= Solution
Write $G=\langle a,b\rangle$ and $c=[a,b]$. The <commutator subgroup> $G\prime$ is the <normal closure> of $c$: imposing $c=1$ makes the two generators commute, so the quotient is abelian. Part (iii), with $N=G$, makes $G\prime$ a <powerfully embedded subgroup>. Part (ii) therefore gives
$$
G\prime=\langle c\rangle.
$$
This proves the hint, but cyclicity of the <commutator subgroup> alone is not yet the required cyclic normal subgroup with cyclic quotient.
If $c=1$, the <group> is abelian and any one of the two generator subgroups is normal with cyclic quotient. Otherwise $c\in G^p$. Part (v) permits taking a $p$th root of $c$. Whenever the chosen root is still in $G^p$, take another $p$th root. The order increases by a factor $p$ at every step, since the power being rooted is nonidentity. Finiteness forces this process to stop. Thus
$$
c=d^{p^r},\qquad d\notin G^p=\Phi(G)
$$
for some $d$. The <cyclic subgroup> $A=\langle d\rangle$ contains $G\prime$, and hence is normal: every conjugate of $d$ differs from $d$ by an element of $G\prime\leq A$. The nonzero image of $d$ in the <Frattini quotient> can be completed to a basis of dimension at most two. By the <Burnside basis theorem>, $G=\langle d,e\rangle$ for a suitable $e$; if the quotient has dimension one, $d$ already generates $G$. Therefore $G/A$ is cyclic. \b[The powerful two-generator finite p-group is metacyclic.]
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