How do we prove that a nonzero nilpotent Lie algebra has a nontrivial center?

[HDB1]

Please, in the book of Introduction to Lie Algebras and Representation Theory J. E. Humphreys p.12, I have a question:

Proposition. (3.2). Let $L$ be a Lie algebra.

(c) If $L$ is nilpotent and nonzero, then $Z(L) \neq 0$.

how we prove this,

Thanks in advance,

 

[HDB1]

dear @fresh_42 , if you could help, I would appreciate that,

 

[fresh_42]

Let's start a little tutoring so that you get used to these calculations. Start with what you have.

$L$ is nilpotent. Means what?

 

[HDB1]

if $\mathfrak{g}^0=\mathfrak{g}, \mathfrak{g}^1=[\mathfrak{g}, \mathfrak{g}], \mathfrak{g}^2=\left[\mathfrak{g}, \mathfrak{g}^1\right], \ldots, \mathfrak{g}^i=\left[\mathfrak{g}, \mathfrak{g}^{i-1}\right]$
A Lie algebra $\mathfrak{g}$ is called a nilpotent if $\mathfrak{g}^n=0$,

 

[fresh_42]

if $\mathfrak{g}^0=\mathfrak{g}, \mathfrak{g}^1=[\mathfrak{g}, \mathfrak{g}], \mathfrak{g}^2=\left[\mathfrak{g}, \mathfrak{g}^1\right], \ldots, \mathfrak{g}^i=\left[\mathfrak{g}, \mathfrak{g}^{i-1}\right]$
A Lie algebra $\mathfrak{g}$ is called a nilpotent if $\mathfrak{g}^n=0$,

Yes. And that means if we write it out
$$
[\mathfrak{g},\underbrace{[\mathfrak{g},[\mathfrak{g},[\mathfrak{g},[\ldots[\mathfrak{g},\mathfrak{g}]\ldots ]]]]}_{=:\mathfrak{g}^{n-1}}]=0
$$
If we pick the smallest $n$ such that $g^{n}=0$ then $g^{n-1}\neq 0$ and we can pick a vector $x\in \mathfrak{g}^{n-1}\backslash \{0\}.$

What can we say about $x$?

 

[HDB1]

Yes. And that means if we write it out
$$
[\mathfrak{g},\underbrace{[\mathfrak{g},[\mathfrak{g},[\mathfrak{g},[\ldots[\mathfrak{g},\mathfrak{g}]\ldots ]]]]}_{=:\mathfrak{g}^{n-1}}]=0
$$
If we pick the smallest $n$ such that $g^{n}=0$ then $g^{n-1}\neq 0$ and we can pick a vector $x\in \mathfrak{g}^{n-1}\backslash \{0\}.$

What can we say about $x$?

$x$ will be in the centre, means $Z(g)\neq 0$?

thanks a lot,

 

[HDB1]

Please, @fresh_42 , what is about this :
If $\mathfrak{g} / Z(\mathfrak{g})$ is nilpotent, then $\mathfrak{g}$ is nilpotent .
I know the idea, but how we can move from to $\mathfrak{g} / Z(\mathfrak{g})$ to $\mathfrak{g}$ to prove it is nilpotent,

Thanks in advance,

 

[HDB1]

Please, @fresh_42 , in general how we can prove that every nilpotent is solvable?

the opposite direction is not true, because the upper triangular matrix is nilpotent but not solvable.

Thanks in advance,

 

[fresh_42]

Please, @fresh_42 , what is about this :
If $\mathfrak{g} / Z(\mathfrak{g})$ is nilpotent, then $\mathfrak{g}$ is nilpotent .
I know the idea, but how we can move from to $\mathfrak{g} / Z(\mathfrak{g})$ to $\mathfrak{g}$ to prove it is nilpotent,

Thanks in advance,

This is again about the quotient's multiplication rules. We have that $\mathfrak{g} / Z(\mathfrak{g})$ is nilpotent, so $\left(\mathfrak{g} / Z(\mathfrak{g})\right)^n=\{\bar 0\}.$ So what is $\bar 0 \in \mathfrak{g}/Z(\mathfrak{g})$? It is the set $\bar 0 =0+ Z(\mathfrak{g})=Z(\mathfrak{g}).$ We also have
$$
[\mathfrak{g}/Z(\mathfrak{g})\, , \,\mathfrak{g}/Z(\mathfrak{g})]=[\mathfrak{g}\, , \,\mathfrak{g}]+Z(\mathfrak{g})
$$
This means that $0+Z(\mathfrak{g})=\bar 0=\left(\mathfrak{g} / Z(\mathfrak{g})\right)^n= \mathfrak{g}^n+ Z(\mathfrak{g})$ or in other words $\mathfrak{g}^n\subseteq Z(\mathfrak{g}).$ But if $\mathfrak{g}^n$ is in the center of $\mathfrak{g}$ then
$$
\mathfrak{g}^{n+1} =[\mathfrak{g}\, , \,\mathfrak{g}^n]\subseteq [\mathfrak{g}\, , \,Z(\mathfrak{g}^n)]=\{0\}
$$
and $\mathfrak{g}$ is nilpotent.

 

[fresh_42]

Please, @fresh_42 , in general how we can prove that every nilpotent is solvable?

the opposite direction is not true, because the upper triangular matrix is nilpotent but not solvable.

Thanks in advance,

We prove that $L^{(n)}\subseteq L^{n}.$ If the RHS gets zero (nilpotency) for some $n$ so does the LHS (solvability).

The proof is formally by induction.