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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.

A topological group module $A$ is a topological abelian group with a jointly continuous action of $G$ by group automorphisms. Its continuous cochains with topological coefficients arewith pointwise addition. The inhomogeneous differential isFor $n=0$, this means $(da)(g)=ga-a$. Continuity of multiplication and of the action ensures that $d$ preserves continuous cochains. Expanding $d(dc)$ gives cancellation in pairs: the two orders of multiplying neighbouring arguments agree by associativity, and the two ways of applying the action agree by its action law. Thus $d^2=0$. The cocycles are $Z^n=\ker (d:C^n\to C^{n+1})$ and the coboundaries are $B^n=\mathrm{im}(d:C^{n-1}\to C^n)$, with $B^0=0$. DefineFor general topological coefficients this is an algebraic quotient of the continuous cochains; its optional quotient topology need not be Hausdorff. For discrete coefficients it is the usual continuous cohomology of a profinite group. In degree zero it is the subgroup of invariant elements.

Give the discrete prime field ${\mathbb{F}}_p$ its trivial $G$-action. The one-cocycle condition becomes $c(gh)=c(g)+c(h)$ and every one-coboundary is zero. HenceEvery such group homomorphism kills powers and group commutators, hence their closure $\Phi (G)$, and every continuous functional on the quotient pulls back. If $G$ is topologically finitely generated, $V=G/\Phi (G)$ is a finite elementary abelian group. To verify finiteness without assuming it, the closure of a subgroup generated by $r$ elements of this abelian exponent-$p$ quotient has at most $p^r$ elements; a finite subgroup is already closed. Thus $V$ is a finite-dimensional vector space. Question 2 says that a generating family of $G$ is minimal precisely when its images give a basis of $V$. The dual space has the same finite dimension. This proves

There is a genuine missing hypothesis if the printed assertion is read for arbitrary pro-p groups and $d(G)$ means the least cardinality of any dense generating family. Here is an explicit infinite-rank failure of the first-cohomology generator formula. Let $I=\{0,1{\}}^{\mathbb{N}}$ and $G={\prod }_{r\in I}{\mathbb{F}}_p$. Enumerate the finite binary strings as $s_1,s_2,\dots$ and define $g_n(r)=1$ if $s_n$ is a prefix of $r$, and $0$ otherwise. For finitely many distinct sequences $r_1,\dots ,r_k$, choose long enough prefixes to distinguish each individually. The corresponding $g_n$ project onto the coordinate unit vectors of ${\mathbb{F}}_p^k$, so $⟨g_n:n\ge 1⟩$ surjects onto every finite coordinate projection and is dense in the product topology. No finite family can generate densely, since a finitely generated abelian exponent-$p$ subgroup is finite and closed. Thus $d(G)={\aleph }_0$. On the other hand, continuity of a linear functional to ${\mathbb{F}}_p$ means its kernel contains a basic neighbourhood, so it depends on finitely many coordinates. The coordinate projections are consequently a basis of the continuous dual space, givingThe unrestricted infinite-rank formula is false. A convention requiring a generating basis tending to the identity is different from the printed unrestricted topological generator count; the finite-rank identity proved above has no such ambiguity.

For finite $X$, a free pro-p group $F_p(X)$ is characterized by the universal property that every map $X\to P$ to a pro-p group extends uniquely to a continuous group homomorphism $F_p(X)\to P$. It is the pro-p completion of the abstract free group on $X$: each map to a finite $p$-group factors through one of the defining finite quotients, and passing to the inverse limit gives the extension to any pro-p group; uniqueness follows from density. This unrestricted universal property can also define the completion on an arbitrary set $X$. When using an infinite basis required to tend to the identity, one instead imposes the universal property only for maps tending to the identity. Only the finite-rank version is needed for the following argument.

A finite pro-p presentation is an isomorphismThere are finitely many generators and finitely many relators, and the denominator is their closed normal closure. The relators are elements of the free pro-p group; they may be limits of ordinary words. That distinction is important when writing conjugation relations below.

