Let $L$ be a Lie algebra over a field $F$ and denote by $U(L)$ the universal enveloping algebra of $L$. Regard $U(L)$ as a Lie algebra with respect to the Lie bracket $[a,b]=abba$ for any $a,b\in U(L)$. If $U(L)$ is solvable as a Lie algebra, is $L$ necessarily abelian?
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If $F$ has characteristic different from 2 the answer is yes. This follows from Corollary 6.1 in the paper by D. Riley  A.Shalev: The Lie structure of enveloping algebras, J. Algebra 162, 4661 (1993). On the other hand, in characteristic 2 this conclusion is false. For instance, if $L$ is a 2dimensional nonabelian Lie algebra or a 3dimensional Heisenberg algebra, then one can see by explicite calculations that $U(L)$ is Lie solvable of derived length 3. 

