Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 5 / 3

For a pro-p group, the lower p-series is the descending sequence of closed characteristic subgroupsThe bar means closure of the subgroup generated by the displayed powers and group commutators. Throughout this question $p$ is odd. For a finitely generated powerful pro-p group, group commutator collection gives $P_i(G)=G^{p^{i-1}}$, each $P_i$ is powerful, and $\Phi (P_i)=P_{i+1}$. All these subgroups are open. A uniform pro-p group is such a group with all subgroup indices $[P_i:P_{i+1}]$ equal to $[G:P_2]$. We use the standard powerful-group torsion criterion: a finitely generated powerful pro-p group is uniform exactly when it is a torsion-free group. For a uniform group of generator number $d$, the indices are $p^d$, and $x\mapsto x^{p^a}$ is a homeomorphism from $G$ onto $G^{p^a}$.

We record explicitly the collection facts used for the operations below. With $U_b=G^{p^b}=P_{b+1}$, one hasThe second statement is the power-congruence property: the $p$th-power map is an isomorphism on adjacent uniform layers, and collection gives $(yu{)}^p\equiv y^p{u}^p(\mathrm{mod}{U}_{b+2})$ for $u\in U_b$. Thus the first nonzero layer of a difference moves down exactly one step under a $p$th power. Iterating gives the asserted equivalence and also justifies taking roots of congruences. These facts follow from powerful-group group commutator collection, using divisibility by $p$ of $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ and the successive-layer power isomorphisms; they are the uniform-group structure results being used here.

The transported addition on a uniform pro-p group isThe product is in $U_n$ and has a unique $p^n$th root in $G$. For the comparison at $n=1$ define $x{+}_{0}y=xy$. The numbered proofs below establish associativity, approximate commutativity and compatibility of these operations.

The intrinsic Lie algebra of a uniform pro-p group has underlying set $G$, zero equal to the group identity, additive inverse $x^{-1}$, and operationsHere the group commutator is $[x,y]=x^{-1}{y}^{-1}xy$. For $\lambda \in {\mathbb{Z}}_p$, $x^{\lambda }$ is the limit of $x^{m_j}$ for integers $m_j\to \lambda$; the pro-$p$ topology makes it well defined. The addition sequence is Cauchy by the compatibility proved below. The group commutator estimate puts $[x^{p^n},y^{p^n}]$ in $U_{2n+1}$, so its displayed root exists and lies in $U_1$. Collection of group commutators shows that successive bracket approximants differ in layers tending to the identity, giving the other limit. The group commutator identities give additivity of the limiting bracket, and the Hall-Witt identity gives its Jacobi identity. Approximate commutativity makes limiting addition commutative. The resulting lattice $L_G$ is a free module over ${\mathbb{Z}}_p$ of rank $d(G)$, and $[L_G,L_G]\le p{L}_G$.

A powerful Lie algebra over p-adic integers, in the convention relevant here, is a Lie algebra $L$ whose underlying module is finite free over ${\mathbb{Z}}_p$ and whose bracket satisfies $[L,L]\le pL$. For $p=2$ the condition is instead $[L,L]\le 4L$. For odd $p$, define multiplication on its underlying set by the convergent Baker--Campbell--Hausdorff formulaAn iterated bracket of degree $m$ lies in $p^{m-1}L$. The BCH coefficient bound loses at most $\lfloor (m-1)/(p-1)\rfloor$ powers of $p$, so all higher-degree terms are integral, and their valuations tend to infinity. This explains both convergence and why multiplying a lattice with small bracket stays inside the lattice. Formal associativity of BCH passes to the convergent series, the identity is $0$, the inverse is $-x$, and every group power satisfies $x^{\ast m}=mx$. The quotients by $p^{a}L$ are finite p-groups. Hence the resulting compact group is a finitely generated torsion-free group and a pro-p group, and its group commutator lies in $pL$. Thus $(L,\ast )$ is a uniform pro-p group, with $P_i(L,\ast )=p^{i-1}L$ and successive index $p^{{\mathrm{rank}}_{{\mathbb{Z}}_p}L}$.

