Finite index subgroup $G$ of $\mathbb{Z}_p$ is open.

Suppose $[\mathbb{Z}_p:G] = n <\infty$. Write $n = p^km$ with $p\nmid m$. The idea is to show that $p^k\mathbb{Z}_p = n\mathbb{Z}_p \subseteq G$, after which I am done, since for any $x\in G$ we have that the open ball $x + p^k\mathbb{Z}_p\subseteq G$, which makes $G$ open.

Unfortunately I am stuck at how to prove that $n\mathbb{Z}_p\subseteq G$. Could someone lend a hand with a hint?

Solution 1:

$n\mathbb{Z}_p\subseteq G$ is just Lagrange's theorem applied to the quotient group $\mathbb{Z}_p/G$ in the form: $|\mathcal G|=n, x\in \mathcal G \implies x^n =1$.

Solution 2:

Just a remark - this does hold in much greater generality. According to a theorem of Nikolay Nikolov and Dan Segal, in any topologically finitely-generated profinite group (that is, a profinite group that has a dense finitely-generated subgroup) the subgroups of finite index are open. In particular, this holds for the $p$-adic integers.