 Solution

Here $p$ is odd. For a finite p-group, write $N^p=⟨n^p:n\in N⟩$. A powerfully embedded subgroup satisfies $[N,G]\le N^p$; this also makes $N$ normal, since every conjugate of an element of $N$ differs from it by an element of $N$. A powerful p-group is one with $[G,G]\le G^p$, namely one powerfully embedded in itself. Use $[a,b]=a^{-1}{b}^{-1}ab$ and $a^b=b^{-1}ab$ throughout.

The lower p-series is defined byEach quotient $G_i/G_{i+1}$ is an elementary abelian p-group, and $[G_i,G]\le G_{i+1}$. These are generated subgroups, without closure bars, because the group is finite.