# Open normal subgroups of a profinite group form a basis for the open neighborhoods of 1

In the following link: An equivalent definition of the profinite group
I'm having trouble understanding the following quote from the answer:

So if you replace each subgroup in a basis by the intersection of its conjugates, you obtain a basis made up of normal subgroups.

How can one prove that these normal subgroups form a basis?

If $$\mathscr H$$ is a local base of $$1$$, then $$\{\bigcap_{x\in G} xHx^{-1}\mid H\in \mathscr H\}$$ is a local base of $$1$$ because for every open neighborhood $$A$$ of $$1$$, $$1\in \bigcap_{x\in G} xHx^{-1}\subset H\subset A$$ for some $$H\in \mathscr H$$. The fact that $$\bigcap_{x\in G} xHx^{-1}\mid H$$ is open is proven in your linked answer.
• Thank you! on the same subject, can you please explain why must every open nbhd subset $A$ of 1 contain a subgroup $H$? – Khal May 13 '19 at 9:22
• @Shmooze Because $\mathscr H$ is a basis. By definition, every open subset of $1$ contains a basis element in $\mathscr H$. – YuiTo Cheng May 13 '19 at 9:26