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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?

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  • $\begingroup$ The first occurrence of automorphism means automorphism of monoids and the second one means automorphism of semigroups, right? $\endgroup$ – J.-E. Pin Aug 19 '14 at 10:19
  • $\begingroup$ 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... $\endgroup$ – guest Aug 19 '14 at 10:23
  • $\begingroup$ Right, sorry... $\endgroup$ – J.-E. Pin Aug 19 '14 at 10:30
  • $\begingroup$ 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 $P$? $\endgroup$ – James Mitchell Aug 19 '14 at 16:43
  • $\begingroup$ @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. $\endgroup$ – guest Aug 19 '14 at 19:46
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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.
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