Proposition 5

Tag 0572, permanent. Cite it as [Bourbaki, Tag 0572].

Algebra, chapter I, § 2. Identity element; cancellable elements; invertible elements

If in a monoid x and x' are inverses and x commutes with y, so does x'.

From $x \top y = y \top x$ , we deduce $x' \top (x \top y) \top x' = x' \top (y \top x) \top x'$ and hence $(x' \top x) \top (y \top x') = (x' \top y) \top (x \top x')$ , that is $y \top x' = x' \top y$ .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Theorem 1, § 2
Corollary 2, § 4
Proposition 1, § 9

Cites

Nothing.