Let \(G\) be a finite \(p\)-group and let \(1\ne N\trianglelefteq G\). Prove that \[ N\cap Z(G)\ne\{e\}. \]
Solution
Let \(G\) act on \(N\) by conjugation. This is well-defined because \(N\) is normal. The fixed points are precisely \[ N\cap Z(G). \] Every non-fixed orbit has size \[ [G:C_G(x)], \] a positive power of \(p\), hence divisible by \(p\).
The class equation for the action gives \[ |N| = |N\cap Z(G)| + \sum \text{(non-fixed orbit sizes)}. \] Since \(N\) is a nontrivial subgroup of a \(p\)-group, \(p\mid|N|\). Therefore \[ p\mid |N\cap Z(G)|, \] so \(N\cap Z(G)\) is nontrivial.