Generating sets for topological groups Let G be a compact topological group.
Suppose G has a subset X and a normal subgroup N such that the subgroup generated by X is dense in N.
Moreover, suppose G has a subset Y such that the subgroup generated by the canonical image of Y in G/N is dense in G/N.

I'm pretty sure that the subgroup generated by X and Y is dense in G. Is this correct?

I know how to proceed in the discrete case and I think I cannot play with words here.
Thank you for any help.
 A: The following proof at least works in the case of compact Lie groups and will transfer over to topological groups iff the natural projection $\pi:G\rightarrow G/N$ is a quotient map in the sense that a subset of $G/N$ is open iff its preimage in $G$ is.  I've only ever studied compact Lie groups, so I simply don't know if this is true of compact topological groups or not.
Let $U$ be an open subset of $G$.  We wish to find a point in the subgroup generated by $X$ and $Y$ in $U$.
We know $\pi(U)$ is open in $G/N$ so that there is some element $y\in Y$ with $\pi(y)\in \pi(U)$.  So there is a $u\in U$ with $\pi(u) = \pi(y)$.  But then $uy^{-1} \in N$.  So $uy^{-1} = n$ for some $n\in N$.
Now, right multiplication by $y^{-1}$, thought of as a map from $G$ to itself is a homeomorphism.  Hence, if $V$ is an open set around $u$ entirely contained in $U$, then $Vy^{-1}$ is an open set around $n$.  Since $X$ is dense in $N$, there is some $x\in X\cap(N\cap Vy^{-1})$.
Hence, $vy^{-1} = x$ for some $v\in V\subseteq U$.  But $v = xy$, so we've found an element in the subgroup generated by $X$ and $Y$ lying in $U$.
Edit  According to wikipedia, the coset space $G/N$ is always given the quotient topology and the natural projection is automatically open, so aparently my proof works for all compact topological groups.
