About automorphisms of commutative semigroups

Question

Suppose that $M$ is a commutative monoid and that the product $P$ of $M$ and the nonnegative integers $\mathbb{N}$ with addition has no nontrivial automorphisms. The set $S$ of pairs $(m,n)$ in $P$ with $n>0$ is closed under addition. Can $S$ have a nontrivial automorphism?

Answer 1

The answer is no, here are the steps in the proof (the details are left to you):

• show that every automorphism $\phi$ of $M$ induces an automorphism $\overline{\phi}$ of $M\times \mathbb{N}$
• deduce that $\operatorname{Aut}(M)$ is trivial
• show that if $\phi\in \operatorname{Aut}(M\times \mathbb{N}\setminus \{0\})$, then $\phi$ maps $\{(m,n):m\in M\}$ to itself for every $n\in \mathbb{N}$
• conclude that $\phi$ is trivial.