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Let $G$ be a profinite group with a basis of neighbourhoods $U_n$ of normal subgroups.

Furthermore let $H\subset G$ be a closed subgroup. Then we can define the open subgroup $ H_n:= H\cdot U_n$.

Is it true that $\cap H_n = H$?

Clearly $H$ is contained in the intersection, but I am not sure how I would prove the converse containment (or if this is even true).

Notone
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1 Answers1

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If $g$ lies in $\bigcap H_n$, write $g=h_nu_n$, for every $n$, where $h_n\in H$, and $u_n\in U_n$. As $G$ is compact and $H$ closed, there is a converging subsequence $h_{n_i}$, whence $u_{n_i}$ also converges, necessarily to $1$. Consequently $$ g=\lim h_nu_n = \lim h_{n_i}u_{n_i}= \lim h_{n_i}\in H.$$

Ruy
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