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.

  • $\begingroup$ Thank you! on the same subject, can you please explain why must every open nbhd subset $A$ of 1 contain a subgroup $H$? $\endgroup$ – Khal May 13 '19 at 9:22
  • 1
    $\begingroup$ @Shmooze Because $\mathscr H$ is a basis. By definition, every open subset of $1$ contains a basis element in $\mathscr H$. $\endgroup$ – YuiTo Cheng May 13 '19 at 9:26

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