Prove that if \(G\) is a \(p\dash\)group, every subgroup \(N\normal G\) intersects the center \(Z(G)\).
Hint: use the class equation.
Solution
Let \(G\) be a finite \(p\)-group and \(1\neq N\normal G\).
1. \(N\) is a disjoint union of \(G\)-conjugacy classes, and the classes of size \(1\) contained in \(N\) are the elements of \(N\cap Z(G)\).
Normality gives \(gng^{-1}\in N\) for \(n\in N\), \(g\in G\), so the \(G\)-conjugacy class of each \(n\in N\) lies in \(N\). The class of \(n\) is \(\{n\}\) exactly when \(n\) commutes with every element of \(G\).
2. Every \(G\)-conjugacy class of size greater than \(1\) has size divisible by \(p\).
The class of \(n\) has size \([G:C_G(n)]\), which divides \(\size G\) and so is a power of \(p\).
3. Q.E.D.
By steps 1 and 2, \(\size N\equiv\size(N\cap Z(G))\pmod p\). Since \(N\neq1\) is a subgroup of a \(p\)-group, \(p\mid\size N\), so \(p\mid\size(N\cap Z(G))\). As \(e\in N\cap Z(G)\), this gives \(\size(N\cap Z(G))\ge p\), so \(N\cap Z(G)\neq1\).