Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 4 (Pro-$p$ groups)

An inverse system of topological spaces over a directed set $I$ consists of spaces $X_i$ and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=1$ and $f_{ik}=f_{ij}{f}_{jk}$. Its inverse limit is a space $X$ with compatible continuous projections ${\pi }_i$ such that every compatible family of continuous maps $u_i:Y\to X_i$ factors uniquely through a continuous map $u:Y\to X$. To construct it, takewith the subspace topology inherited from the product topology. The map $u(y)=(u_i(y))$ has values in $X$ and is continuous by the definition of the product topology; its coordinates force uniqueness. If $X^{\prime }$ is another inverse limit, the two universal properties give continuous maps $X\to X^{\prime }$ and $X^{\prime }\to X$ whose composites have all the same projections as the identity. Uniqueness makes these composites the identity. The inverse limit exists and is unique up to a unique projection-preserving homeomorphism. An arbitrary inverse limit may be empty; nonemptiness requires additional hypotheses.

Use the usual Hausdorff convention for a topological group. An inverse limit of finite discrete groups is a closed subgroup of their product: each compatibility condition is closed because the finite factors are Hausdorff spaces. The Tychonoff theorem gives compactness, and distinct tuples are separated by a clopen coordinate cylinder, so the limit is totally disconnected. Coordinatewise multiplication and inversion are continuous, giving a profinite group.

Conversely, let $G$ be a compact Hausdorff space and a totally disconnected topological group. We use the standard topological fact that a compact Hausdorff space that is totally disconnected has a basis of clopen sets. Given an identity neighbourhood, choose a clopen identity neighbourhood $U$ inside it. Compactness of $U$ and continuity of multiplication give a symmetric identity neighbourhood $V$ with $VU\subseteq U$: cover $U$ by finitely many neighbourhoods on which multiplication by a sufficiently small identity neighbourhood stays inside $U$, then intersect those identity neighbourhoods. Symmetry gives $vU=U$ for $v\in V$. Consequently $H=\{g:gU=U\}$ is a subgroup containing $V$, hence an open subgroup, and $H\subseteq U$ since $1\in U$. Its open cosets cover the compact group, so there are finitely many. The normal core $N={\bigcap }_{g\in G}gH{g}^{-1}$ is a finite intersection of open subgroups, hence an open normal subgroup contained in $U$. Thus the open normal subgroups form an identity-neighbourhood basis and have trivial intersection.

The canonical group homomorphism $G\to {\underleftarrow{\lim }}_{N}G/N$ is continuous and injective. For a compatible tuple of quotient elements, its inverse-image cosets in $G$ are closed and have the finite intersection property: a finite family is refined by the quotient for the intersection of its normal subgroups. Compactness gives an element in all those cosets, proving surjectivity. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The compact totally disconnected and inverse-limit definitions of a profinite group agree.

Index the finite nonempty sets $\mathcal{P}(N)$ by open normal subgroups, with the transition from $M$ to $N$ when $M\subseteq N$ given by image under $G/M\to G/N$. Images of Sylow subgroups under surjections of finite groups are Sylow subgroups. These transitions are surjective: if $S\le G/N$ is a Sylow subgroup, choose a Sylow subgroup in its full inverse image in $G/M$. It maps onto $S$, and its order has the full $p$-part of $|G/M|$, so it is a Sylow subgroup of $G/M$ too. The transitions respect composition, giving an inverse system of finite sets.

Its inverse limit is nonempty. Indeed, finitely many compatibility conditions can be satisfied by choosing one Sylow subgroup in a quotient refining all the involved quotients and taking its images; compactness of the product of the finite sets then gives a compatible family $(P_N{)}_N$. DefineThis is a closed subgroup of $G$. Its image in $G/N$ is exactly $P_N$: for a prescribed $s\in P_N$, each finite collection of conditions defining $P$ together with ${\pi }_N(g)=s$ is satisfiable in a common refining quotient, because the transitions on the chosen Sylow subgroups are surjective. The finite intersection property again supplies $g\in P$. Hence $P\cong {\underleftarrow{\lim }}_N{P}_N$ is a pro-p group and $PN/N=P_N$ for every $N$.

Finally, if a pro-p subgroup $Q$ contains $P$, its image in every finite quotient is a p-group containing the Sylow subgroup $P_N$, so that image equals $P_N$. Thus every $q\in Q$ belongs to every ${\pi }_N^{-1}(P_N)$, whence $Q\subseteq P$. The constructed $P$ is a maximal pro-p subgroup.

