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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).

Matheus Manzatto
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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 $p:g\rightarrow g/s$.

Tsemo Aristide
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  • How do you know that $p_0 (\mathfrak{s})$ is semi-simple? This isn't clear to me. – Matheus Manzatto Sep 21 '18 at 02:12
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    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. – Tsemo Aristide Sep 21 '18 at 02:16
  • Thx very much to your help, I was very stuck on this problem, now I feel invincible. Very clean demonstration. – Matheus Manzatto Sep 21 '18 at 02:21
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    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. – Torsten Schoeneberg Sep 24 '18 at 14:34
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    It seems that instead a dimension argument might finish the proof and indeed show equality, not just isomorphism. – Torsten Schoeneberg Sep 24 '18 at 14:42
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    @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 $\mathfrak{s_0}=\mathfrak{s}$. – Matheus Manzatto Sep 25 '18 at 03:21