Let $\mathfrak{g}$ be a Lie algebra, then Levi's decomposition theorem affirms that we can decompose $\mathfrak{g} = \mathfrak{r(g)}\oplus \mathfrak{s}$, where $\mathfrak{r(g)}$ is the radical of $\mathfrak{g}$, and $\mathfrak{s}$ is a semisimple subalgebra of $\mathfrak{g}$.
Question: If $\mathfrak{s_0}$ is a semisimple ideal of $\mathfrak{g}$ such that $\mathfrak{g}/\mathfrak{s_0}$ is a solvable Lie algebra, is it true that $\mathfrak{s_0}\cong \mathfrak{s} $ (isomorfism of Lie Algebras)?
Can anyone help me?
I was able to conclude that $\mathfrak{g/s_0} = \mathfrak{r (g/s_0)} \cong \mathfrak{r(g)}$ (by this question Radical of a quotient Lie algebra), but I could not imply that $s_0 \cong \mathfrak s$ (the dimensions are the same but this doesn't imply that there is an isomorphism of Lie Algebras).
Best Answer
Yes, it is true, consider $p_0:g\rightarrow g/s_0$, $p_0(s)$ is a semi-simple subalgebra of $g/s_0$, since $g/s_0$ is solvable, we deduce that $p_0(s)=0$ and $s\subset s_0$, a similar argument shows that $s_0\subset s$ by considering $p:g\rightarrow g/s$.