open subgroup of $\mathbb{Z}_p^*$ without torsion

Question

In this paper of Laurent Berger & Pierre Colmez, I have a question in the proof of lemma3.1.1 on page7.

We have a exact sequence of group $1\rightarrow A\rightarrow \widetilde\Gamma_H \rightarrow B\rightarrow1$ where $A$ is a finite group and $B$ is an open subgroup of $\mathbb{Z}_p^*$ without torsion, then why $B$ is topological isomorphic to $\mathbb{Z}_p$ as a topological group? I know open subgroup of $\mathbb{Z}_p$ are all of $p^k\mathbb{Z}_p$, but what about $\mathbb{Z}_p^*$?

Let $C_H$ be the center of $\widetilde\Gamma_H$, why $C_H$ is of finite index in $\widetilde\Gamma_H$? Can we see $\mathbb{Z}_p$ as a subgroup of $\widetilde\Gamma_H$?

Thanks!

Answer 1

I think I have a answer myself.

If $B$ is a open subgroup of $\mathbb{Z}_p^*=\mathbb{Z}/(p-1)\mathbb{Z}\times (1+p\mathbb{Z}_p)$ without torsion, then $B\subset 1+p\mathbb{Z}_p$ since all elements of finite order in $\mathbb{Z}_p$ are $p-1$-th roots of unity. But we also have $1+p\mathbb{Z}_p\cong \mathbb{Z}_p$ and open subgroups in $\mathbb{Z}_p$ are $p^n\mathbb{Z}_p$ where $n\geq0$, so $B\cong \mathbb{Z}_p$.