= Solution
For a <powerful p-group>, $G_2=G^p$, and $G_3=(G^p)^p[G^p,G]$. In $G/G_3$, the image of $G_2$ is central and has exponent dividing $p$, while the <commutator subgroup> is contained in that image. The quotient thus has nilpotency class at most two. Direct collection in such a <group> gives
$$
(xy)^p=x^py^p[y,x]^{\binom p2}\equiv x^py^p\pmod {G_3},
$$
because $p$ is odd. If $z\in G_2$, the same calculation gives $(xz)^p\equiv x^p$ modulo $G_3$, since $z^p\in G_3$. Consequently the power map is well defined on $G/G_2$ and is a <group homomorphism> to $G_2/G_3$. Its image contains all the power generators of that quotient. \b[It is therefore surjective.]
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