All homomorphic images of a solvable Lie algebra are solvable?

Question

Here I read in a book that:

All homomorphic images of a solvable Lie algebra are solvable as well.

How to prove such a statement? Let's say we have a homomorphism $$\phi: \mathcal{G}\rightarrow \mathcal{H},$$ where $\mathcal{G}$ is a solvable Lie algebra and $\phi(0)=\mathcal{I}\in \mathcal{G}$ is the non-trivial kernel of the homomorphism. Then there is $g^{\{i\}}=[g^{\{i-1\}},g^{\{i-1\}}]=0$ for a number $i$. $[\phi(0)\equiv\mathcal{I},\phi(0)\equiv\mathcal{I}]=0$.

@DietrichBurde, My notation indicates that $\mathcal{I}$ is an ideal rather than a specific element. — Wein Eld, Aug 31 '16 at 14:27
But $\phi(0)$ is a specific element, and not an ideal. So $\phi(0)=I$ imeans that $I=0$. Perhaps you mean $\ker(\phi)=I$ ? — Dietrich Burde, Aug 31 '16 at 14:27
@DietrichBurde, Sorry, I see my problem. I mis-consider $\mathcal{I}$ as an ideal of $\mathcal{H}$. Thanks. — Wein Eld, Aug 31 '16 at 14:29
$\ker(\phi)$ is an ideal in $\mathcal{G}$, not in $\mathcal{H}$. — Dietrich Burde, Aug 31 '16 at 14:30

score 4 · Accepted Answer · answered Aug 31 '16 at 14:24

4

Let $L$ be a solvable Lie algebra. Then all homomorphic images $\phi(L)$ are again solvable, because for the derives series we see inductively that $$ \phi(L^ {(n)})=(\phi(L)^ {(n)}. $$ So if we have $L^{(n)}=0$, because $L$ is solvable, then this implies that also $\phi(L)$ is solvable.

answered Aug 31 '16 at 14:24

Dietrich Burde

130,978

Homomorphism have to be surjective homomorphism. – Infinity Aug 20 '23 at 17:04
@Spectrum No, $\phi$ need not be surjective. We always have $\phi([L,L])=[\phi(L),\phi(L)]$. If you say that $f\colon L\rightarrow L_1$ is a surjective homomorphism, then even $\phi(L^ {(n)})=L_1^{(n)}$. – Dietrich Burde Aug 20 '23 at 18:25

All homomorphic images of a solvable Lie algebra are solvable?

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