The normalizer of a proper sub-algebra properly contains the sub-algebra in a nilpotent Lie algebra. So I am just making my way into the theory of Lie Algebras. The question at hand comes from page 14 of Humphreys, Introduction to Lie Algebra and Representation Theory.

Given a finite dimensional nilpotent Lie algebra $L$, and a proper sub-algebra $K$ of $L$, prove that the normalizer of $K$ properly contains $K$.

I keep trying to use Engel's theorem, but I have only been able to show there is a vector such that $[v\ K]$ is in the center, which much be in $K$. Though the condition for $v$ was only that it is in $L$.
A proof or anything else would be helpful. Thank you.
 A: Here is my comment as an answer:
We do this by induction on the length of the ascending central series (which is finite since $L$ is nilpotent).
If $K$ does not contain $Z(L)$ then we are done, as clearly $Z(L)$ normalizes $K$. So we assume that $Z(L)\subseteq K$ and look at the quotient $L/Z(L)$ which is nilpotent and has a shorter ascending central series than $L$.
$K$ corresponds to a subalgebra $\overline{K}$ of $L/Z(L)$ since $Z(L)\subseteq K$, and by induction, the normalizer of $\overline{K}$ strictly contains $\overline{K}$. But anything in the preimage of this normalizer in $L$ will normalize $K$, and we are done.
A: You are right, you need Engel's theorem.
Since $K$ is a proper subalgebra of $L$, we have $L/K$ nontrivial. Then consider the adjoint action of $K$ on $L/K$. By Engel's theorem, there exists an element $\bar{x} \neq \bar{0}$ such that ad$y.\bar{x} \neq \bar{0}$. It means that there exists element $x \notin K$ such that $[y,x]\notin K$ for all $y \in K$. Hence we find such element $x\in N_L(K)$ but $x\notin K$. 
Done.