Now suppose $K{⊴}_{o}G$ has a finite pro-p presentation with generators $k_1,\dots ,k_a$ and relators $R_1,\dots ,R_t$. The quotient $Q=G/K$ is a finite p-group. Choose a finite finite pro-p presentation $Q=⟨{\bar{y}}_1,\dots ,{\bar{y}}_b\mid S_1,\dots ,S_c⟩$. Such a presentation exists, for example by using its entire finite multiplication table. Lift the quotient generators to $y_j\in G$. For every $i,j$, choose elements $W_{ij},V_{ij}$ of the free pro-p group on the $x_i$ that map respectively to $y_j^{-1}{k}_i{y}_j$ and $y_j{k}_i{y}_j^{-1}$ in $K$. Choose $U_{\ell }(x)$ mapping to $S_{\ell }(y)$, which lies in $K$. Form the finite finite pro-p presentationAll listed relations hold in $G$, giving a continuous surjective group homomorphism $P\to G$: its compact image contains $K$ and maps onto $Q$.

Let $X$ be the closed subgroup of $P$ topologically generated by the $x_i$. Both conjugation directions were included, so each $y_j$ normalizes $X$. The $x_i$ also normalize it, and its closed normalizer contains the generators of $P$. Thus $X⊴P$. The relations $R_h$ give a continuous surjection $K\to X$. Its composite with $X\to K$ induced by $P\to G$ fixes every $k_i$, and hence is the identity. Therefore the surjection $K\to X$ is injective as well, so $X\cong K$. On quotienting by $X$, the conjugation and kernel relations disappear, and the remaining presentation is exactly $⟨y_1,\dots ,y_b\mid S_1,\dots ,S_c{⟩}_{\mathrm{pro}\text{-}p}=Q$. Thus $P/X\to G/K$ is an isomorphism. The kernel of $P\to G$ lies in $X$ and meets $X$ trivially, so is trivial. Consequently $P\cong G$, proving

Let $G=G_K$ and choose a profinite Sylow subgroup $P$ at the prime $p$. Its image in every finite quotient of $G$ is a Sylow $p$-subgroup. Put $F=K_s^P$. Every finite intermediate field $K\subseteq L\subseteq F$ has degree prime to $p$, because its corresponding open subgroup contains $P$. Condition (iii), together with the preceding Kummer reduction, gives $H^{n+1}(L,{\mu }_p)=0$ for each such field. The cohomology continuity at closed subgroups property saysfor a discrete $G$-module $M$. Applying it to $M={\mu }_p$ gives $H^{n+1}(P,{\mu }_p)=0$.

The action of the pro-p group $P$ on ${\mu }_p$ is trivial. Indeed it maps continuously to $\mathrm{Aut}({\mu }_p)\cong {\mathbb{F}}_p^{\times }$, whose order $p-1$ is prime to $p$; every finite image of $P$ is a $p$-group, so that image is trivial. Choosing a primitive root identifies this module with the trivial module ${\mathbb{F}}_p$. Thus $H^{n+1}(P,{\mathbb{F}}_p)=0$.

We spell out why this controls arbitrary $p$-primary coefficients. A finite discrete $p$-primary $P$-module has a composition series with trivial ${\mathbb{F}}_p$ factors: its action factors through a finite $p$-group, whose only simple module in characteristic $p$ is the trivial one. Induction through the long exact cohomology sequence therefore gives vanishing of $H^{n+1}$ for every such finite module. Any discrete $p$-primary $P$-module is the filtered union of finite $P$-stable submodules. The orbit of an element is finite by continuity, and its orbit generates a finite abelian $p$-group. Continuous cohomology commutes with these filtered unions, sofor every discrete $p$-primary module $M$. This is the mechanism behind trivial coefficients detect the cohomological dimension of a pro-p group.