A topological group is a group with a topology for which $(x,y)\mapsto xy$ and $x\mapsto x^{-1}$ are continuous; we use the Hausdorff convention. Translations and inversion are then homeomorphisms.

For the final generator-extraction assertion, use the profinite Frattini subgroupFor a finitely generated pro-p group, the Burnside basis theorem says that $G/\Phi (G)$ is a finite-dimensional vector space over ${\mathbb{F}}_p$ and a subset topologically generates $G$ precisely when its image spans that vector space. In particular its dimension is the minimal number $d(G)$ of topological generators. A given $d$-element generating set shows $d(G)\le d$. The image of any other generating set spans the same finite-dimensional vector space, so it contains a basis of $d(G)$ elements. Choose their preimages in that generating set and apply the Burnside basis theorem again. Every generating set contains a generating subset of size $d(G)\le d$. This argument also applies to an infinite generating set.

If $H$ is an open subgroup, each left coset $gH$ is open because translation is a homeomorphism. Its complement is the union of the other cosets, hence is open. Every open subgroup is closed. No compactness is needed.

Let $x,y\in \overline{H}$ and let $W$ be an open neighbourhood of $x{y}^{-1}$. Continuity of $(u,v)\mapsto u{v}^{-1}$ supplies neighbourhoods $U$ of $x$ and $V$ of $y$ with $U{V}^{-1}\subseteq W$. Choose $h\in H\cap U$ and $k\in H\cap V$. Then $h{k}^{-1}\in H\cap W$, so $x{y}^{-1}\in \overline{H}$. The subgroup criterion using $x{y}^{-1}$ proves the closure of a subgroup is a subgroup. The neighbourhood argument avoids assuming the topology is determined by sequences.

For an open subgroup $H$ of a profinite group $G$, the open left cosets form a disjoint cover of a compact space. A finite subcover must include every coset, so $[G:H]$ is finite. Choose representatives $t_1,\dots ,t_m$ for these cosets. The normal core isEach conjugate is open, and the intersection is finite, so $N$ is an open normal subgroup contained in $H$. $[G:H]<\infty$ and $H$ contains an open normal subgroup of $G$.

Choose a finite set $S$ of topological generators of $G$ and let $\Gamma =⟨S⟩$ denote the abstract subgroup it generates. Then $\Gamma$ is dense in $G$. For an open subgroup $H$, the abstract subgroup $K=\Gamma \cap H$ is dense in $H$, since every nonempty open subset of $H$ is open in $G$. Every coset of $H$ meets $\Gamma$, so $[\Gamma :K]=[G:H]=m<\infty$.

Here is the needed Schreier lemma argument explicitly. Choose representatives $t$ in $\Gamma$ for the right cosets $K\backslash \Gamma$, including $1$ for $K$, and write $\overline{\gamma }$ for the representative of $K\gamma$. The finite setis contained in $K$. For $k=s_1\cdots s_r\in K$, set $t_j=\overline{s_1\cdots s_j}$, with $t_0=t_r=1$. Thenand every factor belongs to the displayed finite set. Thus $K$ is finitely generated as an abstract group. Its finite generators topologically generate its closure $H$. Every open subgroup of a finitely generated profinite group is finitely generated.

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.

Put $U_0=G$ and $U_{j+1}=U_j^p$, where powers mean closed generated subgroups. The finite-quotient arguments of Question 3 pass to a powerful pro-p group: in particular $U_j$ is powerfully embedded, the power map is onto $U_{j+1}$, andFor the root assertion, apply the finite result to every finite quotient and use the finite intersection property on the closed sets of possible roots in the compact group. Each $U_j$ is finitely generated by the appropriate powers of a finite generating tuple, and $U_j/U_{j+1}$ is finite and elementary abelian. Thus the $U_j$ are open. They form an identity-neighbourhood basis: in any finite continuous quotient their images form a terminating lower p-series, so some $U_j$ is contained in the kernel. In particular ${\bigcap }_j{U}_j=1$. Collection also givesOne can see the extra index directly by induction from $[U_a,G]\subseteq U_{a+1}$: collect $[v,w^p]$ for $w\in U_b$. Its leading factor $[v,w{]}^p$ advances one layer, and the remaining iterated commutators, with their binomial coefficients, lie at least as deep. Products and limits preserve the inclusion. This is the usual strongly central property of the lower p-series.

