Show that in a finite \(p\dash\)group, every nontrivial normal subgroup intersects the center nontrivially.
Solution
Let \(G\) be a finite \(p\)-group and let \(1\neq N\trianglelefteq G\).
1. The group \(G\) acts on the finite set \(N\) by conjugation.
For \(g\in G\) and \(x\in N\), define \(g\cdot x=gxg^{-1}\). Because \(N\trianglelefteq G\), one has \(gxg^{-1}\in N\), so this is a well-defined action on \(N\).
2. The fixed-point set for this action is exactly \(N\cap Z(G)\).
An element \(x\in N\) is fixed by every \(g\in G\) precisely when \[ gxg^{-1}=x\qquad\text{for all }g\in G, \] which is equivalent to \(x\in Z(G)\). Since already \(x\in N\), the fixed points are exactly \(N\cap Z(G)\).
3. Every non-fixed orbit has cardinality divisible by \(p\).
For \(x\in N\), orbit-stabilizer gives \[ |G\cdot x|=[G:C_G(x)]. \] Since \(G\) is a finite \(p\)-group, every subgroup index is a power of \(p\). If \(x\) is not fixed, then \(C_G(x)\neq G\), so \([G:C_G(x)]\) is a positive power of \(p\), hence divisible by \(p\).
4. Therefore\[ |N|\equiv |N\cap Z(G)|\pmod p. \]
Partition \(N\) into its \(G\)-orbits. By 2, the fixed points contribute \(|N\cap Z(G)|\) singleton orbits. By 3, every remaining orbit has size divisible by \(p\). Summing the orbit sizes gives the congruence.
5. The integer \(|N\cap Z(G)|\) is divisible by \(p\).
Because \(N\) is a nontrivial subgroup of the finite \(p\)-group \(G\), its order is \(p^a\) for some \(a\ge1\); hence \(p\mid |N|\). Apply 4.
6. Hence \(N\cap Z(G)\neq1\).
The subgroup \(N\cap Z(G)\) contains the identity, so its cardinality is positive. By 5 it is divisible by the prime \(p\), hence it has at least \(p>1\) elements. Therefore it contains a nonidentity element.