If $\frak{g}$ is a semi-simple lie algebra then every homomorphic image of $\frak{g}$ is semisimple. 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.
 A: The proof depends on the definitions and results we can use. I will assume that a Lie algebra is semisimple if it does not contain a nonzero solvable ideal.
Let $\mathfrak{g}$ be semisimple, and $\varphi: \frak{g} \rightarrow \frak{h}$ be a Lie algebra homomorphism. Suppose that $\phi(\mathfrak{g})$ had a solvable ideal $I$. Then its pre-image $\phi^{-1}(I)$ would also be a solvable ideal. But since $\mathfrak{g}$ is semisimple, this pre-image would be zero. Hence its image would also be zero, i.e., $I=0$. Hence $\phi(\mathfrak{g})$ is semisimple.
A: 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.
