Bourbaki

Exercise 1, § 1

Algebra, VI

A not necessarily commutative ordered monoid M is a (multiplicatively written) monoid equipped with an ordering such that the relation x ≤ y implies zx ≤ zy and xz ≤ yz for all z ∈ M. Let G be a not necessarily commutative ordered group. Show that: a) If P denotes the set of elements greater than the identity e of G, then P . P = P, P ∩ P 1={e}^1 = \{e\} and aPa−1=PaPa^{-1} = P for all a ∈ G. Prove the converse. Find a condition for G to be totally ordered. b) If one of the elements sup(x, y), inf(x, y) exists, then so does the other, and

sup⁡(x,y)=x(inf⁡(x,y))−1y=y(inf⁡(x,y))−1x. \sup(x, y) = x (\inf(x, y))^{-1} y = y (\inf(x, y))^{-1} x .
c) The elements sup(x, e) and sup(x−1x^{-1}, e) commute. Two coprime elements commute. d) The subgroup G' generated by the irreducible elements of G is commutative. Deduce that Th. 2 of VI, p. 18 is still valid.

Solution

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