A conjugacy class of prime-power size forces a proper nontrivial normal subgroup

Choose $g\in C$. Since $\mathrm{\mid }C\mathrm{\mid }>1$, one has $g\mathrm{\ne }e$. Split the irreducible characters into the trivial character, the set $D$ of nontrivial irreducibles whose degree is divisible by $p$, and the set $N$ of nontrivial irreducibles whose degree is not divisible by $p$. Applying [F1] at $h=e$ gives $0=1+{\sum }_{\chi \in D}\chi (1)\chi (g)+{\sum }_{\chi \in N}\chi (1)\chi (g)$.

Write $\chi (1)=p{m}_{\chi }$ for $\chi \in D$. Because each $\chi (g)$ is an algebraic integer by [F3], the sum $a:={\sum }_{\chi \in D}{m}_{\chi }\chi (g)$ is an algebraic integer. If the sum over $N$ were zero, step 1.1 would give $1=-pa$, so $-1\mathrm{/}p$ would be a rational algebraic integer, contradicting [F4]. Therefore some $\chi \in N$ satisfies $\chi (g)\mathrm{\ne }0$.

For that chosen $\chi$, the class size $\mathrm{\mid }C\mathrm{\mid }=p^a$ is coprime to $\chi (1)$ by the definition of $N$, so [F2] shows that $g$ acts as a scalar on the irreducible representation $V$ affording $\chi$. Every element of the conjugacy class $C$ is conjugate to $g$, so each of them acts by the same scalar on $V$.

Let $M=⟨a{b}^{-1}:a,b\in C⟩$. Because conjugation permutes $C$, the set $\{a{b}^{-1}:a,b\in C\}$ is stable under conjugation, so [F5] makes $M$ a normal subgroup of $G$. By step 3.1 every generator $a{b}^{-1}$ acts trivially on $V$, so $M$ acts trivially on the nontrivial irreducible representation $V$; hence $M\mathrm{\ne }G$.

Since $\mathrm{\mid }C\mathrm{\mid }>1$, choose distinct $a,b\in C$; then $a{b}^{-1}\mathrm{\ne }e$ lies in $M$, so $M$ is nontrivial. Together with step 4.1, this makes $M$ a proper nontrivial normal subgroup of $G$.