Exercise 6, § 3
Algebra, I, 130
Bourbaki marks this one as harder than the rest.
An internal law of composition on is said to determine on a quasi-group structure if, for all , the left and right translations and are bijective mappings of onto itself. A quasi-group is said to be distributive if the law is doubly distributive with respect to itself.
(a) Determine all the distributive quasi-group structures on a set of elements for .
(b) Show that the set of rational numbers, with the law of composition , is a distributive quasi-group.
(c) In a distributive quasi-group , every element is idempotent. Deduce that, if E has more than one element, the law can neither have an identity element nor be associative.
(d) The right and left translations of a distributive quasi-group E are automorphisms of E.
(e) If E is a finite distributive quasi-group, the structure induced on every stable subset of E is a distributive quasi-group structure.
(f) Let E be a distributive quasi-group. If R is an equivalence relation which is left (resp. right) compatible (no. 3) with the law , the classes mod. R are stable subsets of E. If E is finite, all these classes are obtained one from another by left (resp. right) translation; under the same conditions, if R is compatible with the law , the quotient structure on E/R is a distributive quasi-group structure.
(g) Let E be a distributive quasi-group. The set of elements of E which are permutable with a given element a is stable; if E is finite, for all , (note, using (e), that there exists such that ) and when x runs through E the sets are identical with the equivalence classes with respect to a relation compatible with the law .
(h) If E is a finite distributive quasi-group and the law is commutative, the number of elements of E is odd (consider the ordered pairs of elements of E such that , where a is given).
Solution
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