# 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 semisimple Lie algebra is a direct sum of simple Lie algebras. – Lord Shark the Unknown Oct 19 '17 at 18:11

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.