I want to prove the following proposition:
Proposition. If $H$ is a closed subgroup of a pro-$p$ group $G$, then $H$ is pro-$p$
There is a result that maybe can be used in order to prove that.
If $H$ is a closed subgroup of a profinite group $G$, then $H$ is the inverse limite of the open subgroups of $G$ containing $H$.
I prove this result using some lemmas (if it is relevant, I can write the full proof here). By the way, $\color{red}{\text{can this result be used to prove the proposition?}}$ I try some things, but I cannot do something relevant
Also, if no I think there is an easier way to do that. An open subgroup $U$ of $H$ is like $H \cap K$ where $K$ is an open subgroup of $G$. If $HK$ is a subgroup of $G$, then $$p^n = |G:HK||HK:K|$$ from where $$|H:U| = |HK:K| = p^m.$$ Unfortunately, $HK$ does not have to be a subgroup of $G$ (it is a subgroup if $H$ or $K$ is normal) so that the index is undefined. But I suspect that this idea can be improved. For example, $\color{red}{\text{if $U$ is normal in $H$, then $K$ must be normal in $G$?}}$
Write $G= \varprojlim_i G_i$ with each $G_i$ a finite $p$-group; and let $p_i: G\to G_i$ denote the canonical projection.
Can you prove that if $H$ is closed in $G$, then $H\cong \varprojlim_i p_i(H)$ ?