## Every infinite cyclic semigroup is isomorphic to the positive natural numbers under addition

Prove that every infinite cyclic semigroup is isomorphic to the semigroup $(\mathbb{N}^+, +)$ of positive natural numbers under addition.

First, let’s clarify some terminology. If $S$ is a semigroup and $s \in S$, then there is a semigroup homomorphism $\varphi : \mathbb{N}^+ \rightarrow S$ given by $k \mapsto s^k$. The image of $\varphi$ is called the cyclic subsemigroup of $S$ generated by $s$, and if $\varphi$ is surjective, then we say $S$ is cyclic with $s$ as a generator. If $S$ has an (unique) identity element $e$, then the homomorphism $\varphi$ extends to a mapping $\psi : \mathbb{N} \rightarrow S$ by $0 \mapsto e$ and this is also a semigroup homomorphism. The image of $\psi$ is called the cyclic submonoid of $S$ generated by $s$.

First, it is clear that $(\mathbb{N}^+, +)$ is an infinite cyclic semigroup.

Suppose now that $S$ is an infinite cyclic semigroup with $s$ a generator. Now $\varphi : \mathbb{N}^+ \rightarrow S$ given by $k \mapsto s^k$ is a surjective semigroup homomorphism. We only need to show that $\varphi$ is injective. To that end, suppose we have $\varphi(a) = \varphi(b)$ for some $a,b \in \mathbb{N}^+$, so that $s^a = s^b$. Suppose, without loss of generality, that $a < b$; say $b = a+k$, so that $s^{a+k} = s^{a}$. But now for all $t \geq 0$, by the division algorithm we have $t = qk + r$ for some $q$ and $0 \leq r < k$. Then $s^{a+t} = s^{a+qk+r} = s^{a+r}$. Indeed, for all positive natural numbers $\ell$, we have $\varphi(\ell) \in \{s^1, s^2, \ldots, s^a, s^{a+1}, \ldots, s^{a+k-1}\}$. In particular, $\mathsf{im}\ \varphi$, hence $S$, is finite- a contradiction. Thus $a = b$, and so $\varphi$ is injective. So $S \cong \mathbb{N}^+$.