Irreducibility implies semisimplicity?

Question

Let $g \subset so(n)$ be a Lie subalgebra, that acts irreducibly on $\mathbb{R}^n$ with the standard representation of $so(n)$. I have read that $g$ then has to be semisimple. Is this true? If yes, how can we prove it.

Answer 1

I don't know how much theory you are willing to use. Knowing that $g$ is contained in $so(n)$, it has to be a compact Lie algebra and there is a basic general results that says that $g$ is reductive. To prove that $g$ is semisimple, it is then only necessary to prove that the center of $g$ is trivial. But by irrducibility of the representation $\mathbb R^n$ and Schur's lemma, it follows that any element in the center of $g$ must be a multiple of the identity matrix and thus zero since $g\subset so(n)$.