A uniform pro-p group is a finitely generated powerful pro-p group for which all the indices $|U_j:U_{j+1}|$ equal $|G:G^p|$. The preliminary congruence gives surjective linear mapsIf $G$ is uniform, these maps are bijective. If $x\ne y$, write $y=xu$ and choose the unique layer $j$ with $u\in U_j\setminus U_{j+1}$. Bijectivity of ${\theta }_j$ and the power congruence imply $y^p(x^p{)}^{-1}\in U_{j+1}\setminus U_{j+2}$, so $x^p\ne y^p$. Thus the power map is injective. A finite-order element of a pro-p group has $p$-power order; repeated injectivity of the power map excludes every nonidentity such element. Hence a uniform group is a torsion-free group.

Conversely, suppose some ${\theta }_j$ has a nonzero kernel. Choose $x\in U_j\setminus U_{j+1}$ with $x^p\in U_{j+2}$. At step $r\ge j+1$, if the current $x_r$ satisfies $x_r^p\in U_{r+1}$, surjectivity of ${\theta }_r$ supplies $u_r\in U_r$ withSet $x_{r+1}=x_r{u}_r$. The power congruence gives $x_{r+1}^p\in U_{r+2}$. Since the corrections tend to $1$ in the complete profinite group, the $x_r$ converge to $z$, with $z^p=1$ and $z\equiv x(\mathrm{mod}{U}_{j+1})$. In particular $z\ne 1$. A torsion-free group cannot have such an element, so every ${\theta }_j$ is injective and all layer indices are equal. This proves the root correction criterion for uniformity andFor a uniform group the power maps are continuous bijections $U_j\to U_{j+1}$, hence homeomorphisms. Their inverses define the unique roots used below. Comparison at the first differing layer also proves the power congruences in a uniform pro-p group:

The intrinsic Lie algebra of a uniform pro-p group has the same underlying set as $G$, zero equal to the group identity, and operationsHere $x^{\lambda }$ is the limit of integral powers approaching $\lambda$. The first root is taken in $U_n$ and the second in $U_{2n}$; the commutator actually lies in $U_{2n+1}$. The following faithful algebra realization verifies convergence and the Lie algebra axioms, rather than assuming them as part of the definition.

Choose a minimal tuple $a_1,\dots ,a_d$ of topological generators. The layer isomorphisms ${\theta }_j$ show that the images of $a_i^{p^j}$ are bases of $U_j/U_{j+1}$. The ordered-coordinate argument of Question 3 therefore gives a continuous bijectionExistence follows by successive digits and compactness; uniqueness follows by the first digit at which the coordinates differ, because its images are linearly independent in the corresponding power layer. Thus this map is a homeomorphism.

Here are details of the norm needed on the group algebra. In ${\mathbb{Z}}_p[[G]]$ put $b_i=a_i-1$. Its elements have unique ordered expansions ${\sum }_{\alpha }{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To verify this description, identify the ordered coordinates of $G$ with ${\mathbb{Z}}_p^d$. Elements of the completed group algebra are integral measures on this space; their values on the binomial polynomials ${\prod }_i\left(\genfrac{}{}{0ex}{}{{\lambda }_i}{{\alpha }_i}\right)$ are precisely the coefficients $c_{\alpha }$. The finite-difference identity pairs the ordered monomial $b^{\beta }$ with this binomial polynomial as ${\delta }_{\alpha \beta }$. The Mahler expansion of a continuous function therefore proves both existence and uniqueness of the measure with the specified coefficients. This also proves that the ordinary rational group algebra embeds in these expansions.

Choose $1/(p-1)<s<1$ and setThis defines the weighted group algebra of a uniform pro-p group. Its multiplication is submultiplicative in the valuation convention, $v(AB)\ge v(A)+v(B)$. Indeed, $a_i^p-1=p{b}_i+\left(\genfrac{}{}{0ex}{}{p}{2}\right)b_i^2+\cdots +b_i^p$ has valuation $1+s$; collection of ordered coordinates gives the same lower bound for $g-1$ when $g\in U_1$. Interchanging two adjacent $b_i$ in a product introduces a correction coming from $[a_i,a_j]-1$, of valuation at least $1+s>2s$. Iterating the interchange rule never lowers the total weight; the corrections have strictly increasing weight, so infinite collection converges. The same argument with $p^j$-th powers showsThe equality follows from the first nonzero coordinate digit: its leading terms are $p^j{\lambda }_i^{\prime }{b}_i$, at least one with unit ${\lambda }_i^{\prime }$, and cannot cancel in the ordered expansion. Completing the rational group algebra in this norm produces a complete normed algebra $A$ in which $G$ embeds faithfully and with its original topology.

