## Exhibit a semigroup which is not finitely generated and a semigroup which is finitely generated but not cyclic

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 $I$ be a set and let $\{S_i\}_I$ be a family of semigroups indexed by $I$. Then the usual cartesian product of sets $\prod_I S_i$ is a semigroup under the induced operation $(s_i) \cdot (t_i) = (s_it_i)$. Moreover, if each $S_i$ is a monoid with identity $e_i$, then $\prod_I S_i$ is a monoid with identity $(e_i)$, and if each $S_i$ is commutative, then $\prod_I S_i$ is commutative. Proof: We need only show that this operator is associative. Indeed, $((s_i) \cdot (t_i)) \cdot (u_i) = (s_it_i) \cdot (u_i)$ $= ((s_it_i)u_i)$ $= (s_i(t_iu_i))$ $= (s_i) \cdot (t_iu_i)$ $= (s_i) \cdot ((t_i) \cdot (u_i))$ for all $(s_i). (t_i), (u_i) \in \prod_I S_i$ as desired. Now if $e_i$ is an identity in $S_i$ for each $i \in I$, then given $(s_i) \in \prod_I S_i$, we have $(e_i) \cdot (s_i) = (e_is_i) = (s_i)$ and likewise $(s_i) \cdot (e_i) = (s_i)$. Finally, if each $S_i$ is commutative, then $(s_i) \cdot (t_i) = (s_it_i)$ $= (t_is_i)$ $= (t_i) \cdot (s_i)$. $\square$

Lemma 2: Let $I$ be a set and let $\{S_i\}$ be a family of monoids (with identity $e_i \in S_i$) indexed by $I$. Then the set $\{(s_i) \in \prod_I S_i \ |\ s_i = e_i\ \mathrm{for\ all\ but\ finitely\ many}\ i \in I\}$ is a subsemigroup of $\prod_I S_i$, which we denote $\bigoplus_I S_i$. Proof: In view of Lemma 1, we need only show that this set is closed under multiplication. Indeed, if $(s_i), (t_i) \in \bigoplus_I S_i$, then there exist finite sets $A,B \subseteq I$ such that, if $i \notin A$, then $s_i = e_i$, and if $i \notin B$, then $t_i = e_i$. In particular, if $i \notin A \cup B$, then $s_it_i = e_i$. Now $A \cup B \subseteq I$ is finite, and thus $(s_i) \cdot (t_i) \in \bigoplus_I S_i$. So $\bigoplus_I S_i$ is a subsemigroup of $\prod_I S_i$. $\square$

Now consider $\mathbb{N}$ as a semigroup under addition and let $S = \bigoplus_\mathbb{N} \mathbb{N}$. We claim that $S$ is not finitely generated. To see this, suppose to the contrary that $S$ has a finite generating set $A = \{\alpha_1, \ldots, \alpha_k\}$, where $\alpha_j = (a_{j,i})$ and $a_{j,i} \in S_i$ for each $i \in \mathbb{N}$. Now for each $j$, the set of all indices $i$ such that $a_{j,i} \neq 0$ is finite- in particular, it has a largest element with respect to the usual order on $\mathbb{N}$. Let this number be $N_j$. Now let $N = \mathsf{max}_j N_j$. For any index $k > N$, we have $a_{j,i} = 0$.

Since (as we suppose) $S$ is generated by $A$, every element of $S$ can be written as a finite product (or sum, as it were, since $S$ is commutative) of elements from $A$. But in any such sum, the $k$th entry must be 0 for all $k > N$. But $S$ certainly has elements whose $k$th entry is not zero, and so we have a contradiction. So $S$ is not finitely generated.

Now consider $T = \mathbb{N} \times \mathbb{N}$, where again we consider $\mathbb{N}$ to be a semigroup under addition. Using additive notation, $T$ is evidently generated by $\{(0,1), (1,0)\}$ since $(a,b) = a(1,0) + b(0,1)$. Suppose now that $T$ is cyclic- say $T$ is generated by $(a_0,b_0)$. Now every element of $T$ has the form $k(a_0,b_0) = (ka_0,kb_0)$ for some $k$. For example, $(1,1) = (ka_0,kb_0)$, so $1 = ka_0$ and $1 = kb_0$. In $\mathbb{N}$, we then have $a_0 = b_0 = 1$. But then $(1,2) = k(1,1) = (k,k)$, so that $1 = k = 2$– a contradiction. Thus $\mathbb{N}$ is not finitely generated.