# Let <mi mathvariant="fraktur">g be a complex linear Lie algebra. Assume that the center <m

Let $\mathfrak{g}$ be a complex linear Lie algebra. Assume that the center $\mathfrak{z}$ of $\mathfrak{g}$ is trivial
Let $\mathfrak{r}$ be the radical of $\mathfrak{g}$. If $\mathfrak{r}$ is abelian, then $\mathfrak{g}$ is semisimple?
What if $\mathfrak{g}$ is the Lie algebra of an algebraic complex linear group?
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Mekjulleymg
Your question whether $\mathfrak{g}$ is semisimple is equivalent to whether necessarily $\mathfrak{r}=0$. The answer is no. A counterexample is given by
$\mathfrak{g}=\left\{\left(\begin{array}{ccc}a& b& d\\ c& -a& e\\ 0& 0& 0\end{array}\right):a,b,c,d,e\in \mathbb{C}\right\}$
$\mathfrak{r}=\left\{\left(\begin{array}{ccc}0& 0& d\\ 0& 0& e\\ 0& 0& 0\end{array}\right):d,e\in \mathbb{C}\right\}$
is two-dimensional.
Cierra Castillo