Since $v(g-1)\ge s>1/(p-1)$, the seriesconverge on the relevant balls in $A$ and are inverse there. Their convergence follows from $v_p(r!)\le (r-1)/(p-1)$ and $v_p(r)\le (r-1)/(p-1)$; formal one-variable identities prove that they are inverse. Put $X=\log x$ and $Y=\log y$. The Baker--Campbell--Hausdorff formula in $A$ now givesbecause every term of degree at least two tends to zero after this rescaling. Similarly expansion of $\exp (-p^{n}X)\exp (-p^{n}Y)\exp (p^{n}X)\exp (p^{n}Y)$ givesThe leading term is the indicated associative-algebra commutator and all terms of degree at least three tend to zero. These expressions are the logarithms of the unique roots in the intrinsic formulas. The image $\log G$ is compact and hence closed in $A$, so both limits belong to $\log G$, and the inverse logarithm takes the limits back to $G$. Also $\log (x^{\lambda })=\lambda \log x$. Consequently $\log$ identifies the intrinsic operations with addition, scalar multiplication and commutation in $A$. The Jacobi identity follows by expanding the associative products; thus $L_G$ is indeed a Lie algebra over ${\mathbb{Z}}_p$.

The commutator roots in the intrinsic formula belong to $U_1$, so their limit gives $[L_G,L_G]\subseteq \log U_1=p{L}_G$. Iterating the power congruence gives $x^{p^n}{y}^{p^n}\equiv (xy{)}^{p^n}(\mathrm{mod}{U}_{n+1})$. Taking the unique roots shows that intrinsic addition modulo $U_1$ is the original group product modulo $U_1$. Consequently $L_G/p{L}_G$ is the original elementary abelian quotient $G/U_1$, and the images of the $a_i$ form a basis. Successively removing digits expresses every element uniquely as an additive sum ${\sum }_i{\lambda }_i{a}_i$: existence uses completeness, while a nonzero relation would have a first nonzero digit contradicting the basis in $U_j/U_{j+1}$. Hence $L_G$ is a free module of rank $d$ over the p-adic integers and is a powerful Lie lattice.

Conversely a powerful Lie algebra over p-adic integers, for odd $p$, is a finite free ${\mathbb{Z}}_p$-module $L$ with a bilinear alternating Lie bracket satisfying the Jacobi identity and $[L,L]\subseteq pL$. DefineFor completeness, integrality and convergence can be verified without guessing which individual denominators are units. In ${\mathbb{Q}}_p\otimes L$ set $M=p^{-1}L$. Then $[M,M]\subseteq M$, so its integral universal enveloping algebra has an ordered basis by the Poincaré-Birkhoff-Witt theorem. Its coefficientwise $p$-adic norm is submultiplicative, and $L=pM$ consists of elements of valuation at least one. The homogeneous degree-$r$ terms in $\log (\exp X\exp Y)$ have coefficients of valuation at least $-\lfloor (r-1)/(p-1)\rfloor$: expand the logarithm, use $v_p(k!)\le (k-1)/(p-1)$ on each nonconstant exponential factor, and $v_p(\ell )\le (\ell -1)/(p-1)$ on a product of $\ell$ such factors. Therefore evaluating degree $r$ on $X,Y\in pM$ gives valuation at leastwhich tends to infinity and is at least two when $r\ge 2$. Each homogeneous term is a rational Lie polynomial by the formal Baker--Campbell--Hausdorff formula. The ordered enveloping-algebra basis shows that its intersection with ${\mathbb{Q}}_{p}M$ is exactly $M$, so the estimate places the sum in $L$, with the non-linear part in $pL$.

Associativity follows from the identity $\exp (X\ast Y)=\exp X\exp Y$ in the completed universal enveloping algebra. Its identity is $0$, its inverse is $-X$, and $X^{\ast m}=mX$, since the bracket of $X$ with itself is zero. The same coefficient estimate for the group commutator puts it in $pL$. Conjugation in the BCH group is $\exp (\mathrm{ad}X)$; each term $\mathrm{ad}(X{)}^r/r!$ preserves $p^{j}L$, because each bracket adds a factor of $p$ and $v_p(r!)<r$. Thus the submodules $p^{j}L$ are normal subgroups for $\ast$, their intersection is zero, and their finite quotients have $p$-power order. Thus $(L,\ast )$ is a powerful pro-p group, withA module basis topologically generates this group: it generates the quotient $L/pL$, and successive power digits lift that generation through all $p^{j}L$. The BCH construction therefore gives a uniform pro-p group.

