Exercise 16, § 2
Algebra, I, 128
Bourbaki marks this one as harder than the rest.
Let be an associative law on a set and the set of cancellable elements of ; suppose that is non-empty and that every cancellable element is a central element. Let denote the set of subsets of with the following property: there exists such that .
(a) Show that the intersection of two sets of belong to .
(b) Let be the set of functions defined on a set of taking their values in and such that, for , belongs to . Let denote the relation between elements and of : "there exists a set such that the restrictions of and to are identical". Show that is an equivalence relation; let be the quotient set of under this relation.
(c) Let and be two elements of , and the sets where and are respectively defined; there exists and belonging to such that ; if is the restriction of to , show that the mapping belongs to and that its class (mod. ) depends only on the classes of and (and not on ); this class is called the composition of that of and that of ; show that the law of composition thus defined on is associative and has an identity element.
(d) For all , show that the left translation belongs to ; let be its class (mod. ). Show that the mapping is an isomorphism of onto a submonoid of and that, if , is invertible in (consider the inverse mapping of , show that it belongs to and that its class (mod. ) is the inverse of ). Deduce a generalization of Theorem 1 of no. 4.
Solution
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