# The center of a nilpotent Lie algebra intersects each ideal

If $\cal h$ is a nonzero ideal in a nilpotent Lie algebra $\cal g$. How to prove that $\mathcal h\cap Z(\mathcal g)\not =0$, where $Z(\mathcal g)$ is the center of $\mathcal g$?

• Is this not what you asked 45 minutes ago? – Tobias Kildetoft Feb 16 '16 at 19:46
• yes, I think this question is more general than the last one – Ronald Feb 16 '16 at 19:51
• This would follow immediately from an affirmative answer to your last question. Please at least point out the connection between them. – Tobias Kildetoft Feb 16 '16 at 19:52
• Should I delete one of them? – Ronald Feb 16 '16 at 20:00
• They are not the same question, so no. But you should link them up. Anyway, I think the answer to the other one is "no" as it is equivalent to the last non-zero term in the central series being the entire center, which I don't think is always the case. – Tobias Kildetoft Feb 16 '16 at 20:02

Let $I$ be a nontrivial ideal of $L$. Then $L$ acts on $I$ by the adjoint action, because $I$ is an ideal. Then by Engel's theorem (or lemma to it) there exists a $v\neq 0$ in $I$ with $0=L.v=[L,v]$, because $L$ is nilpotent. But this just means that $$v\in I\cap Z(L),$$ so that the intersection is nontrivial. In particular, the center of a nilpotent Lie algebra itself is nontrivial.

• This seems to show that the center is non-trivial, rather than it intersecting all non-zero ideals. – Tobias Kildetoft Feb 16 '16 at 19:58
• Yes, thank you, I was thinking of the nontrivial center, indeed. – Dietrich Burde Feb 16 '16 at 20:06
• Do you happen to know the answer to his earlier question, about intersecting the last non-zero term of the central series? I think there should be a counterexample (as I mentioned in the comments above), but it is eluding me right now. – Tobias Kildetoft Feb 16 '16 at 20:09
• Well, if the Lie algebra has a direct abelian factor it cannot be true. – Dietrich Burde Feb 16 '16 at 20:44