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I could not get the following, could someone give me a hint?

Let $\mathfrak{H}$ be a Cartan subalgebra of a simple Lie algebra $\mathfrak{L}$. Show that $\mathfrak{H}$ is abelian.

So, we need to prove that $[\mathfrak{H},\mathfrak{H}]=0$. It seems that I should find a proper ideal of $\mathfrak{L}$ containing $[\mathfrak{H},\mathfrak{H}]$ but then I can not get the way.

Thanks in advance.

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    $\begingroup$ The definition of Cartan subalgebra is not standard. What is your definition? $\endgroup$ – PAD Nov 2 '12 at 11:06
  • $\begingroup$ Definition : $\mathfrak{H}$ is a Cartan subalgebra if $\mathfrak{H}$ is nilpotent and self-normalizing, i.e., if $[x,y]\in \mathfrak{H}$ for all $x\in\mathfrak{H}$ then $y\in\mathfrak{H}$. $\endgroup$ – 9999 Nov 10 '12 at 8:29
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Here is the proof from Knapp's "Lie groups beyond an introduction":

Since $\mathfrak{h}$ is nilpotent, it is solvable and ad$\mathfrak{h}$ is solvable as a subalgebra of transformations of $\mathfrak{g}$. By Lie's Theorem it is triangular in some basis. For any three triangular matrices $A,B,C$ we have tr(ABC)=tr(BAC). Therefore $$tr(ad[H_1, H_2]ad H)=0 \ \ \ (1)$$ for $H_1, H_2, H \in \mathfrak{h}$.
Next, let $\alpha$ be a generalized non-zero weight, let $X$ be in $\mathfrak{g}_{\alpha}$ and let $H$ be in $\mathfrak{h}$. Then $ad H adX$ carries $\mathfrak{g}_{\beta}$ to $\mathfrak{g}_{\alpha+\beta}$. Since $\mathfrak{g}=\oplus \mathfrak{g}_{\alpha}$ we get $$tr(adH adX)=0 \ \ \ (2) $$ Specializing (2) to $H=[H_1, H_2]$ we have that $k([H_1, H_2], X)=0$ where $k$ is the Killing form. Together with (1) we get that $k([H_1, H_2], Y)=0$ for all $Y \in \mathfrak{g}$. Since $k$ is non-degenerate for $\mathfrak{g}$ simple we conclude that $[H_1, H_2]=0$.

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