Having problems with the following statement:
If $N$ is a nontrivial normal subgroup of a nilpotent group $G$ then $N \cap Z(G) \neq \langle e \rangle$
Here $Z(G)$ denotes the center of $G$.
I've solved the finite case, where I reduced the problem to proving that a p-$group$ has this property. The infinite case resists my attempts though.
Since $G$ is nilpotent, the ascending central series ends with $G$ itself:
$\langle e\rangle \subseteq Z_1(G) \subseteq \ldots \subseteq Z_n(G) = G$ where $Z_1(G) = Z(G)$. Since $N \subseteq G$, there must be a smallest $i$ such that $N \cap Z_i(G) \neq \langle e \rangle$. I would like to show that $i = 1$ in this case, or perhaps just show that $N \cap Z_i(G)$ commutes with elements of $G$. Could somebody give me a tip in the right direction?
Best Answer
From the discussion in the comments two possible solutions arised, here is the first:
And a perhaps a bit simpler one, relying on an alternative characterisation of nilpotent groups: