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Let $L$ be a Lie algebra and Let $R$ be the radical and $N$ be the nilradical of $R$. Let $Z$ be the center of $N$. Is $Z$ an ideal of $L$?

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    $\begingroup$ $N$ is the nilradical of $L$, so no need to introduce $R$ to formulate the question $\endgroup$ – YCor Apr 7 '18 at 9:47
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Yes, at least assuming $L$ finite-dimensional (for infinite-dimensional Lie algebras, it's unclear what would be meant by "nilradical").

Actually, for every Lie algebra $L$, ideal $N$, and $Z$ the center of $N$, it is true that $Z$ is an ideal in $L$. Indeed for $g\in L$, $n\in N$, $z\in Z$, we have

$$[n,[g,z]]=-[g,[z,n]]-[z,[n,g]];$$ $[z,n]=0$, and since $[n,g]\in N$, we also have $[z,[n,g]]=0$. So $[n,[g,z]]=0$. Hence $[g,z]\in Z$ for all $g\in L$ and $z\in Z$.

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Consider $R$ the non commutative Lie algebra of dimension $2$, it is solvable and it's nilradical has dimension $1$ and is not in the center of $R$.

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