# 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 ?

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A straightforwards way to show there is only one subgroup of order $m$ in a cyclic group of order $n$ (assuming $m\mid n$) is to show that the set of elements whose order divides $n/m$ is a subgroup, for it is then the unique such subgroup. – Mariano Suárez-Alvarez May 8 '12 at 21:08
Or, alternatively, show that in $\mathbb{Z}_{n}$, we have $\langle m \rangle = \langle \textrm{GCD}(m,n) \rangle$. Then the cyclic subgroups are in bijection with the divisors of $n$. – coolpapa Jun 1 '14 at 21:28

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 $G$.
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 $H=f(m \mathbb{Z}) = H'$.