Condition (i) can be removed, as already observed by Daniel. In positive characteristic you can find a counterexample in the book "J. Humphreys: Introduction to Lie algebras and representation theory" (Chapter 2, Section 4, page 20, Exercise 4), so condition (ii) cannot be relaxed. Finally, let $H={\mathbb F}x+{\mathbb F}y+{\mathbb F}z$ be the Heisenberg algebra with basis $x,y,z$, where $z$ is central in $H$ and $[x,y]=z$. Consider the ring of polynomials ${\mathbb F}[t]$ as a right left $H$-module with $x$ acting by multiplication by as $t$, d/dt$, $y$ acting as by multiplication by $d/dt$ t$, and $z$ acting as the identity. Now consider the split extension $L=H\ltimes {\mathbb F}[t]$. Then $L$ is solvable of derived length 3, but the derived subalgebra $[L,L]={\mathbb F}z+{\mathbb F}[t]$ is not nilpotent. Thus condition (iii) cannot be removed.
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Condition (i) can be removed, as already observed by Daniel. In positive characteristic you can find a counterexample in the book "J. Humphreys: Introduction to Lie algebras and representation theory" (Chapter 2, Section 4, page 20, Exercise 4), so condition (ii) cannot be relaxed. Finally, let $H={\mathbb F}x+{\mathbb F}y+{\mathbb F}z$ be the Heisenberg algebra with basis $x,y,z$, where $z$ is central in $H$ and $[x,y]=z$. Consider the ring of polynomials ${\mathbb F}[t]$ as a right $H$-module with $x$ acting by multiplication by $t$, $y$ acting as $d/dt$ and $z$ acting as the identity. Now consider the split extension $L=H\ltimes {\mathbb F}[t]$. Then $L$ is solvable of derived length 3, but the derived subalgebra $[L,L]={\mathbb F}z+{\mathbb F}[t]$ is not nilpotent. Thus condition (iii) cannot be removed. |
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Condition (i) can be removed, as already observed by Daniel. In positive characteristic you can find a counterexample in the book "J. Humphreys: Introduction to Lie algebras and representation theory" (Chapter 2, Section 4, page 20, Exercise 4), so condition (ii) cannot be relaxed. Finally, let $H={\mathbb F}x+{\mathbb F}y+{\mathbb F}z$ be the Heisenberg algebra with basis $x,y,z$, where $z$ is central in $H$ and $[x,y]=z$. Consider the ring of polynomials ${\mathbb F}[t]$ as $H$-module with $x$ acting by multiplication by $t$, $y$ acting as $d/dt$ and $z$ acting as the identity. Now consider the split extension $L=H\ltimes {\mathbb F}[t]$. Then $L$ is solvable of derived length 3, but the derived subalgebra $[L,L]={\mathbb F}z+{\mathbb F}[t]$ is not nilpotent. Thus condition (iii) cannot be removed. |
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