I'm trying to prove that If $H$ is a subgroup of $\mathbb Z_p$ of a finite index, then $H = p^i\mathbb Z_p$ for some $i\in\mathbb N$.
Here, $\mathbb Z_p$ is the p-adic integers.
My problem is showing the following:
If $[\mathbb Z_p:H]=n$ then $n\mathbb Z_p<H$
I've encountered the accepted answer here: Any finite index subgroup of $\mathbb Z_p$ is open
In which the writer states that this result follows from applying Lagrange's theorem to the quotient group $\mathbb Z_p/H$ but I couldn't quite understand how.
(Though they're kinda old posts, I tried commenting here Finite index subgroup $G$ of $\mathbb{Z}_p$ is open. to get clarifications but with no response..)
If $H$ is a nontrivial subgroup of $\mathbb{Z}_p, \ $ take any nonzero $x$ in $H$, then note that $ \ x \mathbb{Z}< H$ and $ \ x \mathbb{Z}$ is dense in $ \ x\mathbb{Z}_p$ as $\mathbb{Z}$ is dense in $\mathbb{Z}_p$.
So H contains $x\mathbb{Z}_p$.
Now $x\mathbb{Z}_p= p^{v(x)}\mathbb{Z}_p$, where $v$ is p-adic valuation, so $H$ is between $p^n \mathbb{Z}_p$ and $\mathbb{Z}_p$ for some nonnegative integer $n$.
So $H$ is of the form $p^i\mathbb{Z}_p$ for some $i \in \mathbb{N}$ i.e., $$H=p^i \mathbb{Z}_p, \ \ i \in \mathbb{N}.$$
This answers your question.