Compact subgroups of a p-adic field Definition: A p-adic field is a finite extension of $Q_p$.
Question: Let $E$ be a p-adic field, $G$ is a nontrivial additive compact subgroup of $E$, how to prove: $G$ is isomorphic to $Z_p^n$ for some positive integer $n$. This isomorphism is not only a topological group isomorphism but also a $Z_p$ module isomorphism.
I guess we can prove it using non-archimedean analysis, but I still don't know how to prove it.
Thanks for any answers!

$G$ is compact, thus closed in $E$. It is stable by multiplication by an element of $\mathbb{Z}$ (dense in $\mathbb{Z}_p$), thus is a $\mathbb{Z}_p$-submodule of $E$. 
Note that there exists a finite $\mathbb{Q}_p$-base of the vector subspace $V$ spanned by $G$, with vectors $a_1, \ldots, a_n$. 
Now, for each $v \in V$, denote $v_i \in \mathbb{Q}_p$ to be the coordinate of $v$ in the direction $a_i$. 
Then $v \longmapsto v_i$ is a linear form, thus is continuous, hence $G_i=\{g_i,\,g \in G\} \subset \mathbb{Q}_p$ is compact. Therefore, there is a $N>0$ such that for each $i$, $p^NG_i \subset \mathbb{Z}_p$. 
Now, let $$G’=\bigoplus_{i=1}^n{\frac{a_i}{p^N}\mathbb{Z}_p}.$$
$G’$ is a finitely generated $\mathbb{Z}_p$-module, thus is Noetherian, and since $G$ is a submodule of $G’$, $G$ is finitely generated over $\mathbb{Z}_p$. 
Since $\mathbb{Z}_p$ is principal and $G$ has no torsion, and all the nontrivial quotients of $\mathbb{Z}_p$ are finite, $G$ is a power of $\mathbb{Z}_p$. 
