Are normal subgroups of a profinite group with finite index closed?

Question

Let $G$ be a profinite group and $H \subseteq G$ be a normal subgroup with $[G:H] < \infty$, i.e. $H$ has finite index.

Question: Is $H$ closed?

Problems:

I have a lot of trouble to understand these profinite groups in general. I know that they are inverse limits where every finite group is endowed with the discrete topology. Then we say that this inverse limit is endowed with the product topology of these discrete groups which I find hard already since I only worked with finite products before. I really have no clue how open/closed sets look like.

The only proof I understood is that in a compact topological space, every open subgroup is exactly the closed subgroups of finite index. I heard that there is some relation to the normal subgroups, but I did not really understand it. I need something to understand the proof in this post.

Could you please explain this concept to me? Thank you in advance!

Answer 1

Please check out this paper page 4

https://arxiv.org/pdf/1406.6837.pdf

It says that the group $A_5^\mathbb{N}$ is in "class 2", and they define class 2 to be compact groups having nonclosed subgroups of finite index.

I can't find a concrete subgroup of $A_5^\mathbb{N}$ which satisfies this but at least you know it's false.

Answer 2

Let me answer to your question based on my limited knowledge on this subject. According to the definition, any normal subgroup of finite index is open. Also, as multiplication is continuous, all its left cosets are open. Now the union of all left cosets of $H$ other than itself must be open, and hence $H$ must be closed.