This an exercise in Algebraic Number Theory written by Jürgen Neukirch. It is in chapter $2$, section $5$. The question is as follows:
For a $\mathfrak{p}$-adic number field $K$, every subgroup of finite index in $K^*$ is
both open and closed.
Here is what I thought:
I think that we may use proposition $(5.7)$ on page $140$, if we denote the subgroup as $G$, then there is a subgroup corresponding to $G$ in $$\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}/p^\alpha\mathbb{Z}\oplus\mathbb{Z}_p^d$$ with finite index too, as $Z$, $Z/(q-1)Z$, and $Z/p^\alpha Z$ have discrete topology, and the only subgroups with finite index of $Z_p$ are of the form $p^n Z_p$ for some integer n, which is both open and closed in $Z_p$, so I think we may use these facts to solve this question.
I am not sure if this is right, so I hope we can discuss it and I could learn more.
Thank you for seeing and answering it.
Here is the proposition $(5.7)$ on page $140$ in Algebraic Number Theory written by Jürgen Neukirch:
$(5.7)$ Proposition: Let $K$ be a local field and $q = p^f$ the number of
elements in the residue class field. Then the following hold.(i) If $K$ has characteristic $0$, then one has (both algebraically and topologically)
$$K^*\cong\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}/p^\alpha\mathbb{Z}\oplus\mathbb{Z}_p^d$$
$\qquad$ where $a>=0$ and $d$=[$K$:$Q_p$].(ii) If K has characteristic p, then one has (both algebraically and topologically)
$$ K^*\cong\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}_p^N $$
Actually my basic idea is to analyze the same topological structure problem on a homeomorphism but easier algebraic group as $\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}/p^\alpha\mathbb{Z}\oplus\mathbb{Z}_p^d$ or $\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}_p^N $. And we know $Z$, $Z/(q-1)Z$ and $Z/p^\alpha Z$ all have discrete topology and their subgroups are open and closed, since the quotient group is discrete. To $Z_p$, its only subgroup of finite index is of the form $p^nZ_p$ for some $n$, and this subgroup is both open and closed in $Z_p$, so every subgroup with finite index of $\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}/p^\alpha\mathbb{Z}\oplus\mathbb{Z}_p^d$ or $\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}_p^N $ is both open and closed. So every subgroup with finite index of $K^*$ is both open and closed.
Best Answer
Here is a couple hints. Since the group is finite index it must contain some $K^{*n}$ for some $n$ so it suffices to show that $K^{*n}$ is open for all $n.$ To do this we just need to show that $K^{*n}$ contains some neighbourhood of the identity or equivalently that it contains $U^{(m)} = 1+\mathfrak{p}^m$ for some $m.$ My final hint on how to do this is to use log and exp. There are other ways but this is a great application of these functions.
If you still have trouble I can post a more detailed answer.
EDITED:
Here is some more detail. Note that I wrote this up a while ago for an assignment so I apologize for mistakes.
For sufficiently large $m$ we note that $U^{(m)} \cong \mathfrak{p}^m$ by the mutually inverse homeomorphisms (and isomorphisms), $\exp $ and $\log.$ In particular, choose $m$ large enough so that this isomorphism is true for $U^{(m - v_p(n))}.$ Then if $\alpha \in U^{(m)},$ $\log \alpha \in \mathfrak{p}^{m}$ and $\frac{\log \alpha}{n} \in \mathfrak{p}^{m - v_p(n)}.$ But then $\exp(\frac{\log \alpha}{n}) \in U^{(m - v_p(n))}$ and in addition $\exp(\frac{\log \alpha}{n})^n = \exp(\log(\alpha)) = \alpha.$ Thus $\exp(\frac{\log \alpha}{n})$ is an $n$-th root of $\alpha$ so $\alpha \in K^{\ast n}.$ Since $\alpha$ was arbrirtary this shows that $U^{(m)}\subset K^{\ast n}$ so we are done.