Could someone give me a suggestion to solve this 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.

  • 1
    $\begingroup$ A semisimple Lie algebra is a direct sum of simple Lie algebras. $\endgroup$ – Lord Shark the Unknown Oct 19 '17 at 18:11

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.


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