# About automorphisms of commutative semigroups

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?

• The first occurrence of automorphism means automorphism of monoids and the second one means automorphism of semigroups, right? – J.-E. Pin Aug 19 '14 at 10:19
• Thank you for your comment. I don't understand what the difference is. A semigroup automorphism will always send the neutral element to itself if the semigroup is a monoid... – guest Aug 19 '14 at 10:23
• Right, sorry... – J.-E. Pin Aug 19 '14 at 10:30
• What is $m$ allowed to be in the definition of $S$? If it is any element in $M$, then what's the difference between $S$ and $P$? – James Mitchell Aug 19 '14 at 16:43
• @James m is anything in M. The difference is in the other component: n has to be positive in S and can be zero in P. – guest Aug 19 '14 at 19:46

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.