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An Iwasawa algebra is a completed group algebra such as ${\mathbb{Z}}_p[[G]]$ or ${\mathbb{F}}_p[[G]]$ for a compact p-adic analytic group $G$.

Choose a topological generator $\gamma$ of the Galois group of a Zp-extension and set $T=\gamma -1$. The completed group algebra becomes the formal power series ring ${\mathbb{Z}}_p[[T]]$, a two-dimensional complete regular local integral domain with maximal ideal $(p,T)$. Its finite-layer quotient is $\Lambda /((1+T{)}^{p^n}-1)\cong {\mathbb{Z}}_p[\Gamma /{\Gamma }^{p^n}]$.

Use $[g]$ for the basis element of the group algebra $A={\mathbb{F}}_p[{\mathbb{Z}}_p]$, so $[g][h]=[g+h]$ and $[0]=1$. The kernel of the augmentation map is the augmentation ideal $I$. This algebra must not initially be identified with a Laurent polynomial ring: the additive group ${\mathbb{Z}}_p$ is not abstractly generated by the integer $1$.

For $r\ge 1$, map $A$ onto ${\mathbb{F}}_p[{\mathbb{Z}}_p/p^r{\mathbb{Z}}_p]$ and call its kernel $J_r$. It is generated by $[g]-1$ for $g\in p^r{\mathbb{Z}}_p$: differences of basis elements in one coset have the form $[h]([g]-1)$. Since each such $g=p^{r}h$,by the characteristic-$p$ binomial identity. Thus $J_r\subseteq I^{p^r}$.

The quotient group is a cyclic group of order $p^r$, generated by the image of $1$. Its group algebra isThe image of $I$ is $(T)$ and its $p^r$th power vanishes, proving the reverse inclusion. HenceThis is cyclic finite quotients of the p-adic augmentation filtration.

For any $n\ge 0$, choose $p^r>n+1$. The finite quotient then identifies $I^n/I^{n+1}$ with $(T^n)/(T^{n+1})$, a one-dimensional space generated by the image of $([1]-1{)}^n$. The degree-one symbol $X$ of $[1]-1$ consequently gives a surjective graded map ${\mathbb{F}}_p[X]\to {\mathrm{gr}}_{I}A$, and no power of $X$ vanishes in its degree. ThusThe same finite quotients show that the I-adic completion is ${\mathbb{F}}_p[[T]]$, and its associated graded ring is identical. The proof therefore covers the completed group algebra interpretation too, while proving the literal ordinary-algebra statement without assuming completion. Applying part ii to a ring would require its actual completeness, not just this polynomial associated graded ring.

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.

The mod-$p$ Iwasawa algebra is the completed group algebrawhere $U$ runs through open normal subgroups and the transition maps come from quotient maps. This is the mod-$p$ algebra, distinguished from the integral Iwasawa algebra ${\mathbb{Z}}_p[[K]]$.

Use the stated complete separated negative filtered ring structure, writing $F_{-n}={\mathfrak{m}}^n$ and $\nu (x)=\max \{n:x\in F_{-n}\}$ for a nonzero element. In the usual odd-prime congruence-group situation,The dimension-three assertion records the three independent entries of a trace-zero matrix; the argument below only needs the given graded polynomial-algebra property. Completeness and separatedness are with respect to this filtration and follow from the inverse-limit topology for the standard filtration. If $x,y\ne 0$, their leading symbols are nonzero, so their product in the graded domain is nonzero. Therefore $\nu (xy)=\nu (x)+\nu (y)$ and $xy\ne 0$. This proves the domain assertion by filtered-graded transfer for complete rings.

For a left ideal $I$, its leading symbols form a homogeneous left ideal $\mathrm{gr}I$ of the polynomial ring. The Hilbert basis theorem gives finitely many homogeneous generators; choose lifts $a_1,\dots ,a_s\in I$. Given $x\in I$, express its leading symbol using these generators, lift the homogeneous coefficients, and subtract the resulting $\sum r_j{a}_j$. The remainder is still in $I$ and has strictly larger valuation. Repeat. At a stage where the remainder has valuation $n$, the coefficient contribution for $a_j$ has valuation at least $n-\nu (a_j)$. Thus each accumulated coefficient is a Cauchy series, converging to an element of ${\Omega }_K$. The remainder tends to zero, and separatedness gives $x={\sum }_j{r}_j{a}_j$. Hence $I$ is algebraically generated by these finitely many lifts; no prior assumption that $I$ is closed was used. This proves $\boxed{{\Omega }_K\text{ is left Noetherian and a domain}}$ under the stated filtration hypothesis. The same proof for right ideals also gives right Noetherianity.

For the classical fractions, a left Noetherian domain satisfies the left Ore condition. Here is the common-multiple argument. If two nonzero principal left ideals $Ra,Rb$ had zero intersection, the ideals $Ra{b}^i$, $i\ge 0$, would form a direct sum: from a finite relation, its first term lies in $Ra\cap Rb$, and then cancellation of the nonzero $b$ proves the assertion inductively. Their partial sums give an infinite ascending chain, impossible in a left Noetherian ring. Thus $Ra\cap Rb\ne 0$, giving common left multiples for every two nonzero denominators. The right-handed argument also applies. Ore localization at all nonzero elements therefore produces the classical division ring of fractions, with expressions $s^{-1}a$ or $b{s}^{-1}$; every nonzero fraction is invertible.

There is a genuine prime restriction behind the auxiliary hypothesis. For $p=2$, the matrix $z=-I$ is a nonidentity element of $K$ of order two. In ${\mathbb{F}}_2[[K]]$, the nonzero element $z-1$ satisfies $(z-1{)}^2=z^2-2z+1=0$. It is nonzero already in a finite quotient separating $z$ from one. Thus the actual $p=2$ algebra is not a domain, and cannot have the supplied separated graded-domain filtration. The intended conclusion applies for odd $p$; at $p=2$ it instead applies to a suitable deeper torsion-free congruence subgroup, such as the kernel modulo four.

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.

Choose ordered minimal topological generators $a_1,\dots ,a_d$ of a uniform pro-p group for odd $p$, put $b_i=a_i-1$, and choose $1/(p-1)<s<1$. Ordered monomials $b^{\alpha }=b_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ give power-series coordinates in the integral completed group algebra. The displayed Non-Archimedean absolute value filtration gives a complete normed ${\mathbb{Q}}_p$-algebra after completion of the rational group algebra. The power relation raises valuation by one, and interchanging $b_i,b_j$ adds terms of valuation at least $1+s>2s$, so multiplication does not decrease total valuation. The p-adic logarithm and p-adic exponential are inverse on the relevant small balls. In particular $\log G$ is a faithful realization of the Lie algebra of a uniform pro-p group, useful for verifying its intrinsic limit operations and the Baker--Campbell--Hausdorff formula without assuming the correspondence in advance.