By Finkenstadt B. F.

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2). It is also continuous, and has the continuous inverse s 7! 7. Let S be a locally compact semigroup with a completely simple kernel K. x; gyx 0 g 0 ; y 0 /: Then K is closed and topologically isomorphic to the Rees product X defined. G Y just 28 1 Semigroups Proof. 1. e0 se0 / 1 s/; s 2 S then «0 , the restriction of «00 to K, is its inverse. Let us first note that X and Y are closed subsets of S and G is algebraically a group. The set e0 Se0 D e0 Ke0 is closed, hence locally compact and a topological group by Ellis’ Theorem.

The usual inclusion relation is a partial order on F . Furthermore, any linearly ordered subfamily of F has a minimal element, since sets in F are compact. By Zorn’s lemma, there exists at least one minimal element (with respect to inclusion) in F . Call this minimal closed left ideal I0 . It remains to prove that I0 is a minimal left ideal of S . Suppose that I1 I0 , I1 ideal of S . Then for a 2 I1 , I1 so Sa is a closed ideal of S included in I . Hence, Sa D I0 and I0 D Sa I1 D I0 . The last argument also shows that I0 must be equal to S x for any x 2 I0 .

Let a D xay. By iterating this equality, we obtain a D x n ay n . For some n, x n belongs to a subgroup of S , indeed of S . Let e ¤ 0 be the identity element of this subgroup. 22 1 Semigroups Suppose S is not completely 0-simple. 15, e is the identity element of a bicyclic subsemigroup of S with generators p and q, say (pq D e; qp ¤ e). For some n, p n belongs to a group with identity f . Let r be the inverse of p n in that group: rp n D p n r D f . We have f e D fp n q n D p n q n D e and f e D rp n e D rp n D f , so f D e.