In Lang's Algebra book in page 25, we can find the proof of proposition (iv) which says "Let $G$ be a cyclic group of order $n$. Let $d$ be a positive integer dividing $n$. Then there exists a unique subgroup $G$ of order $d$."
The proof states that we are considering that $d|n$, $m = d/n$. Then, we also have the surjective homomorphism $f: \mathbb{Z} \rightarrow G$. Then, we know that $f(m\mathbb{Z}$ is a subgroup of G. However, the text claims that we have the isomorphism $\mathbb{Z}/m\mathbb{Z} \cong G /f(m\mathbb{Z}) $. Could someone explain me where this isomorphism comes from? I am quite lost in understanding where it comes from. Thanks!
Any cyclic group of order $n$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$. Consider the given map $f:\mathbb{Z} \to G$. $f(m\mathbb{Z})$= subgroup of $G$ generated by $[m]$, that is a cyclic subgroup of $G$ of order d. Hence $G/f(m\mathbb{Z})$ is a cyclic group of order $m$, and hence the isomorphism.