How could I prove the above statement?
For $G$ nilpotent and $N \lhd G$, how do I show that $N \cap Z(G) \neq 1$?
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For the Lie algebra analogue see here. – Dietrich Burde Feb 17 '16 at 19:18
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Take the upper central series of the group
$$1=Z_0\le Z_1\le\ldots\le Z_n=G$$
Since $\;1\neq N\lhd G\;$ there exists $\;1\le k< n\;$ such that $\;N\cap Z_k=1\;$ but $\;N\cap Z_{k+1}\neq 1\;$ , so take commutator groups:
$$[G, N\cap Z_{k+1}]\le[G,N]\cap[G,Z_{k+1}]\le N\cap Z_k=1$$
because normality: a subgroup $\;N\le G\;$ is normal iff $\;[G,N]\le N\;$ , and because centrality of the upper central series: $\;[G,Z_i]\le Z_{i-1}\;$
Thus, $\;[G, N\cap Z_{k+1}]=1\iff N\cap Z_{k+1}\subset Z_1=Z(G)$
DonAntonio
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No need to use an upper central series for this proof to work: $[G,N\cap Z_{k+1}]=1$ implies $N\cap Z_{k+1}\subseteq Z(G)$ anyway. – Leandro Caniglia Jan 02 '24 at 21:43