Take an algebraic group $G$ defined over a finite field $K$. Suppose its Lie algebra $\mathfrak{g}$ is simple. It should follow that $G$ is almost-simple. (By this I mean not that $G(K)$ is simple -- I've heard there is a tricky though in some sense inessential counterexample -- but rather that $G$ has no normal algebraic subgroups of dimension greater than $0$ and less than $ \text{dim}\, G(K)$.)

Does anybody know of a concise and straightforward proof of this fact? One can't proceed as for $K=\mathbb{R}$, since $\exp$ is not defined (except on nilpotent subalgebras of $\mathfrak{g}$). I feel this should be simpler than studying the simplicity of $G(K)$. Is this so?

Edit: I just realized that I meant to ask the converse of this question. If $G$ is almost-simple in the sense of algebraic groups, how do you prove that $\mathfrak{g}$ is simple? I agree that the other direction is very easy.