Exhibit a semigroup which is not finitely generated and a semigroup which is finitely generated but not cyclic.
We begin with some lemmas of a more general nature.
Lemma 1: Let be a set and let be a family of semigroups indexed by . Then the usual cartesian product of sets is a semigroup under the induced operation . Moreover, if each is a monoid with identity , then is a monoid with identity , and if each is commutative, then is commutative. Proof: We need only show that this operator is associative. Indeed, for all as desired. Now if is an identity in for each , then given , we have and likewise . Finally, if each is commutative, then .
Lemma 2: Let be a set and let be a family of monoids (with identity ) indexed by . Then the set is a subsemigroup of , which we denote . Proof: In view of Lemma 1, we need only show that this set is closed under multiplication. Indeed, if , then there exist finite sets such that, if , then , and if , then . In particular, if , then . Now is finite, and thus . So is a subsemigroup of .
Now consider as a semigroup under addition and let . We claim that is not finitely generated. To see this, suppose to the contrary that has a finite generating set , where and for each . Now for each , the set of all indices such that is finite- in particular, it has a largest element with respect to the usual order on . Let this number be . Now let . For any index , we have .
Since (as we suppose) is generated by , every element of can be written as a finite product (or sum, as it were, since is commutative) of elements from . But in any such sum, the th entry must be 0 for all . But certainly has elements whose th entry is not zero, and so we have a contradiction. So is not finitely generated.
Now consider , where again we consider to be a semigroup under addition. Using additive notation, is evidently generated by since . Suppose now that is cyclic- say is generated by . Now every element of has the form for some . For example, , so and . In , we then have . But then , so that – a contradiction. Thus is not finitely generated.