Uniqueness of subgroups of a given order in a cyclic group 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!  
 A: What you're missing is that the homomorphism $f$ is fixed. Every subgroup $H$ of $G$ of order $d$ is the group $f[m\Bbb Z]$, so they're all the same subgroup of $G$.
A: Let $H, H'$ be two subgroups of $G$ of order $d$. Let $m \mathbb{Z} = f^{-1}(H)$ and $m' \mathbb{Z} = f^{-1}(H')$. Then $H = f(m \mathbb{Z})$ and $H' = f(m' \mathbb{Z})$. But $G/H, G/H'$ have the same order. Also by the canonical isomorphism given at the bottom of p. 17, $G/H$ is isomorphic to $\mathbb{Z} / m \mathbb{Z}$ and similarly $G/H'$ is isomorphic to $\mathbb{Z} / m' \mathbb{Z}$. Thus $\mathbb{Z} / m \mathbb{Z}$ has the same order with $\mathbb{Z} / m' \mathbb{Z}$, thus $m=m'$. Hence $H=f(m \mathbb{Z}) = H'$.
A: Another way is to use the fact that a cyclic group is the same thing as a quotient of $\mathbf{Z}$.  If $G$ is a cyclic group, let $a$ be a generator, and define a homomorphism $\mathbf{Z} \rightarrow G$ by $n \mapsto a^n$.  This is surjective, so the first isomorphism applies.  Now you can classify all subgroups of $G$ like this:
Lemma 1: Every subgroup of $\mathbf{Z}$ is of the form $n\mathbf{Z} = \{na : a \in \mathbf{Z}\}$ for a unique $n \geq 0$.
Proof: For different $n$, the subgroups $n\mathbf{Z}$ are clearly distinct, since for $n \geq 1$, $n$ is the smallest positive element of $n\mathbf{Z}$.  Let $H$ be any subgroup of $\mathbf{Z}$.  If $H = 0$, then $H = 0 \mathbf{Z}$.  Otherwise, there is a nonzero, hence a positive element of $H$.  Let $n$ be the smallest one.  Since $H$ is a subgroup, $n\mathbf{Z} \subseteq H$.  Conversely, let $x \in H$.  By the division algorithm, there exist $q, r \in \mathbf{Z}$, with $0 \leq r \leq n-1$, such that 
$$x = qn + r$$
Then $r = x - qn \in H$, so $r$ must be zero: otherwise, $n$ would not be the smallest positive element of $H$.  This shows that $x \in n\mathbf{Z}$.
Lemma 2: Let $n\mathbf{Z}, m \mathbf{Z}$ be subgroups of $\mathbf{Z}$.  Then $n\mathbf{Z} \subseteq m \mathbf{Z}$ if and only if $m$ divides $n$.
Lemma 3: Let $G$ be a group, $N$ a normal subgroup.  If $H$ is a subgroup of $G$ containing $N$, then $H/N = \{ hN : h \in H \}$ is a subgroup of $G/N$, and $H \mapsto H/N$ gives a bijection between the set of subgroups of $G$ containing $N$, and the set of all subgroups of $G/N$.
Now for $G = \mathbf{Z}/n\mathbf{Z}$, Lemma 3 says that the subgroups of $G$ are of the form $H/n\mathbf{Z}$ for $H$ a subgroup of $\mathbf{Z}$ containing $n\mathbf{Z}$.  But by Lemmas 1,2 the only subgroups of $\mathbf{Z}$ containing $n\mathbf{Z}$ are of the form $m \mathbf{Z}$, where $m$ divides $n$.  Thus the subgroups of $G$ are in one to one correspondence with the positive divisors of $n$.
