Characterize the left ideal of a semigroup generated by a subset

Let S be a semigroup and let A \subseteq S be a nonempty subset. Recall that the left ideal of S generated by A is the intersection L(A) of all the ideals which contain A, and similarly for right- and two-sided- ideals.

Prove that the left ideal generated by A is A \cup SA and that the two-sided ideal generated by A is A \cup SA \cup AS \cup SAS.


Note that if L is a left ideal containing A, then SA \subseteq L. In particular, A \cup SA is contained in every left ideal which also contains A, and thus is contained in the left ideal generated by A. On the other hand, S(A \cup SA) = SA \cup SSA \subseteq SA \subseteq A \cup SA, and thus A \cup SA is a left ideal of S which certainly contains A. So the left ideal generated by A is contained in A \cup SA, so that L(A) = A \cup SA.

Similarly, if I is an ideal of S containing A, then A \cup SA \cup AS \cup SAS \subseteq I. So A \cup SA \cup AS \cup SAS is contained in the ideal \langle A \rangle generated by A. And on the other hand, since S(A \cup SA \cup AS \cup SAS) \subseteq A \cup SA \cup AS \cup SAS, this set is a left ideal, and likewise a right ideal. So A \cup SA \cup AS \cup SAS is a two sided ideal containing A. Hence \langle A \rangle = A \cup SA \cup AS \cup SAS.

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