Is there any example of a Lie Algebra, who has nontrivial radical but contains no abelian ideal? 
Is there any example of a Lie Algebra, who has nontrivial radical but contains no abelian ideal?

Here, the radical of a Lie algebra means its maximal solvable ideal.
This question occurs in the proof of the theorem which states that "A Lie algebra is semsimple if and only if its killing form is nondegenerate." In the proof, it is written that "to prove that $L$ is semisimple, it will suffice to prove that every abelian ideal of $L$ lies in the radical of the killing form." So I am wondering what if $L$ contains no abelian ideal whiling being semisimple.
Is it possible?
Many thanks.
 A: If the radical $\mathfrak r$ of a Lie algebra $\mathfrak g$, then $\mathfrak r$ is a solvable Lie algebra. It follows that either $[\mathfrak r, \mathfrak r]$ is zero, so that $\mathfrak r$ is abelian, or $[\mathfrak r, \mathfrak r]$ is a non-trivial nilpotent ideal in $\mathfrak r$. In the last case, then the center of $[\mathfrak r,\mathfrak r]$, which is not trivial because of nilpotency, is an abelian ideal of $\mathfrak g$.
A: Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{r}$ its radical, and $\mathfrak{a}$ the center of $\mathfrak{r}$ (the set of $x \in \mathfrak{r}$ s.t. for all $y \in \mathfrak{r}$, $[x,y]=0$).
By the Jacobi identity, and since $\mathfrak{r}$ is an ideal of $\mathfrak{g}$, $\mathfrak{a}$ is an ideal of $\mathfrak{g}$.
Moreover a nontrivial solvable Lie algebra has nontrivial center.
EDIT: This last sentence is indeed false (thank you Matt E). So it's easier to formulate it this way: by Jacobi, whenever $\mathfrak{a}$ and $\mathfrak{b}$ are ideals of a Lie algebra, $[\mathfrak{a},\mathfrak{b}]$ is also an ideal, so the last non-trivial $\mathcal{D}^n \mathfrak{r}$ is an abelian ideal of $\mathfrak{g}$.
