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This is about the fact that for a short exact sequence of Lie algebras $$0\rightarrow K\rightarrow L\rightarrow Q\rightarrow 0$$ we have that L is sovable iff K and Q are solvable. In essence i would like to know why the following argument doesent work: Consider the s.e.s. $$0\rightarrow K\xrightarrow[]{\text{f}} L\xrightarrow[]{g} Q\rightarrow 0$$ with K and Q nilpotent. As Q is nilpotent, we find an m s.t. $C^mQ=0$. Then $$g(C^mL)=C^mg(L)=C^mQ=0$$ so $C^mL\subset im(f)$. As the image of f is nilpotent we find an n , s.t. $C^n im(f)=0$. Thus $C^{n+m} L=0$. But my professor stated that this is wrong, in the sense that only solvable Lie algebras have this property but in general nilpotent Lie algebras dont. Can somebody spot the Error and maybe give a counter example? Thanks!

Adronic
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  • Is $C^mL$ supposed to denote the mth term in the lower central series? I am not familiar with your notation. – Callum Jan 10 '24 at 13:47
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    Assuming that notation is the lower central series, you seem to be saying $C^mL \subset \mathrm{im}(f)$ plus $C^n\mathrm{im}(f) = 0$ implies that $C^{n+m}L=0$ but this is not true. Note that $C^n(C^m L) \not\subset C^{n+m}L$ in general. After all, $C^2L$ will always be nilpotent for $L$ a solvable Lie algebra so $C^n(C^2 L) = 0$ for some $n$ but if $L$ is solvable but not nilpotent then $C^mL \neq 0$ for all $m$ – Callum Jan 10 '24 at 14:00
  • Thank you so much!! I understand that fully:) You saved my day! – Adronic Jan 10 '24 at 14:05
  • A good counter example, I think, would be $L$ the set of upper triangular matrices, $K$ the set of strictly upper triangular matrices and $Q$ the set of diagonal matrices with $f$ and $g$ the obvious inclusion and projection maps respectively. – Callum Jan 10 '24 at 14:13

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