The standard uniform-group/Lie-lattice correspondence identifies these constructions as mutually inverse: $L_{(L,\ast )}\cong L$ and $(L_G,\ast )\cong G$. To see why the inverse limits recover the lattice operations, the addition approximant for a BCH group is $p^{-n}\mathrm{BCH}(p^{n}x,p^{n}y)$: its linear term is $x+y$ and every higher term tends to zero. Similarly its group commutator approximant has limiting term $[x,y]$. In the other direction, a local logarithm in the complete normed group algebra identifies the intrinsic operations with algebra addition and commutator and identifies group multiplication with BCH; the uniform power maps extend the identification from a sufficiently small power subgroup to all of $G$. Continuous group homomorphisms respect powers, unique roots and limits, while Lie algebra homomorphisms respect every BCH term, so the equivalence also preserves morphisms.

For a powerful pro-p group, $\Phi (P_i)=P_{i+1}$. The Burnside basis theorem therefore givesIf $G$ is uniform, all these indices equal $[G:P_2]=p^{d(G)}$. Hence $d(P_i)=d(G)$ for every $i\ge 1$, proving (i)$\Rightarrow$(ii).

Conversely, if all $d(P_i)$ equal $d(G)$, the same Frattini-quotient identity gives $[P_i:P_{i+1}]=p^{d(G)}$ for every $i$. This is exactly the index definition of a uniform pro-p group. Thus (ii)$\Rightarrow$(i), completing the equivalence of the first two conditions.

Suppose first that $G$ is uniform and $H$ is a powerful open subgroup. The torsion criterion stated above makes $G$ a torsion-free group; so $H$ is a torsion-free group too. It is finitely generated, for example by the Schreier lemma, and hence is uniform by the same criterion. Put $d=d(G)$ and $e=d(H)$, and choose $k$ with $P_k(G)\le H$. For every $n\ge 0$,Uniformity of the two groups givesTaking logarithms and dividing by $n$, then letting $n\to \infty$, proves $e=d$. This proves (i)$\Rightarrow$(iii) without assuming that $H$ is a power subgroup. Conversely every $P_i(G)$ is powerful and open, so (iii) applied to these subgroups gives (ii). Therefore all three conditions are equivalent.

The remaining nested parts establish the stated properties of the transported operations ${+}_n$ defined in the root solution.

The map ${\sigma }_n:G\to U_n$, $x\mapsto x^{p^n}$, is a bijection, and the operation ${+}_n$ is exactly multiplication transported through it. Explicitly,Injectivity of ${\sigma }_n$ proves $(x{+}_{n}y){+}_{n}z=x{+}_n(y{+}_{n}z)$. The approximate operation need not be commutative; its associativity is exact.

The uniform group commutator estimate givesThus $x^{p^n}{y}^{p^n}$ and $y^{p^n}{x}^{p^n}$ are congruent modulo $U_{2n+1}$. Taking their unique $p^n$th roots and using the power-congruence property moves the error layer back by $n$, giving

Put $a=x^{p^{n-1}}$ and $b=y^{p^{n-1}}$, both in $U_{n-1}$. The collection formula givesIndeed the weight-two group commutator lies in $[U_{n-1},U_{n-1}]\le U_{2n-1}$, and its coefficient $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ is divisible by $p$ since $p$ is odd, moving it into $U_{2n}$. A group commutator of weight $j\ge 3$ in $a,b$ lies in $U_{jn-1}\le U_{2n}$; all higher terms are therefore harmless as well. This also covers $n=1$.

Let $v=x{+}_{n-1}y$ and $u=x{+}_{n}y$. Their powers satisfy $v^{p^n}=(ab{)}^p$ and $u^{p^n}=a^p{b}^p$. Taking $p^n$th roots of the collected congruence gives $u\equiv v(\mathrm{mod}{U}_n)$. Since $U_n=P_{n+1}\le P_n$, this proves the requested, slightly weaker, resultFor $n=1$ the definition ${+}_0=$ ordinary multiplication makes the comparison meaningful.