Let $\cal T$ be a locally compact topology on a group $G$ and $(x,y)\mapsto xy^{-1}$ be continuous at $(1,1)$ and for every $a\in G$, $x\mapsto xa$ be continuous everywhere with respect to $\cal T$. Let $C$ be a connected component of $(G,\cal T)$ containing the neutral element $1$.

If $C$ is a normal(=invariant) subgroup of $G$ then $C$ is the intersection of all open subgroups of $G$. How if $C$ is not a normal subgroup? then is $C$ the intersection of all open subgroups of $G$?

  • $\begingroup$ Can this happen? Can you give examples? $\endgroup$ – user1688 Jan 19 '16 at 11:02
  • $\begingroup$ @Anton It seems trivial that $C$ is a subgroup. However there is no clue for normality. No I have not any example for a not normal $C$. $\endgroup$ – nano Jan 19 '16 at 13:38
  • $\begingroup$ It seems trivial that $C$ is a normal subgroup. $\endgroup$ – Tom De Medts Jan 19 '16 at 17:02
  • $\begingroup$ @TomDeMedts: Let $H$ be a not normal subgroup of $G$. Let $\mathcal F$ be the set of all subsets of $G$ which contain $H$. Let $\cal T$ be a topology on $G$ with $\cal F$ the set of all neighborhoods of $1$ and the maps $x\mapsto xa$ continuous everywhere. $(G,\cal T)$ is a topology as defined above. $H$ is the connected component containing $1$. $\endgroup$ – nano Jan 19 '16 at 17:18
  • $\begingroup$ Oh, I'm sorry, I incorrectly assumed that you were considering a topological group. I should have read your title more carefully :) $\endgroup$ – Tom De Medts Jan 20 '16 at 8:17

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