Group generated topologically

Question

Let $G$ be a topological group and $X,Y$ subsets such that $N = \overline{\langle X \rangle}$ and $G/N = \overline{\langle \pi(Y) \rangle}$. Show that $G=\overline{\langle X,Y \rangle}$.

In that context, $\pi: G \to G/N$ is given by $\pi(Y) = Y/(Y \cap N)$ (which is an open map)

I stuck in this problem. I am able to show that $G$ is topologically generated by $\overline{\langle X \rangle}$ and $\overline{\langle Y \rangle}$, but it do not solves the problem. I'm tried the property of continuity in the product of elements to get some closer to the problems goal, but I'm not successful. I'm think some hints are enough.

Thanks in the advance!

Answer 1

It can be proved this way.

1. If $H$, $N$ are closed subgroups of group $G$ and $N\subset H$, then $H/N$ is closed in $G/N$. (This follows from the fact that the quotient map $\pi:\, G \rightarrow G/N$ is open.)

2. Let $N=\overline{\langle X\rangle}$ and $H=\overline{\langle X,Y\rangle}$. It follows from 1 that $H/N$ is closed in $G/N$.

3. Since $Y\subset H$, $H/N$ is closed in $G/N$ and $\overline{\pi(Y)}=G/N$ hence $\pi(Y)\subset H/N$, $\overline{\pi(Y)}\subset H/N$ and $H/N=G/N$. So $G=H$.