We know that a finite abelian group generated by two elements is isomorphic to $C_r \times C_s$, where $C_r$ and $C_s$ are cyclic group of order $r$ and $s$. I am interested to know what is the structure of a commutative finite semigroup generated by two elements.
Does someone have any idea, when commutative semigroup $S = \langle a, b \rangle$ will be the monogenic semigroup?
Let $M$ be a finite commutative monoid generated by $a$ and $b$. Then $M$ is a quotient of the monoid $\langle a \rangle \times \langle b \rangle$. Indeed, the map $h: \langle a \rangle \times \langle b \rangle \to M$ defined by $h(a^k,b^\ell) = a^kb^\ell$ is a monoid morphism, since, by commutativity, $(a^kb^\ell) (a^{k'}b^{\ell'}) = a^{k+k'}b^{\ell +\ell'}$.
Now suppose that $\langle a \rangle = \langle a \mid a^n = a^{n+r} \rangle$ and $\langle b \rangle = \langle b \mid b^m = b^{m+s} \rangle$. Then $h$ will be an isomorphism if and only if the elements of $M$ of the form $a^pb^q$ with $0 \leqslant p \leqslant n+p-1$ and $0 \leqslant q \leqslant m+s-1$ are all distinct.
This is not always the case. For instance, the monoid $$ M = \langle a, b \mid a^2 = a^3, b^2 = b^3, ab = ba, ab^2 = a^2b \rangle $$ is a quotient of $\langle a \mid a^2 = a^3\rangle \times \langle b \mid b^2 = b^3 \rangle$. However $M$ has 7 elements and is not monogenic.