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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 $\mathcal g$?

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  • $\begingroup$ Is this not what you asked 45 minutes ago? $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 19:46
  • $\begingroup$ yes, I think this question is more general than the last one $\endgroup$ – Ronald Feb 16 '16 at 19:51
  • $\begingroup$ This would follow immediately from an affirmative answer to your last question. Please at least point out the connection between them. $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 19:52
  • $\begingroup$ Should I delete one of them? $\endgroup$ – Ronald Feb 16 '16 at 20:00
  • $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 20:02
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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.

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  • $\begingroup$ This seems to show that the center is non-trivial, rather than it intersecting all non-zero ideals. $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 19:58
  • $\begingroup$ Yes, thank you, I was thinking of the nontrivial center, indeed. $\endgroup$ – Dietrich Burde Feb 16 '16 at 20:06
  • $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 20:09
  • $\begingroup$ Well, if the Lie algebra has a direct abelian factor it cannot be true. $\endgroup$ – Dietrich Burde Feb 16 '16 at 20:44

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