If $\frak{g}$ is a semi-simple lie algebra then every homomorphic image of $\frak{g}$ is semisimple.

Question

Could someone give me a suggestion to solve this problem?

PROBLEM:

If $\frak{g}$ is a semi-simple lie algebra then every homomorphic image of $\frak{g}$ is semisimple.

I was trying to prove that if $\varphi: \frak{g} \rightarrow \frak{h}$ is a homomorphism of lie algebras then the radical of
$\varphi(\frak{g})$ is zero, but I could not.

Answer 1

We know that $\phi(\mathfrak g)=\mathfrak g / ker \phi$.

Since $ker\phi$ is an ideal of $\mathfrak g$, we can write $\mathfrak g=ker\phi \oplus (ker\phi)^\perp$. Here $ker\phi $ and $(ker\phi)^\perp$ are both ideals of $\mathfrak g$. (This is proved on page 23 of 'Introduction to Lie Algebras and Representation Theory-Humphreys'.) So, $\phi(\mathfrak g)=\mathfrak g / ker \phi=(ker\phi)^\perp$. Since every ideal of semisimple is semisimple, $\phi(\mathfrak g)=(ker\phi)^\perp$ is semisimple.