Show that if \(G\) is a group of order \(p^n\), where \(p\) is a prime, and \(N\) is a nontrivial normal subgroup, then \(N\) intersects the center of \(G\) nontrivially.
The hypothesis that \(N\) is nontrivial is necessary. For \(N=\{1\}\), one has \(N\cap Z(G)=\{1\}\), so the conclusion in the unqualified source statement fails.
Solution
1. Conjugation by \(G\) defines an action on \(N\), whose fixed points are exactly \(N\cap Z(G)\).
Normality of \(N\) ensures that \[ gng^{-1}\in N \qquad(g\in G,\ n\in N), \] so conjugation restricts to an action of \(G\) on \(N\). An element \(n\in N\) is fixed by every \(g\in G\) precisely when \[ gng^{-1}=n \qquad\text{for all }g\in G, \] which is equivalent to \(n\in Z(G)\). Thus the fixed-point set is \(N\cap Z(G)\).
2. Every nontrivial orbit in this action has cardinality divisible by \(p\).
For \(n\in N\), orbit-stabilizer gives \[ |G\cdot n|=[G:C_G(n)]. \] Since \(|G|=p^n\), every subgroup index in \(G\) is a power of \(p\). If the orbit is not a singleton, its size is therefore a positive power of \(p\), hence is divisible by \(p\).
3. The fixed-point set contains a nonidentity element.
Partition \(N\) into conjugation orbits. Using 1 and 2 gives a class equation of the form \[ |N|=|N\cap Z(G)|+\sum_i p^{a_i}, \qquad a_i\ge1, \] where the sum runs over the nontrivial orbits. Therefore \[ |N|\equiv |N\cap Z(G)|\pmod p. \]
Because \(N\) is a nontrivial subgroup of the finite \(p\)-group \(G\), its order is a positive power of \(p\) and hence is divisible by \(p\). It follows that \(p\) divides \(|N\cap Z(G)|\). The latter subgroup contains the identity, so its order is positive; divisibility by \(p\) forces \[ |N\cap Z(G)|\ge p>1. \] Thus \(N\cap Z(G)\) is nontrivial.