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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 $p$.

The correspondence concerns uniform pro-p groups and powerful Lie lattices, with their natural morphisms. Put $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. A powerful pro-p group is a pro-p group whose closed commutator subgroup lies in its closed $p^{ϵ}$-power subgroup. It is uniform when it is topologically finitely generated and all successive indices in its lower p-series are equal. Equivalently, a finitely generated powerful pro-p group is uniform precisely when it is a torsion-free group. A powerful Lie algebra over p-adic integers means a Lie algebra over ${\mathbb{Z}}_p$ that is free of finite rank as a module and satisfiesHere a Lie algebra over a ring is a module with an alternating bilinear Lie bracket satisfying the Jacobi identity. Finite freeness is an essential part of the lattice convention. The zero-bracket module ${\mathbb{F}}_p$ satisfies the bracket inclusion but has torsion, and its additive group is not uniform. Thus allowing arbitrary torsion modules would make the categorical assertion false.

The basic structural facts behind the construction can be seen from the power layers. Write $P_n=G^{p^n}$ with $P_0=G$. Collection for a powerful pro-p group givesFor odd $p$, the power lifting is exactly the successive-correction argument of Question 3, followed by compactness; at $p=2$ the stronger fourth-power commutator condition makes the corresponding collection corrections work. The maps between successive layers are surjective linear maps. Uniformity makes their finite dimensions equal, so these maps are isomorphisms. Comparing the first layer in which two elements differ now shows that the power map $P_n\to P_{n+1}$ is injective, while the power-set formula makes it surjective. Thus all the roots used below exist uniquely in the indicated layer. If $g_1,\dots ,g_d$ lift a basis of $G/P_1$, the images of $g_i^{p^n}$ form a basis of $P_n/P_{n+1}$. Successively choosing their coefficients modulo $p$ gives the unique ordered expressionExistence follows by lifting one layer at a time and taking a limit; uniqueness follows by comparing the first nonzero coefficient layer. In particular $[G:P_n]=p^{nd}$ and these layers form a neighbourhood basis. These are the power-layer and coordinate facts needed in the proof, rather than an assumption that an arbitrary pro-p group has unique roots.

On the underlying set of $G$, set the additive zero to $1$, the additive negative of $x$ to $x^{-1}$, and defineThe p-adic integers act through limits of integer powers in the closed cyclic subgroup generated by $x$. In the second formula the product belongs to $P_n$; in the third the group commutator belongs to $P_{2n+ϵ}$. The symbols $p^{-n}$ and $p^{-2n}$ denote the unique group roots, not unrestricted division in a group. We next justify convergence and the Lie algebra laws together.

A useful concrete tool is a complete normed group algebra containing $G$. In its completed group algebra, put $b_i=g_i-1$. Ordered-coordinate uniqueness gives unique expansions ${\sum }_{\alpha \in {\mathbb{N}}^d}{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To see why, first work modulo $p$ and $P_n$: the ordered products of the $g_i$ with exponents below $p^n$ form the regular basis, and the change to the corresponding products of $g_i-1$ is triangular with diagonal entries one. Take inverse limits and then lift the coefficient digits successively in powers of $p$. Define the weightIt is a genuine separated, submultiplicative weight. Indeed reordering $b_j{b}_i$ produces a group-commutator correction in $P_{ϵ}$; the binomial expansion of a $p^{ϵ}$th power gives that correction weight at least $2ϵ$, so collection never decreases total weight. Also $v(pc)=1+v(c)$. Inverting $p$ and completing therefore gives a Banach algebra $B$ over ${\mathbb{Q}}_p$. For $g\in G$, ordered coordinates showwith value infinity at $g=1$. Thus the induced topology on $G$ is exactly its original topology. This construction supplies the ambient algebra without presupposing the uniform pro-p group and powerful Lie lattice correspondence. The weighted construction, including the case $p=2$, is described in Equivariant D-modules on rigid analytic spaces, Section 6.5.

Since $ϵ>1/(p-1)$, the p-adic logarithm and p-adic exponential converge on the relevant balls in $B$:The estimates $v_p(k!)\le k/(p-1)$ and $v_p(k)\le {\log }_{p}k$ show that term weights tend to infinity. Formal substitution is legitimate there and proves that the two maps are inverse. They preserve the leading weight. In particular $\log$ embeds the compact set $G$ homeomorphically into $B$, and $\log (g^{p^n})=p^n\log g$.

Put $a=\log x$ and $b=\log y$. Expanding the convergent Baker--Campbell--Hausdorff formula in $B$ givesThe first expansion has linear part $p^n(a+b)$ and all other terms have degree at least two in $p^{n}a,p^{n}b$. The second has leading term $p^{2n}(ab-ba)$ and all remaining terms have degree at least three. After division by the displayed powers of $p$, the extra factor of $p^n$ forces those errors to zero. The normalized expressions are logarithms of the unique roots already specified. Since $\log G$ is compact and closed in $B$, both limits belong to $\log G$. Scalar multiplication also remains in $\log G$ by continuity of powers. Consequently $\log G$ is a module over ${\mathbb{Z}}_p$, closed under the associative-algebra Lie bracket $ab-ba$. Bilinearity, alternation and the Jacobi identity follow by expanding products in the associative algebra $B$. Transporting these operations back gives exactly $L_G$ above.

