I am currently studying Serge Lang's book "Algebra", on page 25 it is proved that if $G$ is a cyclic group of order $n$, and if $d$ is a divisor of $n$, then there exists a unique subgroup $H$ of $G$ of order $d$.
I have trouble seeing why the proof (as explained below) settles the uniqueness part.
The proof (as I understand it) goes as follows:
First we show existence of the subgroup $H$, given any choice of a divisor $d$ of $n$.
So suppose $n = dm$. Obviously, one can construct a surjective homomorphism $f : \mathbb{Z} \to G$, and it is also clear that $f(m\mathbb{Z}) \subset G$ is a subgroup of $G$. The resulting isomorphism $\mathbb{Z}/m\mathbb{Z} \cong G/f(m\mathbb{Z})$ leads us to conclude that the index of $f(m\mathbb{Z})$ in $G$ is $m$ and so the order of $f(m\mathbb{Z})$ must be $d$.
Ok, so we have shown that a subgroup having order $d$ exists.
The second part is then to show uniqueness - and here is where I am lost as I don't understand why the following argument serves this end:
Suppose $H$ is any subgroup of order $d$. Looking at the inverse image of $f^{-1}(H)$ in $\mathbb{Z}$ we know it must be of the form $k\mathbb{Z}$ for some positive integer $k$ (since all non - trivial subgroups in $\mathbb{Z}$ can be written in this form). Now $H = f(k\mathbb{Z})$ has order $d$, and $\mathbb{Z}/k\mathbb{Z} \cong G/H$, where the group on the right hand side has order $n/d = m$. From this isomorphism we can therefore conclude that $k = m$. Here Lang ends by saying ".. and H is uniquely determined".
But why is this ? Does he mean uniquely determined up to isomorphism ? Because, what I think I have shown is that any subgroup of order $d$ must be isomorphic to $m\mathbb{Z}$ - yet this gives me uniqueness only up to isomorphism.. what am I missing ?
Thanks for your help!