Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 5 / 3 / ii / Solution

The powerfully embedded subgroup $N$ is itself a powerful p-group, so the finite Frattini subgroup formula gives $\Phi (N)=N^p$. Conjugation by $G$ acts trivially on $N/N^p$, because $[N,G]\le N^p$. Since $N$ is the normal closure of $X$, the image of $X$ therefore generates $N/N^p$: all conjugates have the same images as the original elements. The Burnside basis theorem now gives $N=⟨X⟩$. Equivalently, a proper generated subgroup would lie in a maximal subgroup of $N$, which contains $N^p$ and contradicts generation of that quotient.