I know how to prove that $o(4)$ is not a simple Lie algebra, which is finding explicitly two subalgebras that commute with each other. I can show for the Lorentz group that the analogous combination of the infinitesimal generators doesn't give us two invariant subalgebras. Of course, this doesn't prove that the Lorentz group is simple.
I have not found any useful theorem about the conditions for a Lie group to be simple (as the Sylow theorem for finite groups). So far I see the resolution can only be based on the definition of a simple group:
Let G be a group of $r$ parameters. Let $X_{\rho}$ be the generators of its Lie algebra. Let the structure constant $c_{\kappa\lambda}^{\rho}$ be defined as $[X_{\kappa},X_{\lambda}] = c_{\kappa\lambda}^{\rho} X_{\rho}$ (using Einstein notation).
If there is no way to choose the basis elements $X_{\rho}$ of the algebra such that the structure constant $c_{\kappa\lambda}^{\rho}=0$ for $\kappa=1,\ldots,p$ and $\rho=p+1,\ldots,r$ $\Rightarrow$ G is simple.
Does anyone have an idea on how to approach this problem?