Subalgebras and Ideals of a Lie algebra

Question

If $A$ is a vector subspace of $\mathfrak{g}$(Which is a Lie algebra), and $N=\{x\in\mathfrak{g}:[x,A]\subseteq A\}$

So if $N=A$, then $A$ is a subalgebra, and if $N=\mathfrak{g}$ then, $A$ is an ideal, correct?

What can I conclude about $N$?

Answer 1

The space $N$ is called the normalizer of $A$ in $\mathfrak{g}$, and it is the largest subalgebra of $\mathfrak{g}$ containing $A$ as an ideal. If $A$ is already an ideal in $\mathfrak{g}$, then $N_{\mathfrak{g}}(A)=\mathfrak{g}$ of course. If $N_{\mathfrak{g}}(A)=A$, then we only know that $A$ is self-normalising.