Exercise 23, § 4
Algebra, I, 137
Bourbaki marks this one as harder than the rest.
Consider on a set E an internal law, not everywhere defined, written multiplicatively and satisfying the following conditions: (1) if one of the compositions is defined, so is the other and they are equal; (2) if are such that and (resp. and ) are defined and equal, then ; (3) for all , there exist three elements and such that , , ; is called the left unit of , the right unit of , the inverse of (by an abuse of language). (a) Show that the compositions are defined and that . (b) Every idempotent of E (§ 1, no. 4) is a left unit for all the such that is defined and right unit for all the such that is defined. (c) For the composition to be defined, it is necessary and sufficient that the right unit of be the same as the left unit of (to see that the condition is sufficient, use the relation ); if , then (the compositions appearing on the left-hand sides of these relations being defined). (d) For any two idempotents of E, let denote the set of such that is left unit and is right unit of . Show that, if is non-empty, it is a group under the law induced by that on E. E is called a groupoid if it also satisfies the following condition: (4) for all idempotents , is non-empty. (e) In a groupoid E, if , show that is a bijective mapping of onto a bijective mapping of onto and an isomorphism of the group onto the group . (f) Show that the law defined in § 1, Exercise 4(b) determines a groupoid structure on the set , if all the sets of the family are equipotent.
Solution
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