I am reading something about abstract algebra. Can anyone please tell me if the following statements are saying: a cyclic group can only be isomorphic to either $\mathbb{Z}/n\mathbb{Z}$ if it is finite or $\mathbb{Z}$ if it is infinite?
Thanks a lot.
The group $\mathbb{Z}$ is the only infinite cyclic group, up to isomorphism. The group $\mathbb{Z}/n\mathbb{Z}$ is the only cyclic group of order $n$, up to isomorphism
Suppose that $G$ is a cyclic group, and let $g$ be a generator of $G$. Define a group homomorphism $\phi:\mathbb{Z}\to G$ by $\phi(n)=g^n$.
$\phi$ is surjective since $g$ generates $G$, and $\ker(\phi)$ is a subgroup of $\mathbb{Z}$, hence either $\ker(\phi)=\{0\}$ or $\ker(\phi)=n\mathbb{Z}$ for some natural number $n$.
In the first case we get $G\simeq \mathbb{Z}$, and in the second $G\simeq \mathbb{Z}/n\mathbb{Z}$.
Yes, your statement is correct with the 2 given statements in mind.
Related: Link 1
Also worth checking out (See Theorem 9.8) - Link 2
Yes, that is correct. If a group $G$ is cyclic, then $G$ is generated by (the positive and negative powers of) a single element $a$, i.e there exists an $a \in G$ such that $G=\{a^n: n \in \mathbb{Z} \}$. If $G$ is finite, then there exists an isomorphism from $G$ to $C_n$, and if $G$ is infinite, then there exists an isomorphism from $G$ to $\mathbb{Z}$.
