Trouble with "trivial" example of a semi-simple Lie algebra

Question

If the radical of a Lie algebra is zero, we call it semi-simple. In the lecture notes that I'm following its stated that for any arbitrary Lie algebra (over a field with characteristic zero and finite dimensional) $\mathfrak{g}$ we have, $\mathfrak{h}:=\mathfrak{g}/\mathrm{rad}(\mathfrak{g})$ semi-simple. I have trouble seeing why.

I need to show that $\mathrm{rad}(\mathfrak{h})=0$. I think we can use the fact that if $\mathfrak{I}$ is a solvable ideal of $\mathfrak{g}$ such that $\mathfrak{g}/\mathfrak{I}$ is solvable, then $\mathfrak{g}$ is solvable, but I'm not really sure how...

Answer 1

$\mathfrak{g}$ is semisimple if it has no non-zero abelian ideals.

Let $\pi: \mathfrak{g} \rightarrow \mathfrak{h}$ be the quotient map.

if $\mathfrak{h}$ is not semisimple, then $(0)\subsetneq rad(\mathfrak{h})$

therefore $\pi^{-1}(0) = rad(\mathfrak{g}) \subsetneq \pi^{-1}(rad(\mathfrak{h}))$

But $\pi^{-1}(rad(\mathfrak{h}))$ is an abelian ideal, this is a contradiction of the maximality of $rad(\mathfrak{g}).$

Answer 2

HINT:

Suppose for contradiction that $\mathfrak{h}$ is not semi-simple. Write down exactly what this means.

You can now utilize a particular fact, and show that $\mathfrak{g}$ would have to therefore have a non-zero solvable ideal. This gives a contradiction.