Isomorphism of Lie algebra after modding out by a semisimple ideal.

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).

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 $$p:g\rightarrow g/s$$.
• How do you know that $p_0 (\mathfrak{s})$ is semi-simple? This isn't clear to me. Commented Sep 21, 2018 at 2:12
• Let $I$ be a simple ideal, $I=[I,I], p(I)=p([I,I])=[p(I),p(I)]$, since $g/s$ is solvable, $p(I)$ is solvable, we deduce that $p(I)=0$. $s$ is a sum of solvable ideal. Commented Sep 21, 2018 at 2:16
• This seems to show that if $\mathfrak{s}$ is an ideal, then it is equal (not only isomorphic) to $\mathfrak{s}_0$. The question only assumes $\mathfrak{s}$ to be a subalgebra, so the quotient in your very last step is no Lie algebra and you cannot conclude as that. Commented Sep 24, 2018 at 14:34
• @TorstenSchoeneberg I agree with you. Using the first part of the argumentation we conclude that $\mathfrak{s_0} \subset \mathfrak{s}$, using that $\text{dim}(\mathfrak{s_0})= \text{dim}(\mathfrak{s})$, imply that $\mathfrak{s_0}=\mathfrak{s}$. Commented Sep 25, 2018 at 3:21