Let $G$ be a profinite group (or equivalently a compact and totally disconnected topological group ) with the property that all of its normal subgroups of finite index are open sets.

Does this imply that all of its subgroups of finite index are open sets ? (if all subgroups of finite index from $G$ are open sets, than $G$ is called strongly complete ; this motivates the title of this post)



Lemma: Let $H$ be a subgroup of finite index in a group $G$. Then $H$ contains a normal subgroup of finite index, namely $\bigcap_{g \in G} gHg^{-1}$.

Proof. $G$ acts on the left cosets $G/H$ by translation. Since $|G/H|$ is finite, the kernel of this action has finite index (dividing $|G/H|!$), and it is precisely the above intersection. $\Box$

So every subgroup of finite index is a union of cosets of a normal subgroup of finite index. Hence if the latter are open, then so are the former.

  • $\begingroup$ The kernel of this action is contained in $H$ (because $g \in G$ fixing all cosets implies $g$ fixing the coset $H$), so it should have bigger index than $|G/H|$... right? still, it should have finite index because $G/H$ is a finite set, so we can take representatives and write $\bigcap_{g \in G} gHg^{-1}$ as a finite intersection of subgroups of $G$ of finite index $H_i \overset{def}= g_i H g_i^{-1}$, from which we deduce $$ |G/(H_1 \cap \cdots \cap H_n)| \le |G/H_1| \cdots |G/H_n| = |G/H|^n < \infty. $$ I always apply the Orbit-Stabilizer theorem upside down, too... $\endgroup$ – Patrick Da Silva Nov 9 '14 at 21:08
  • $\begingroup$ @Patrick: did you not see the factorial symbol? The kernel of the action has index dividing $|G/H|!$ because that's the size of the group of permutations of $|G/H|$. $\endgroup$ – Qiaochu Yuan Nov 9 '14 at 21:37
  • $\begingroup$ Ohhh!!! I didn't realize it was a factorial, I thought it was an exclamation mark because I was sticking to the Orbit-Stabilizer theorem. Sorry! Now I think my proof is wrong, because the reasoning might not be correct... but anyway, I understand your proof. +1 $\endgroup$ – Patrick Da Silva Nov 9 '14 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.