Nilpotent/ Quotient

Question

I am little fuzzy on how can we conclude that if $L$ is a Lie algebra, and if $L/Z(L)$ is nilpotent, then so is L. Where $Z(L)$ is the center of $L$, namely $Z(L) = \{z \in L \vert [xz] = 0 \ for \ all \ x \in L\}$.

The argument I saw was $L^n \subset Z(L)$, why is that?

Thank you.

Answer 1

For $L$ to be nilpotent, $L^n=0$ for some $n$. But we are given $L/Z$ is nilpotent, where $Z=Z(L)$ is the centre of $L$. Thus $(L/Z)^n=0$. What does this mean for $L^n$? In the natural projection map $\pi:L\to L/Z$ then $(L/Z)^n$ is the image of $L^n$, i.e., $\pi(L^n)=(L/Z)^n=0$. This means that $L^n$ is contained in the kernel of $\pi$, that is $L^n\subseteq Z$.

That implies $L^{n+1}=[L^n,L]\subseteq[Z,L]=0$ as $Z$ is the centre of $L$, so $L$ is nilpotent.

Answer 2

Because, since $L/Z(L)$ is nilpotent, $\bigl(L/Z(L)\bigr)^n=\{0\}$ for some $n$, and this is the same thing as asserting that $L^n\subset Z(L)$. So, $L^{n+1}=\{0\}$.