Show that every compact Lie group contains a finitely generated dense subgroup. I'm trying to show that the connected component has a finite number of distinct maximal tori. So the group generated by its generators must be dense. But I don't know if it is really true.
 A: We begin with a Lemma.
Lemma:  Suppose $H\subseteq G$ are compact connnected Lie groups with $H\neq G$.  Then there is a subgroup $S^1\subseteq G$ for which $S^1\cap H$ is finite.
Proof:  Look on the Lie algebra level.  We have $\mathfrak{h}\subseteq \mathfrak{g}$, and $\mathfrak{h}\neq \mathfrak{g}$ because $H$ and $G$ are connected and $H\neq G$.  Then $\mathfrak{g}\setminus \mathfrak{h}$ is an open dense subset of $\mathfrak{g}$, so contains a vector $v$ for which $\exp(tv)$ closes up.  Then $\{\exp(tv)\} = S^1$ is the desired $S^1$.
To see this, note that if $S^1\cap H$ is infinite, then it has an accumulation point (since $G$ is compact).  Now, by using the group multiplication, we may assume this accumulation point is the identity.  It follows that $\exp(t_n v) \in H$ for a decreasing sequence $t_n\rightarrow 0$.  This, then, implies that $v\in\mathfrak{h}$, a contradiction. $\square$
Now, we prove the theorem.  Suppose $G$ is any connected Lie group.  Let $H = T\subseteq G$ be a maximal torus.  Under the identification $T\cong \mathbb{R}^n/\mathbb{Z}^n$, if we pick an element $x=(x_1,...,x_n)\in T$ for which $\operatorname{span}_\mathbb{Q}\{1,x_1, x_2,...,x_n\}$ has dimension $n+1$, it follows that $x$ generates a dense subgroup of $T$.
If $T = G$, we are done.  Otherwise, using the lemma, pick $S^1\subseteq G$ with $S^1\cap T$ finite.  We pick $y\in S^1$ which generates a dense subgroup of $S^1$.
Let $\langle\langle x,y\rangle\rangle$ denote the closure of the subgroup generated by $x$ and $y$. Then clearly $\langle\langle x,y\rangle\rangle$ contains $H$ and $S^1$.  On the Lie algebra level, the Lie algebra of $\langle \langle x,y\rangle\rangle$  is a subspace containing $\mathfrak{t}$ and $v$, so has dimension at least that of $\dim T + 1$.
Now, we induct.  If $\langle \langle x,y\rangle\rangle \neq G$, we use the lemma to pick $z\in G\setminus \langle\langle x,y\rangle \rangle$.  By the same argument as above, $\langle \langle x,y,z\rangle\rangle$ has larger dimension that $\langle \langle x,y,\rangle \rangle$.  Continuing in this way, since the dimension of the generated subset increases at every stage, the process must stop after finitely many steps.
