Suppose to have a Lie algebra L with a reductive lie subalgebra G. Let l an element of L such that [l,g] is in G for every g in G, is it true that l is an element of G?if not, there are some restriction on G that makes it true?

Sorry, this started as a comment, but got too long. If $G$ is semisimple, then every derivation of $G$ is inner, so that the normalizer $N_L(G)=C_L(G)+G$ where $C_L(G)$ is the centralizer. In this situtaion Michele's condition holds if and only if the centralizer is trivial. However the case where $G$ is reductive isn't so easy. A reductive $G$ is the direct sum of an Abelian and a semisimple Lie algebra. The Abelian part can have (if at least twodimensional) nontrivial derivations. These will lift to noninner derivations of $G$. If we let $L$ be the semidirect product of $G$ with a onedimensional Lie algebra using this derivation, then $L$ normalizes $G$, $L\ne G$ but $C_L(G)$ is trivial. 


EDIT: Realized I had misread the question. The set of $l$ such that $[l,g]\in G$ for all $G$ is called the normalizer of $G$. It's not particularly common for the normalizer to be the same as G, though it does happen sometimes. There are a few theorems I know about specific Lie algebras being selfnormalizing, such as Cartan and Borel subalgebras in reductive Lie algebras, but it's not very common and I don't know any general condition which guarantees it. 

