# Direct product of two finite monogenic semigroup

Let $S_1 = \langle a \rangle = M(m_1, r_1) , S_2 = \langle b \rangle = M(m_2, r_2)$ are two finite monogenic semigroup, where $m_1, m_2$ are the index of $S_1,$ $S_2$ and $r_1, r_2$ are the period of $S_1, S_2$. We know that the direct product finite cyclic group $\mathbb Z_m \times \mathbb Z_n$ is cyclic if and only if $(m, n) = 1$.

What is the condition that $S_1 \times S_2$ is monogenic semigroup.

## 1 Answer

It is clear that $\mathbf S_1 \times \mathbf S_2$ is monogenic if either of them is trivial (that is, if either of them has only one element).
In case both are groups, then the result about cyclic groups that you mentioned applies.
In any other situation (that is, if $|S_1|, |S_2| > 1$ and at least of of $\mathbf S_1$ and $\mathbf S_2$ is not a group), then $\mathbf S_1 \times \mathbf S_2$ is not monogenic.

Indeed, suppose $\mathbf S_1 = \langle a\rangle$ is not a group (that is, $m_1 > 1$). Then $a^k=a$ iff $k=1$.
So, if $\mathbf S_1 \times \mathbf S_2$ were monogenic, it would be generated by $(a,b^s)$ for some $s$.
But then $(a,b^k) \notin S_1\times S_2$ if $b^k\neq b^s$, a contradiction because $|S_2|>1$.