= 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 $[N,G]\leq N^p$. 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 \b[$N=\langle X\rangle$]. Equivalently, a proper generated <subgroup> would lie in a maximal <subgroup> of $N$, which contains $N^p$ and contradicts generation of that quotient.
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