This question already has an answer here:

How could I prove the above statement?

For $G$ nilpotent and $N \lhd G$, how do I show that $N \cap Z(G) \neq 1$?


marked as duplicate by Dietrich Burde, Community Feb 17 '16 at 19:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ For the Lie algebra analogue see here. $\endgroup$ – Dietrich Burde Feb 17 '16 at 19:18

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, $\;[G, N\cap Z_{k+1}]=1\iff N\cap Z_{k+1}\subset Z_1=Z(G)$


Not the answer you're looking for? Browse other questions tagged or ask your own question.