TITLE: Solvable Engel Subalgebras
QUESTION [7 upvotes]: I am reading through Humphrey's Introduction to Lie Algebras and Representation Theory on my own and I am currently stumped by one of the exercises, namely Exercise 2 from Section 15.
Let $L$ be a semisimple Lie algebra over an algebraically closed field of characteristic zero. We want to prove that the only solvable Engel subalgebras of $L$ are necessarily Cartan subalgebras. The hint is to use a previous exercise, in which we proved that given a Cartan subalgebra $H\subseteq L$ and $h\in H$, the centralizer $C_L(h)$ of $h$ is reductive.
My idea was to show that our Engel subalgebra $K$ contains a Cartan subalgebra $H$. This is clear. I then tried to show that $K=C_L(h)$ for some $h\in H$, which I can't quite prove. I am only able to prove this if $K=L_0(\text{ad }x)$ for $x\in K$ semisimple. If I can show this in general however then I know where to go from here.
My question can be distilled to the following: is every Engel subalgebra of the form $C_L(h)$? I suspect not but it would be nice if true. If it's not true, I don't know how I should tackle this exercise.

REPLY [2 votes]: Since answer to exercise is given above, this answer proceeds the hint Humphreys stated. I tried to prove it in the following; the proof is not so elegant as above since it is very much set theoretic. But, it is by the way Humphreys suggested. May be some dropping/jumping possible.
(0) Let $H$ be Engel and solvable. So $H=L_0(ad_x)$ for some $x\in L$.
(1) Write $x=x_s+x_n$ (Jordan decomposition); then $L_0(ad_{x_s})\subseteq L_0(ad_x)$ [Cor. of 15.3].
(2) Since $x_s$ is semisimple so $L_0(ad_{x_s})=C_L(x_s)$ [see Cor. of 15.3].
(3) Since $x_s$ is semisimple part of $x$, so $C_L(x)\subseteq C_L(x_s)$. We have got 
$$C_L(x)\subseteq C_L(x_s)\subseteq H.$$
(4)  $x_s$ is semisimple, let $K$ a maximal toral containing $x_s$. Then $K$ is abelian, so $K\subseteq C_L(x_s)$.
(5) $L$ is semisimple so $C_L(x_s)$ is reductive; so 
$C_L(x_s)=Z(C_L(x_s))+ \underbrace{[C_L(x_s), C_L(x_s)]}_{semisimple}$ 
(6) Since $C_L(x_s)$ is reductive and contained in solvable $H$, its semisimple part in (5) is zero. So 
$$C_L(x_s)=Z(C_L(x_s)).$$
(7) Since Center of Centralizer=Centralizer of Centralizer, so 
$$Z(C_L(x_s))=C_L(C_L(x_s))\subseteq C_L(K)=K.$$
where inclusion is by inclusion in (4) and last equality is property of maximal toral in semisimple $L$.
(8) By (3), (6) and (7) we get $C_L(x)\subseteq C_L(x_s)\subseteq K $; so $x\in K$ i.e. $x$ is semisimple. Thus 
$$H=L_0(ad_x)=L_0(ad_{x_s})=C_L(x_s)\subset K \,\,\,\,\Rightarrow\,\,\,\, H=K.$$