Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 4 / 3

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,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 $N^p$ is powerfully embedded in $G$.

For $H=⟨N,x⟩$, normality of $N$ gives $H=N⟨x⟩$. 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. ThereforeThus $⟨N,x⟩$ 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 isIndeed, 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 $[u,x{]}^{\left(\genfrac{}{}{0ex}{}{p}{2}\right)}$, 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.

Initially $G_1=G$ is powerfully embedded in itself. If $G_i$ is powerfully embedded in $G$, then $[G_i,G]\subseteq G_i^p$ and consequently $G_{i+1}=G_i^p$. The preliminary result says that $G_i^p$ is again powerfully embedded in $G$, completing the induction.

For a finite p-group $A$, the Frattini subgroup is $\Phi (A)=A^p[A,A]$. Here $[G_i,G_i]\subseteq [G_i,G]\subseteq G_i^p$, soSince the Frattini subgroup of a nontrivial finite p-group is proper, this series eventually reaches $1$. Thus $P_j=G^{p^j}$, with iterated power subgroups, terminates.

Modulo $G_3$, the subgroup $G_2=G^p$ is central because $[G_2,G]\subseteq G_3$, and has exponent dividing $p$ because $G_2^p=G_3$. Thus the group $G/G_3$ has nilpotency class at most two. Collection givessince $[y,x]\in G_2$ and $p\mid \left(\genfrac{}{}{0ex}{}{p}{2}\right)$. Also replacing $x$ by $xu$ with $u\in G_2$ leaves its $p$-th power unchanged modulo $G_3$. Hence the assignment is well-defined and is a group homomorphism. Its image contains all the generators $x^p{G}_3$ of $G_2/G_3$, so it is onto:

First, the images of $a_1^p,\dots ,a_n^p$ span $G_2/G_3$ by the surjective group homomorphism in (ii). Part (i) identifies $G_3$ with $\Phi (G_2)$, so the Burnside basis theorem givesThis proves generation by the indicated powers; it does not yet prove that an arbitrary product of powers is itself a power.

For that stronger assertion, let $y\in P_1=G^p$. Part (ii) supplies $x$ with $x^p\equiv y(\mathrm{mod}{P}_2)$. Suppose at some stage $x^p\equiv y(\mathrm{mod}{P}_{j+1})$, where $j\ge 1$. Apply (ii) to the powerful p-group $P_j$: its first two power layers are $P_{j+1},P_{j+2}$. We can therefore choose $u\in P_j$ withThe preliminary power congruence gives $(xu{)}^p\equiv y(\mathrm{mod}{P}_{j+2})$. Replacing $x$ by $xu$ improves the root by one layer. The lower p-series terminates, so finitely many corrections yield $x^p=y$ exactly. Every element of $G^p$ is a $p$-th power in $G$.

Apply (iii) successively to the powerful p-groups $G_i$. Starting with $G_1=⟨a_1,\dots ,a_n⟩$, if $G_i=⟨a_1^{p^{i-1}},\dots ,a_n^{p^{i-1}}⟩$, thenThus

For each $j\ge 0$, the elementary abelian group $P_j/P_{j+1}$ is spanned by the images of $a_i^{p^j}$. We construct exponents $e_i$ so that $g$ equals the ordered product $a_1^{e_1}\cdots a_n^{e_n}$ to successively greater precision. Suppose this product agrees with $g$ modulo $P_j$; initially this is automatic for $j=0$. Its error modulo $P_{j+1}$ can be expressed using the images of $a_i^{p^j}$. Add the corresponding digits $p^j{t}_i$ to the exponents $e_i$. Since $[P_j,G]\subseteq P_{j+1}$, moving each new factor $a_i^{p^j{t}_i}$ past the other factors changes nothing modulo $P_{j+1}$. The new ordered product therefore agrees with $g$ modulo $P_{j+1}$. Termination of the lower p-series provesNo uniqueness of this factorization is asserted for a redundant generating tuple.

The printed converse is false without an additional condition. For any odd $p$, take the Heisenberg group $G={\mathrm{UT}}_3({\mathbb{F}}_p)$ withEvery element is uniquely $x^a{y}^b{z}^c$, with $a,b,c\in {\mathbb{F}}_p$, so $G$ is a product of three cyclic subgroups. For $A$ strictly upper triangular, $A^3=0$ andin characteristic $p$, including $p=3$. Therefore $G^p=1$, whereas $[G,G]=⟨z⟩\ne 1$. This finite p-group is not powerful, disproving the literal request in the authoritative PDF.

A precise nearby correction is the minimal cyclic factorization criterion for powerful p-groups: require a product of $d(G)$ cyclic subgroups, where $d(G)$ is the minimal generator number. To prove it, put $d=d(G)$ and $H=G/G^p$. The quotient has exponent $p$ and still has minimal generator number $d$, because $G^p\subseteq \Phi (G)$. Each cyclic factor has image of order at most $p$, so $|H|\le p^d$. But $|H/\Phi (H)|=p^d$ by the Burnside basis theorem. Consequently $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. A product of exactly $d(G)$ cyclic subgroups is powerful. A direct product of cyclic groups is also powerful, since it is abelian; neither stronger hypothesis appears in the printed converse.