb=ab=ba=aand froma < b < cwe obtainac=a·bc=ab·c=bc=c.
Therefore<really is a (sharp) ordering ofG.
Let again a, b, c ∈ G be pair-wise distinct and with a = bc. If e < a, then ec=e·ab=ea·b=ab=c. Consequently, we have
(6) Leta=bc,b6=a6=candb6=c. Ife < a, thene < bande < c.
Conversely, suppose thata < e. Then b6=e 6=c, eb =ea·b =e·ab =ec and eb·c =e·bc= ea=e. Fromeb =c it followsec =c, e < c and by (2) and (6) e < a. Thereforeeb6=c. Ifeb,c,eare pair-wise distinct, thena < eimplies by (6) thata < c, a contradiction. It followseb=eand we obtain
(7) Leta=bc,b6=a6=candb6=c. Ifa < e, thenb < eandc < e.
For a, b ∈ G put (a, b) ∈ r iff a 6= b and a 6= ab 6= b, and denote by ∼ the least equivalence containing the relationr. From (6) and (7) we get by induction immediately
(8) Leta, b, e∈Gand leta∼b. Thena < eiffb < e, ande < aiffe < b.
Denote byEthe set of equivalence classes of∼. By the definitions of∼and<we have eithera∼b, ora < b, orb < afor anya, b∈G. Hence it follows from (8) that
<induces a linear ordering of E. Suppose that (E, <) has no maximum element.
Then fora∈Gwe can chooseb∈Gwitha < b,a2< b. Thenb·aa=b·a2=b= ba=ba·a, a contradiction.
LetU ∈ E be the maximum element of (E, <) and suppose thata, b∈U,a6=b.
Thena6=ab6=b, and we obtaina2> aby (3) and (5). This is a contradiction, and henceU contains exactly one element, sayu.
Foru6=a∈ Gwe have ua=u, and thusu=ua·a6=u·a2 provides a2 =u.
Thereforea < b < u would implya·bb=au=u=bb=ab·b, which contradicts our hypothesis. It follows that the equivalence∼has exactly two classes and by (7) we haveab /∈ {a, b, u} for anya, b∈G,a6=b, a6=u6=b. Moreover,u2 =aa·u6=
a·au=u.
Groupoids determined by their associativity triples 201 Put nowu= 0 and defineG(+) bya+ 0 = 0 +a=afor anya∈Ganda+b=ab fora, b∈G,a6= 06=b. Clearly,a+ (b+c) = (a+b) +cwhenever 0∈ {a, b, c}, and by (2) also whena=borb=c. Similarly,a+(b+ab) =a+a=ab+ab=ab+(a+b) fora, b∈G,a6=b,a6= 06=b. Finally,a+ (b+c) =a+bc=a·bc=ab·c=ab+c= (a+b) +cwhena, b, c∈Gare pair-wise distinct andc6=ab. It follows thatG(+) is a 2-elementary Abelian group and we see thatG(·) =Gu2.
References
[1] Dr´apal A., Kepka T.,Sets of associative triples, Europ. J. Combinatorics6(1985), 227–231.
[2] Dr´apal A.,Groupoids with non-associative triples on the diagonal, Czech. Math. Journal35 (1985), 555–564.
Faculty of Mathematics and Physics, Sokolovsk´a 83, 186 00 Praha 8, Czech Republic (Received September 11, 1992)
