Let $\mathfrak{g}$ be a finite dimensional simple Lie Algebra over an algebraically closed field $K$. I'm having trouble to show that always exists Cartan subalgebras $\mathfrak{h}_1,\mathfrak{h}_2$ such that $\mathfrak{h}_1\cap\mathfrak{h}_2=0$.
In general, if all (or one) cartan subalgebras of a finite dimensional Lie Algebra $\mathfrak{g}$ are abelian, then $\bigcap_{\mathfrak{h}\text{ cartan}}\mathfrak{h}=\mathfrak{z(g)}$. For this, call $\mathfrak{h}'=\bigcap\mathfrak{h}$. Since $\mathfrak{z(g)}\subset\mathfrak{n(h)}=\mathfrak{h}$, for all $\mathfrak{h}$ cartan subalgebra, we have $\mathfrak{z(g)}\subset\mathfrak{h}'$. Now, let $X\in\mathfrak{h}'$. Since each $\mathfrak{h}$ is abelian, $X$ commutes with all $Y\in\mathfrak{\bar g}=\{\text{regular elements of }\mathfrak{g}\}$, so, $\mathfrak{z}(X)\supset\mathfrak{\bar g}$. Since $\mathfrak{z}(X)$ is an subalgebra (and so a vector subspace), we have $\mathfrak{z}(X)=\mathfrak{g}$, so $X\in\mathfrak{z(g)}$.
From this, since any cartan subalgebra of a simple algebra is abelian, we have that the intersection of all subalgebras is $0$, since in this case $\mathfrak{z(g)}=0$. But I have no idea how to show that only two is sufficient...
Any help will be appreciated!