An infinite cyclic group has a unique subgroup of index $m$ for any integer $m \geq 1$.

Question

Let $G$ be an infinite cyclic group. Show that for any integer $m \geq 1$, there exists a unique subgroup of index $m$. My attempt is consider $G=(\mathbb{Z},+)$. Consider a subgroup $m\mathbb{Z}$ of $\mathbb{Z}$. Then the index $[\mathbb{Z}:m\mathbb{Z}]=m$. This shows the existence part. Bt I'm stuck at uniqueness part. Anyone can help ?

Answer 1

Yes, you've got existence, and you're only a step away from uniqueness.

TIP:
Recall that all cyclic groups of infinite order are isomorphic to $\mathbb{Z}.$ It follows that all subgroups $H\leq \mathbb{Z}$ with finite index $m \ge 1$ are of the form $m\mathbb{Z}$. You can classify all subgroups of $\mathbb{Z}$ by noting that for $H \leq \mathbb{Z}, H\neq 0$, and considering the smallest $m\in H\cap \mathbb{N}$ and compare $H$ to $m\mathbb{Z}$.

So you can prove uniqueness by showing that no other subgroup besides $m\mathbb{Z}$ has index $m$.

E.g., to make explicit the uniqueness of $H = m\mathbb{Z}$, assume, then, that there exists another subgroup $H' \leq G$ with index $m \geq 1$ such that $[G \cong \mathbb{Z}:H']= m$. Then we have that $H' = m\mathbb{Z} = H$.

Hence...

Answer 2

Hint: $G$ is isomorphic to $\mathbb{Z}$. Show that all subgroups of $\mathbb{Z}$ are of the form $m\mathbb{Z}$.