Cartan subalgebra of compact group as "annihilator" of a single element Let $G$ be a compact Lie group. The Cartan subalgebra $\mathfrak{h}$ can be defined to be a maximal abelian subalgebra of the Lie algebra $\mathfrak{g}$ of $G$. I know this is not the standard definition for more general Lie groups, but this is the definition I have when $G$ is compact.
Now I am reading page 100 of Sepanski's Lie groups and he proves in Lemma 5.7  $\mathfrak{h}$ (all hypotheses as above) that there is an element $X \in \mathfrak{h}$ such that 
$$\mathfrak{h} = \{ Y \in \mathfrak{g} : [X,Y]= 0\}.$$
However the proof seems kinda confusing to me so I ask can we prove this directly? If so, how does one approach such a problem?
 A: Here is a possible method. We'll show that almost all $X$'s have the property. First of all, the connected Lie subgroup $H\subset G$ generated by $\mathfrak h$ is closed. If not, then $\bar H$ is bigger, still connected and Abelian, so it's Lie algebra would be still Abelian and bigger. So $H$ is a compact Abelian Lie group, i.e. a torus, i.e. (isomorphic to) $\mathbb R^n/\mathbb Z^n$. Now take any $X\in\mathfrak h =\mathbb R^n$ such that the 1-parametric group $L$ it generates is dense in $\mathbb R^n/\mathbb Z^n$ (in other words, the coefficients of $X\in\mathbb R^n$ should be linearly independent over $\mathbb Q$). If $Y\in\mathfrak g$ is such that $[X,Y]=0$ then $Ad_g Y=Y$ for every $g\in L$, hence $Ad_gY=Y$ for every $g\in\bar L=H$, hence $[Z,Y]=0$ for every $Z\in\mathfrak h$, hence (by maximality of $H$) $Y\in\mathfrak h$.
edit Here is a proof with roots.  Take any $X\in\mathfrak h$ such that $\alpha(X)\neq0$ for every root $\alpha$. Now use the decomposition $\mathfrak g=\mathfrak h\oplus\bigoplus_\alpha R_\alpha$, where $R_\alpha=\{Z\in\mathfrak g|[Y,Z]=\alpha(Y)Z\,\forall Y\in\mathfrak h\}$ is the $\alpha$-root space, to see that if $[X,Y]=0$ for some $Y\in\mathfrak g$ then all the $R_\alpha$-components of $Y$ vanish, i.e. $Y\in\mathfrak h$.
