Show that every nontrivial normal subgroup of a finite \(p\)-group meets the center nontrivially.
Hint: let \(G\) act on the normal subgroup by conjugation and count the orbits.
Solution
Let \(1\neq N\trianglelefteq G\), where \(G\) is a finite \(p\)-group.
1. Conjugation gives an action of \(G\) on \(N\).
Normality of \(N\) gives \(gng^{-1}\in N\) for all \(g\in G\) and \(n\in N\), so \[ g\cdot n=gng^{-1} \] defines the action.
2. The fixed-point set of this action is \(N\cap Z(G)\).
An element \(n\in N\) is fixed by every \(g\in G\) exactly when \(gng^{-1}=n\) for every \(g\), which is equivalent to \(n\in Z(G)\).
3. Every nontrivial orbit has cardinality divisible by \(p\).
For \(n\in N\), orbit-stabilizer gives \[ |G\cdot n|=[G:C_G(n)]. \] This index divides the order of the \(p\)-group \(G\), so it is a power of \(p\). If the orbit has more than one element, its cardinality is therefore divisible by \(p\).
4. The intersection \(N\cap Z(G)\) has cardinality divisible by \(p\).
The orbit decomposition of \(N\) gives \[ |N|=|N\cap Z(G)|+\sum_i |\mathcal O_i|, \] where the \(\mathcal O_i\) are the nontrivial orbits. Since \(N\) is a nontrivial subgroup of a finite \(p\)-group, \(p\mid |N|\), and by 3 each summand in the sum is divisible by \(p\). Hence \[ |N\cap Z(G)|\equiv0\pmod p. \]
5. Therefore \(N\cap Z(G)\) is nontrivial.
4. Thus \(N\cap Z(G)\neq\{1\}\).