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

Here $p$ is odd. For a finite p-group, write $N^p=⟨n^p:n\in N⟩$. A powerfully embedded subgroup satisfies $[N,G]\le N^p$; 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 $[G,G]\le G^p$, 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 byEach quotient $G_i/G_{i+1}$ is an elementary abelian p-group, and $[G_i,G]\le G_{i+1}$. These are generated subgroups, without closure bars, because the group is finite.

Since $N$ is normal in $G$, it is normal in $H=⟨x,N⟩$, and $H/N$ is a cyclic group. More specifically, quotienting $H$ by $[N,H]$ makes the image of $N$ central. The quotient is generated by this central image and the single element $x$, and is therefore abelian. HenceThus $H$ is a powerful p-group. Merely saying that $H/N$ is cyclic would give only $[H,H]\le N$; centralizing $N$ is the extra step needed here.

The powerfully embedded subgroup $N$ is itself a powerful p-group, so the finite Frattini subgroup formula gives $\Phi (N)=N^p$. Conjugation by $G$ acts trivially on $N/N^p$, because $[N,G]\le N^p$. Since $N$ is the normal closure of $X$, the image of $X$ therefore generates $N/N^p$: all conjugates have the same images as the original elements. The Burnside basis theorem now gives $N=⟨X⟩$. Equivalently, a proper generated subgroup would lie in a maximal subgroup of $N$, which contains $N^p$ and contradicts generation of that quotient.

Use the allowed reduction $[N,G,G,G]=1$. Set $C=[N,G]$ and $D=[C,G]$. Then $[D,G]=1$, so $D$ is central. For $n\in N$ and $g\in G$, put $c=[n,g]$. The identity $[ab,g]=[a,g{]}^b[b,g]$ and centrality of $[c,n]\in D$ give, by induction on $k$,At $k=p$, both factors belong to $C^p$: the first is a $p$th power in $C$, and the second is a $p$th power because $p\mid \left(\genfrac{}{}{0ex}{}{p}{2}\right)$ and $[c,n]\in D\le C$. The subgroup $C^p$ is normal in $G$. Commuting products of the generators $n^p$ with $g$ and using the same identity consequently yieldsSince $C=[N,G]\le N^p$, it follows that $[C,G]\le C^p$. Thus $[N,G]$ is powerfully embedded. This proves the reduced case, and the reduction permitted in the question gives the general assertion.

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 givesbecause $p$ 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. It is therefore surjective.

Set $P_0=G$ and $P_{i+1}=P_i^p$. The permitted power-embedding result, applied repeatedly, gives $[P_i,G]\le P_{i+1}$, and each $P_i$ is a powerful p-group. These subgroups eventually reach $1$: if $P_i\ne 1$, its Frattini subgroup $P_i^p$ is proper. In fact these are the lower p-series layers, with their indices shifted by one.

For $x\in G$ and $y\in P_i$, the commutator $[y,x]$ lies in $P_{i+1}$, is central modulo $P_{i+2}$, and has exponent dividing $p$ there. The class-two collection calculation from part (iv) therefore givesApplying part (iv) inside $P_i$, whose third lower-series term is contained in $P_{i+2}$, shows that the map $y\mapsto y^p$ is onto $P_{i+1}/P_{i+2}$.

Given $a\in P_1$, first choose $x$ with $x^p\equiv a(\mathrm{mod}{P}_2)$. Suppose $x^p\equiv a(\mathrm{mod}{P}_{i+1})$. Choose $y\in P_i$ with $y^p\equiv x^{-p}a(\mathrm{mod}{P}_{i+2})$. The displayed collection congruence then gives $(xy{)}^p\equiv a(\mathrm{mod}{P}_{i+2})$. Replacing $x$ by $xy$ improves the approximation by one layer. Since the filtration on a group terminates, a finite sequence of corrections gives $x^p=a$ exactly. The reverse inclusion is the definition of $G^p$. Hence

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.