Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 5 / 3 / vi / Solution

Write $G=⟨a,b⟩$ 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 givesThis 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. Thusfor some $d$. The cyclic subgroup $A=⟨d⟩$ contains $G\prime$, and hence is normal: every conjugate of $d$ differs from $d$ by an element of $G\prime \le A$. 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, $G=⟨d,e⟩$ for a suitable $e$; if the quotient has dimension one, $d$ already generates $G$. Therefore $G/A$ is cyclic. The powerful two-generator finite p-group is metacyclic.