If I have topological groups $H \le K \le G$, I could prove that if $H \trianglelefteq K$ then $\overline{H} \trianglelefteq K$ but is it also true that $H \trianglelefteq \overline{K}$?
Asked
Active
Viewed 40 times
2
-
You consider closures in $G$ or in $\bar G$? – markvs Jul 14 '20 at 21:10
-
closures in G I guess – roi_saumon Jul 14 '20 at 21:11
-
What do you mean by "I guess"? – markvs Jul 14 '20 at 21:13
-
2$G$ is the topological group I consider (with subgroups $H$ and $K$). So I thought $G$ is closed and open so $\overline{G}=G$ – roi_saumon Jul 14 '20 at 21:16
1 Answers
1
If $H$ is a normal subgroup of $G$, then it is normal in any intermediate subgroup $K'$, since it is closed under conjugation by any element $g\in K'$.
In particular, this holds for $K'=\bar K$.
So, the following statements are true in this case: $$H\trianglelefteq K,\quad H\trianglelefteq \bar K,\quad \bar H\trianglelefteq \bar K\,.$$ However, the forth statement, $\bar H\trianglelefteq K$ only holds if $\bar H\subseteq K$, which is not guaranteed in general, for example if $K$ is assumed to be closed.
Berci
- 90,745
-
Thank you. While reading your answer I realised I made a typo in the question. It should be $H \trianglelefteq K$ then $\overline{H} \trianglelefteq K$ (as you point out, provided $\overline{H}\subset K$) – roi_saumon Jul 14 '20 at 22:17
-
Ok, then your solution is probably fine. I think also in this case $H\trianflelefteteq\bar K$ follows, though that's a different question. – Berci Jul 14 '20 at 22:22