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I'm working through Fulton and Harris's book on Representation theory, and I've just done the exercise where I had to show:

If $\mathfrak{g}$ is a reductive Lie algebra (defined as $Z(\mathfrak{g}) = rad(\mathfrak{g})$) then $[\mathfrak{g},\mathfrak{g}]$ is semisimple.

The way I did this was to say that $\frac{\mathfrak{g}}{rad(\mathfrak{g})} = \frac{\mathfrak{g}}{Z(\mathfrak{g})}$ is semisimple. So then $\frac{[\mathfrak{g},\mathfrak{g}]}{Z(\mathfrak{g})}$ is also semisimple, and if $\mathfrak{a}$ is any abelian ideal of $[\mathfrak{g},\mathfrak{g}]$ we have that $\frac{\mathfrak{a}}{Z(\mathfrak{g})}$ is an abelian ideal of $\frac{[\mathfrak{g},\mathfrak{g}]}{Z(\mathfrak{g})}$, so must be zero.

I want to show $[\mathfrak{g},\mathfrak{g}] \cap Z(\mathfrak{g})$ is zero, so that $\mathfrak{a}$ must be zero.

If anyone can hep me with this step or provide an alternative proof I'd be really grateful, thanks!

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For a reductive Lie algebra $\mathfrak{g}$ the adjoint representation is semisimple. Hence also its restriction to $[\mathfrak{g},\mathfrak{g}]$ is semisimple, so that there is a complement $U$ to $[\mathfrak{g},\mathfrak{g}]\cap Z(\mathfrak{g})$, i.e., we have $U\oplus ([\mathfrak{g},\mathfrak{g}]\cap Z(\mathfrak{g}))=[\mathfrak{g},\mathfrak{g}]$. But we have $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]+Z(\mathfrak{g})$, because the adjoint module has a complement, so that $$ [\mathfrak{g},\mathfrak{g}]=[\mathfrak{g},[\mathfrak{g},\mathfrak{g}]+Z(\mathfrak{g})]=[\mathfrak{g},[\mathfrak{g},\mathfrak{g}]]=[\mathfrak{g},U]\subseteq U. $$ Together with $U\oplus ([\mathfrak{g},\mathfrak{g}]\cap Z(\mathfrak{g}))=[\mathfrak{g},\mathfrak{g}]$ this means $[\mathfrak{g},\mathfrak{g}]\cap Z(\mathfrak{g})=0$.

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