Intersection of Radical With Semisimple Subalgebra Suppose $\mathfrak{g}$ is a Lie algebra with radical $\text{Rad }\mathfrak{g}$ and let $\mathfrak{a}\subseteq\mathfrak{g}$ be a semisimple subalgebra. Is it necssarily the case that $\text{Rad }\mathfrak{g}\cap\mathfrak{a}=\{0\}$?
I am trying to prove that maximal semisimple subalgebras are Levi subalgebras, but I can't get past this step. If I know this to be true however, I can finish off the rest of the proof.

Yes. $\text{Rad }\mathfrak{g}\cap\mathfrak{a}$ is an ideal of $\mathfrak{a}$, because $\mathfrak{a}$ is a subalgebra and $\text{Rad }\mathfrak{g}$ is an ideal of $\mathfrak{g}$. Also, $\text{Rad }\mathfrak{g}\cap\mathfrak{a}$ is solvable, because it is a subalgebra of the solvable $\text{Rad }\mathfrak{g}$. But since $\mathfrak{a}$ is semisimple, its only solvable ideal is $\lbrace 0 \rbrace$.
