Suppose $L_1$ and $L_2$ are two Lie algebras, and that $f: L_1\to L_2$ is a Lie algebra homomorphism. If $L_1$ is nilpotent, does it follow that $f(L_1)$ is nilpotent?
Remark 1. The corresponding statement for groups is true. In other words, homomorphic image of a nilpotent group is nilpotent.
Remark 2. If $L_1$ is solvable Lie algebra, then $f(L_1)$ is solvable. This is also well-known.
On
I found the answer after googling. This is Proposition 3.2 in Humphreys' Lie algebras and Representation Theory. I will not delete the post for the benefit of other readers.
Proposition. Let $L$ be a Lie algebra.
$\textbf{(a)}\,$ If $L$ is nilpotent, then so are all subalgebras and homomorphic images of $L$.
$\textbf{(b)}\,$ If $L|Z(L)$ is nilpotent, then so is $L$.
$\textbf{(c)}\,$ If $L$ is nilpotent and nonzero, then $Z(L)\ne0$.
Unless I'm missing something, isn't it obvious? No need to quote a book. Take any sequence $f(x_1), f(x_2), ..., f(x_r)$ in $f(L_1)$, then if the sequence is large enough $ad_{x_1} ... ad_{x_r}$ is zero on any $x \in L_1$, and since $f$ is a Lie algebra morphism this implis that $ad_{f(x_1)} ... ad_{f(x_r)}$ is zero on $f(x)$. This is the definition of nilpotency.