= Solution
Here $p$ is odd. For a <finite p-group>, write $N^p=\langle n^p:n\in N\rangle$. A <powerfully embedded subgroup> satisfies $[N,G]\leq N^p$; 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 $[G,G]\leq G^p$, 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 by
$$
G_1=G,\qquad G_{i+1}=G_i^p[G_i,G].
$$
Each quotient $G_i/G_{i+1}$ is an <elementary abelian p-group>, and $[G_i,G]\leq G_{i+1}$. These are generated <subgroups>, without closure bars, because the <group> is finite.
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