Prove that a cyclic semigroup is either finite or isomorphic to $\langle \mathbb{N},+ \rangle$

Question

A semigroup $G$ is cyclic if $G$ is generated by a single element.

I know that a finite cyclic group generated by $a$ is necessarily abelian, and can be written as $\{1, a, a^2, . . . , a^n−1\}$ where $a^n = 1$, or in additive notation, $\{0, a, 2a, . . . , (n − 1)a\}$, with $na = 0$. Thus a finite cyclic group with $n$ elements is isomorphic to the additive group $\mathbb{Z}_n$ of integers modulo $n$. If $G$ is an infinite cyclic group generated by $a$, then $G$ must be abelian and can be written as $\{1, a^±1, a^±2, . . . \}$, or in additive notation as $\{0,±a,±2a, . . . \}$. In this case, $G$ is isomorphic to the additive group $\mathbb{Z}$ of all integers.

Q: Prove that a cyclic semigroup is either finite or isomorphic to $\langle \mathbb{N},+ \rangle$.

Answer 1

Hint and observations The standard proof of cyclic group is either finite or isomorphic to $\mathbb{Z} $ will work fine.

You have to note that a "cyclic semigroup" (denoted $ S $) can't be uncountable, just because there exists a generator $ g \in S$ together with the operation "power" ^ $: \mathbb{N} \to S \ \ n \mapsto g^n$ which generate all the elements. So they can't be more than countable.

Answer 2

In a semigroup, an element does not (necessarily) have an inverse, so it does not make sense to write $a^{-1}$. You have constructed an isomorphism from a semigroup ($G$) to a group ($\mathbb{Z}$) which is not possible unless $G$ is actually a group.

Now assume that $G$ is cyclic and infinite. Since it's cyclic, there is a $g \in G$ such that for each $h \in G$ there is $n \in \mathbb{N}$ such that $h = g^n$. Now define $\phi: \mathbb{N} \to G$ by $$\phi(n) = g^n.$$ This is a homomorphism since $\phi(n+m) = g^{nm} = g^n g^m$, so to prove that $\mathbb{N} \cong G$ we need to show that $\phi$ is bijective. From $G$ being cyclic with generator $g$, we already know that it is surjective (given $h \in G$ we know that there is an $n \in \mathbb{N}$ such that $\phi(n) = g^n = h$).

To prove injectivity, assume for contradiction that there is $n,m \in \mathbb{N}$ with $n \neq m$ such that $phi(n) = \phi(m)$, ie. $g^n = g^m$. Now without loss of generality we may assume that $m >n$ and we let $m = n + k$ for some $k \in \mathbb{N}$. Then

$$g^n = g^{n+k} = g^n g^k.$$

Note that multiplying by $g^k$ on the left hand side gives us $g^n g^k = g^n$, and on the right hand side gives us $g^n g^k g^k = g^n g^{2k}$. Thus we can multiply with $g^k$ as many times as we want without changing the left hand side, so for any integer $q \geq 0$ we have

$$g^n = g^n g^{qk} = g^{n + qk}.$$

Now by integer division with residue we have for each $l \geq 0$ that there are $q \geq 0$ and $r = 0,1,\ldots,q-1$ such that $l = qk + r$, so

$$g^{n+l} = g^{n + qk + r} = g^{n+qk} g^r = g^r.$$

Now this proves that $\phi(n+l) \in \{g^0, g^1, \ldots, g^{q-1}\}$ for all $l \geq 0$, so the image of $\phi$ is finite. We have already proved that $\phi: \mathbb{N} \to G$ is surjective, so this must imply that $G$ is finite which is a contradiction. So $\phi$ is injective and hence an isomorphism.