Solution (source code)

= Solution

Throughout this question $p$ is odd, $A^p$ means the subgroup generated by the $p$-th powers of elements of $A$, and the <commutator of group elements> convention is $[x,y]=x^{-1}y^{-1}xy$. A subgroup $N$ is <powerfully embedded> in $G$ when $[N,G]\subseteq N^p$. In particular $N$ is normal, since $n^g=n[n,g]\in N$. A <finite p-group> $G$ is a <powerful p-group> when $[G,G]\subseteq G^p$.

First put $M=N^p$. It is normal in $G$. By the reduction permitted in the question, it suffices to prove $[M,G]\subseteq M^p$ after factoring out $[M,G,G]$. In that quotient, $[M,G]$ is central. For $n\in N$ and $g\in G$, put $c=[n,g]\in M$. Then $[c,n]$ is central, and the elementary identity $[ab,g]=[a,g]^b[b,g]$ gives, by induction,
$$
[n^p,g]=c^{n^{p-1}}c^{n^{p-2}}\cdots c
=c^p[c,n]^{p(p-1)/2}.
$$
Both $c$ and $[c,n]$ belong to $M$, and $p$ divides $p(p-1)/2$ because $p$ is odd. Hence $[n^p,g]\in M^p$. The subgroup $M^p$ is normal in $G$, so the commutator identity extends this inclusion from the generating elements $n^p$ to every element of $M$. Applying the allowed reduction proves \b[$N^p$ is powerfully embedded in $G$.]

For $H=\langle N,x\rangle$, normality of $N$ gives $H=N\langle x\rangle$. Its quotient by $N$ is cyclic. Modulo $[N,H]$, the image of $N$ is central and the remaining generator is $x$, so that quotient is abelian. Therefore
$$
[H,H]\subseteq[N,H]\subseteq[N,G]\subseteq N^p\subseteq H^p.
$$
Thus \b[$\langle N,x\rangle$ is powerful.]

The <lower p-series> is $G_1=G$, $G_{i+1}=G_i^p[G_i,G]$. For the later calculations write $P_j=G_{j+1}$, so $P_0=G$. Part (i) will show $P_{j+1}=P_j^p$ and $[P_j,G]\subseteq P_{j+1}$. A useful <power lifting in a powerful p-group> congruence is
$$
(xu)^p\equiv x^pu^p\pmod {P_{j+2}}\qquad(x\in G,\ u\in P_j).
$$
Indeed, modulo $P_{j+2}$ the <commutators> $[u,x]$ are central, lie in $P_{j+1}$ and have exponent dividing $p$. Collecting the $p$ copies of $xu$ therefore gives only the correction $[u,x]^{\binom p2}$, which is trivial. This argument also covers $j=0$. The congruence is the reason roots and coordinates can be corrected one power layer at a time.