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Let $L$ be a Lie algebra.

(a) If $L$ is solvable, then so are all subalgebras and homomorphic images of $L$.

(b) If $I$ is a solvable ideal of $L$ such that $L/I$ is solvable, then $L$ itself is solvable.

Proof of (b): Say $(L/I)^{(n)}=0$. Applying part (a) to the canonical homomorphism $\pi:L\rightarrow L/I$, we get $\pi(L^{(n)})=0$, or $L^{(n)}\subset I=\operatorname{Ker}\pi$.

How does $(L/I)^{(n)}=0$ imply $\pi(L^{(n)})=0$? I don't see why this follows from part (a).

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Because $$\pi(L^{(n)}) \subseteq (L/I)^{(n)}.$$

This statement is pretty easy to prove by induction, and a good exercise, so I leave it to you.

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  • $\begingroup$ Thank you, Jim. Actually, using the base case $\pi(L)=L/I$, don't we have equality $\pi(L^{(n)})\subseteq(L/I)^{(n)}$? $\endgroup$
    – PJ Miller
    Aug 18, 2013 at 10:56
  • $\begingroup$ Probably, but I didn't want to think through whether $I \nsubseteq L^{(n)}$ was an issue or not. $\endgroup$
    – Jim
    Aug 19, 2013 at 2:06

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