On p.25 of the revised third edition of Serge Lang´s Algebra, he proves that any cyclic group $G$ of order $n$ has a unique subgroup of order $d$ for each $d\mid n$ and his proof is as follows:
Proof. Let $m=n/d$. Let $f:\mathbb{Z}\to G$ be a surjective homomorphism. Then $f(m\mathbb{Z})$ is a subgroup of $G$, and from the isomorphism $\mathbb{Z}/m\mathbb{Z}\simeq G/f(m\mathbb{Z})$ we conclude that $f(m\mathbb{Z})$ has index $m$ in $G$, whence $f(m\mathbb{Z})$ has order $d$. Conversely, let $H$ be a subgroup of order $d$. Then $f^{-1}(H)=\ell\mathbb{Z}$ for some positive integer $\ell$, so $H=f(\ell\mathbb{Z})$, $\mathbb{Z}/\ell\mathbb{Z}\simeq G/H$, so $n=\ell d$, $\ell=n/d$ and $H$ is uniquely determined.
My Question:
I understand the part of uniqueness, but I don't understand why $\mathbb{Z}/m\mathbb{Z}\simeq G/f(m\mathbb{Z})$. ¿Why $f^{-1}(f(m\mathbb{Z}))$ (which is the kernel of the homomorphism $\mathbb{Z}\to G/f(m\mathbb{Z})$ ) is exactly $m\mathbb{Z}$?, since $f$ is not necessarily injective, i don't see why this is true. It is clear that $f^{-1}(f(m\mathbb{Z}))=\ell \mathbb{Z}$ for some positive integer $\ell$ and that $m\mathbb{Z} \subseteq \ell\mathbb{Z}$, so $\ell \mid m$. I want to conclude that $\ell=m$ but i don't know how to proceed.
Before we justify why the kernel of $\mathbb{Z} \to G \to G/f(m\mathbb{Z})$ is $m\mathbb{Z}$, it is worth pointing out that $\ker f = n\mathbb{Z}$. To see this, we quickly observe that we must have $\ker f = k\mathbb{Z}$, for some $k$ and we will have $\mathbb{Z}/k\mathbb{Z} \cong G$ since $f$ is surjective. Comparing cardinalities we conclude $k = n$.
As you have pointed out, we already have $m\mathbb{Z} \subseteq f^{-1}(f(m\mathbb{Z}))$. Consider any $u \in f^{-1}(f(m\mathbb{Z}))$, then $f(u) = f(v)$ for some $v \in m\mathbb{Z}$. Since $f(u - v) = 0$, we get $u-v \in n\mathbb{Z}\subset m\mathbb{Z}$ and therefore $u\in m\mathbb{Z}$.
Thus, $m\mathbb{Z} = f^{-1}(f(m\mathbb{Z}))$, and $\mathbb{Z}/m\mathbb{Z} \cong G/f(m\mathbb{Z})$.