# 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.

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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.

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I may be missing something, but where do you use compactness? And yes, you'll always want to give the quotient the quotient topology. – t.b. Jun 28 '11 at 2:43
@Theo: Yes, no compactness is needed. And V can be taken as U. @Jason: Thank you! – da Fonseca Jun 28 '11 at 2:48
@Theo: I was thinking the same thing - I never used compactness. Of course you want to give the quotient the quotient topology that was dumb of me! On the other hand, the issue of whether or not the projection map is open is (potentially) an issue, so I'm glad I was overcautious rather than undercautious. – Jason DeVito Jun 28 '11 at 3:04
This is a general fact: If a group $G$ acts on a space $X$ by homeomorphisms then the projection $p:X \to X/G$ is open because $V \subset X/G$ is open if and only if $p^{-1}(V)$ is open and for $U \subset X$ we have $p^{-1}p(U) = GU = \bigcup_{g \in G} gU$. If $U$ is open, so is $p^{-1}p(U)$ as a union of open sets, hence $p(U)$ is open. – t.b. Jun 28 '11 at 3:08
@Jason: Reading carefully I note that you missed some $\langle$'s and $\rangle$'s. In fact $y\in \langle Y\rangle$ and $x\in \langle X\rangle$. – da Fonseca Jun 28 '11 at 3:15