 Isaacs' character-degree criterion for p-nilpotency

The integer $h=|G:G^{\prime }{|}_p$ divides $s={\sum }_{\chi \in {\mathrm{Irr}}_{p^{\prime }}(G)}\chi (1{)}^2$. The group is p-nilpotent exactly when $p\nmid s/h$. To see this, let $N=O^p(G)$. Invariant p-prime-degree characters of $N$ extend, and each has exactly $h$ extensions of p-prime degree. Conjugation by the p-group $G/N$ then gives $s/h\equiv |N|(\mathrm{mod}p)$. Thus the quotient is prime to $p$ precisely when $N$ is a normal p-complement.