The commutator-layer inclusion places every normalized bracket root in $P_{ϵ}$. Since $\log P_{ϵ}=p^{ϵ}\log G$ and this set is compact, passage to the limit proves $[L_G,L_G]\subseteq p^{ϵ}{L}_G$. Addition modulo $p{L}_G$ is the original multiplication modulo $P_1$: collection shows $(x^{p^n}{y}^{p^n}{)}^{p^{-n}}\equiv xy(\mathrm{mod}{P}_1)$, as the nonlinear errors lie at least one layer deeper. Thus $L_G/p{L}_G$ has dimension $d$, with basis $e_i=\log g_i$. Given $a\in \log G$, write its residue as ${\sum }_i{\lambda }_{0i}{e}_i$ and repeat on $(a-{\sum }_i{\lambda }_{0i}{e}_i)/p$. This iterative digit lifting converges because the remainders stay in the bounded compact set $\log G$. It gives $a={\sum }_i{\lambda }_i{e}_i$ with ${\lambda }_i\in {\mathbb{Z}}_p$. A nontrivial relation among the $e_i$ could be divided by the smallest power of $p$ among its coefficients, contradicting their independence modulo $p$. Hence $L_G$ is free of rank $d$, proving that it is a powerful Lie lattice.

Conversely, let $L$ be a powerful Lie lattice. In $L{\otimes }_{{\mathbb{Z}}_p}{\mathbb{Q}}_p$, defineThe formal Baker--Campbell--Hausdorff formula and its coefficient estimate are the algebraic ingredients used here. More precisely, the degree-$m$ component can be expressed as an integral Lie polynomial divided by a number of $p$-adic valuation at most $\lfloor (m-1)/(p-1)\rfloor$. One way to obtain this estimate is to bound the associative coefficients of $\log (\exp X\exp Y)$ and use an integral ordered basis of the free Lie algebra. The denominator bound $v_p(m!)+\lfloor {\log }_p{s}_p(m)\rfloor$, where $s_p(m)$ is the base-$p$ digit sum, is at most $\lfloor (m-1)/(p-1)\rfloor$. A precise primary reference for that denominator estimate is Smallest common denominators for the homogeneous components of the Baker-Campbell-Hausdorff series. A length-$m$ Lie bracket in $L$ lies in $p^{ϵ(m-1)}L$. Thus the degree-$m$ term belongs toThese exponents are nonnegative and tend to infinity; for $m\ge 2$ they are at least one. This proves integral convergence, continuity and $x\ast y\equiv x+y(\mathrm{mod}pL)$. Associativity follows from the formal identity $\exp (\mathrm{BCH}(X,Y))=\exp X\exp Y$ in the completed universal enveloping algebra. Substitution of three elements is justified by the same valuation estimates, so the two associative expressions agree. The identity is $0$, the inverse is $-x$, and $x^{\ast k}=kx$ because every nonlinear term involving only $x$ vanishes.

For powerfulness, formal conjugation givesThe difference $\delta =y-e^{-\mathrm{ad}x}y$ lies in $p^{ϵ}L$: its $k$th term has valuation at least $kϵ-v_p(k!)\ge ϵ$ for $k\ge 1$. The group commutator is therefore $(-y+\delta )\ast y$. Its linear term is $\delta$. Every nonlinear Lie term contains $\delta$, since it vanishes when $\delta =0$, and the extra divisibility makes all those terms lie in $p^{ϵ}L$ too. Hence $[(L,\ast ),(L,\ast )]\subseteq p^{ϵ}L$. The ideals $p^{n}L$ are open normal subgroups for $\ast$, and their finite quotients have order $p^{nd}$. The group is consequently a pro-p group. Its power sets are exactly $p^{n}L$, its lower-series indices are constantly $p^d$, and a basis of $L$ topologically generates it by the same successive digit lifting. The BCH group $(L,\ast )$ is uniform.

To check the inverse constructions, powers in the Baker--Campbell--Hausdorff formula group are ordinary multiples in $L$. Thereforeand the similarly normalized group commutator tends to $[x,y]$. The limit constructions recover the original Lie algebra exactly. In the other direction, the ambient Banach algebra construction gives $\log (xy)=\mathrm{BCH}(\log x,\log y)$, so the recovered operation is the original multiplication of $G$.

Finally a continuous group homomorphism between uniform pro-p groups preserves integer powers and powers with exponents in the p-adic integers, preserves the unique roots in the power layers, and commutes with their limits. It therefore becomes a linear Lie algebra homomorphism. Conversely a linear Lie algebra homomorphism between finite free modules is continuous and preserves every term of the convergent Baker--Campbell--Hausdorff formula, hence is a continuous group homomorphism for $\ast$. The constructions preserve identity maps and composition, and the inverse identifications are natural. They are mutually inverse functors, giving the stated category correspondence with the finite-free lattice convention.