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Let $\mathfrak{g}$ be a Lie algebra with Cartan subalgebra $\mathfrak{h}$ and root system $\Phi$. Show that $C_\mathfrak{g}(h)$ is reductive, that is $Z(C_\mathfrak{g}(h))=Rad(C_\mathfrak{g}(h))$, for all $h\in\mathfrak{h}$.

For brevity, put $C=C_\mathfrak{g}(h)$, $Z=Z(C)$, and $R=Rad(C)$. Obviously $Z$ is a solvable ideal in $C$, so I just need to show it's maximal. Is there a "nice" way of showing maximality?

Bey
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1 Answers1

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In case you're still interested, here's a suggestion: note that $\mathfrak h \subseteq \mathrm C_{\mathfrak g}(h)$. Look at the root system of $\mathrm C_{\mathfrak g}(h)$, and observe that it is symmetric. Then prove that this is enough to make $\mathrm C_{\mathfrak g}(h)$ reductive.

LSpice
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