In the uniform pro-p group $(L,\ast )$, the $p^n$-th power of $X$ is exactly $p^{n}X$, and its unique $p^n$-th root in the appropriate layer is division by $p^n$. The intrinsic addition is thereforeThe linear term is exact; every non-linear term tends to zero by the convergence estimates above. The intrinsic Lie bracket issince the leading term is $p^{2n}[X,Y]$ and the terms of degree at least three vanish after rescaling. Finally the intrinsic scalar operation sends $(\lambda ,X)$ to $\lambda X$. All underlying sets are identical, and all three operations agree. Thus

Use the faithful logarithm realization from the root Solution: $\log :G\to A$ identifies $L_G$ with the Lie algebra $\log G$. For $x,y\in G$, the convergent exponential identity in $A$ givesEvery Lie polynomial and its convergent sum are preserved by this realization of $L_G$. Therefore the BCH product on the underlying set of $L_G$ has exactly the same logarithm as the original product $xy$. Injectivity of the logarithm proves equality of the two products, with the same topology, identity and inverses:This proves the reverse direction of the uniform pro-p group and powerful Lie lattice correspondence directly; the forward identity alone would not imply it.

The proof of the linearity of a uniform pro-p group has three distinct steps: pass to a finite-dimensional Lie algebra, integrate a faithful representation on a sufficiently small open subgroup, and extend it to the entire group. Working first on a small subgroup is essential because a faithful Lie algebra representation need not send the original integral lattice into the convergence domain of the matrix exponential.

Let $G$ be a uniform pro-p group and let $L_G$ be its intrinsic powerful Lie lattice. Thenis a finite-dimensional Lie algebra over a field of characteristic zero. The Ado theorem, the classical Lie-theoretic ingredient in this proof, supplies a faithful finite-dimensional Lie algebra representationOne must use a faithful representation rather than just the Adjoint representation, which annihilates the center of a Lie algebra.

Choose a p-adic lattice $\Lambda \subset V$. Since $L_G$ is finitely generated as a module, the entries of the representing matrices of a module basis have bounded denominators. Hence for sufficiently large $k$,For odd $p$, the matrix exponential and p-adic logarithm converge and are inverse between $p\mathrm{End}(\Lambda )$ and $1+p\mathrm{End}(\Lambda )$. Put $H=G^{p^k}$; under the uniform pro-p group and powerful Lie lattice correspondence it is the BCH group on $p^k{L}_G$. DefineHere ${\exp }_G$ denotes the inverse of the intrinsic logarithm identification, not an assumed matrix representation of $G$. The Lie algebra homomorphism $\rho$ respects the convergent Baker--Campbell--Hausdorff formula, soThus $\tau$ is a continuous group homomorphism. If $\tau ({\exp }_{G}X)=1$, taking the matrix logarithm gives $\rho (X)=0$, whence $X=0$. Therefore $\tau$ is faithful and all its matrices preserve $\Lambda$. This proves linearity of the open normal subgroup $H$.

To finish, use the extension of a faithful representation from an open normal subgroup. Choose representatives $t_1=1,t_2,\dots ,t_r$ of the finite left cosets $G/H$, where $r=[G:H]$. On the finite free module $W={⨁}_{j=1}^r\Lambda$, define the action of $g\in G$ by writing uniquelyand mapping the $j$-th block to the $i$-th block by $\tau (h_{ij})$. If $g^{\prime }{t}_j=t_{\ell }{h}^{\prime }$ and $g{t}_{\ell }=t_{i}h$, then $g{g}^{\prime }{t}_j=t_{i}h{h}^{\prime }$; this proves the representation identity because $\tau (h{h}^{\prime })=\tau (h)\tau (h^{\prime })$. Each block matrix is invertible over ${\mathbb{Z}}_p$, and the map is continuous because the finite coset permutation is locally constant and the remaining matrix entries are obtained continuously from $\tau$.

If $g\notin H$, its permutation of $G/H$ moves the identity coset, so its block representation is not the identity. If $g\in H$ and its block representation is the identity, the block corresponding to $t_1=1$ is $\tau (g)=1$, and faithfulness of $\tau$ gives $g=1$. Thus this induced representation is faithful. Choosing a basis of $W$ provesThe continuous injection is a topological embedding, since $G$ is compact and the target is Hausdorff. The trivial group is immediate. For uniform groups at $p=2$, the same proof uses the usual stronger powerfulness condition and the convergence ball $4\mathrm{End}(\Lambda )$ instead; the preceding questions concern odd $p$.