Now let $M$ be any discrete $p$-primary $G$-module and $a\in H^{n+1}(G,M)$. Its restriction to $P$ is zero. Continuity at the closed subgroup $P$ makes its restriction zero in some open $U\supseteq P$. The index $m=[G:U]$ is prime to $p$, and the restriction-corestriction identity in group cohomology for the corestriction map in group cohomology givesBut $a$ has $p$-power order, so multiplication by $m$ is invertible on the cyclic group it generates. Hence $a=0$. This proves $H^{n+1}(G,M)=0$ for all discrete $p$-primary $M$. Dimension shifting through an acyclic coinduced module gives vanishing in every higher degree, which is exactly ${\mathrm{cd}}_p(K)\le n$.

Therefore (iii) implies (i), completing the equivalence of all three conditions. The key reason prime-to-$p$ extensions suffice is that they approximate a pro-$p$ Sylow subgroup, where the cyclotomic module becomes trivial; prime-to-$p$ transfer then detects every $p$-primary cohomology class.

Write $\Gamma =\mathrm{Gal}(K_{\infty }/K)\cong {\mathbb{Z}}_p$. Its only finite subgroup is trivial. In particular, a real place cannot acquire complex inertia in this Zp-extension: the possible nontrivial inertia at a real place has order two. Thus every infinite place splits in the tower.

Suppose that no finite prime ramified. Every finite layer would then be an abelian everywhere unramified extension of number fields, including splitting at real places. Every layer would lie in the ordinary Hilbert class field of $K$, a finite extension. Their degrees are unbounded, a contradiction. Therefore some finite prime must ramify.

Let $v$ have residue characteristic $\ell \ne p$. Apply local class field theory to the corresponding decomposition group, a closed subgroup of $\Gamma$. The image of the unit group ${\mathcal{O}}_{K_v}^{\times }$ is the inertia group. Its first principal-unit subgroup is a pro-l group, so its continuous image in the pro-p group $\Gamma$ is trivial. The residue-unit quotient is the finite group ${\mathbb{F}}_{q_v}^{\times }$. Its image is finite, and is therefore also trivial in the torsion-free group ${\mathbb{Z}}_p$. The entire inertia group is trivial. HenceEquivalently, the tame ramification relation with a Frobenius element would force a tame inertia generator to satisfy ${\tau }^{q_v-1}=1$, which is impossible nontrivially in ${\mathbb{Z}}_p$.

For the final assertion, the cyclotomic Zp-extension of $K$ is $K\mathbb{Q}[p^{\infty }]$. Fix $v\mid p$ and put $E=K_v$, a finite extension of ${\mathbb{Q}}_p$. Let $C_n$ be the completion of $\mathbb{Q}[p^n]$ at its unique prime over $p$. This is a totally ramified extension of ${\mathbb{Q}}_p$ of degree $p^n$. In a compatible local algebraic closure,Here the inequality follows from the tower through $C_n$. The right side is unbounded. These composita occur among the completions of the cyclotomic tower over $K$, so $v$ has unbounded ramification index, in particular nontrivial inertia group. ThusThis argument allows an arbitrary finite intersection between $K$ and the rational cyclotomic tower; it does not assume disjointness.

Let $\mathcal{G}=\mathrm{Gal}(M_{\infty }/F)$. The Galois group sequence isFor $g\in \Gamma$ and $x\in X$, choose a lift $\tilde{g}\in \mathcal{G}$ and defineThis is independent of the lift because $X$ is abelian. Write $X$ additively. As an abelian pro-p group, it is a compact ${\mathbb{Z}}_p$-module, with $ax$ defined by the limit of integer powers, and the conjugation action is continuous and ${\mathbb{Z}}_p$-linear. It extends to the Iwasawa algebraIndeed, use the finite quotients of $X$ obtained by intersecting it with open normal subgroups of $\mathcal{G}$; on each quotient the action factors through one of these finite group rings. This defines the natural compact Galois module structure, even when $X$ is not finitely generated. Choosing $\gamma$ identifies $\Lambda$ with ${\mathbb{Z}}_p[[T]]$ through $T=\gamma -1$.

The subgroup $(\gamma -1)X$ is closed: it is the image of the compact group $X$ under a continuous map. Every element $(\gamma -1)x$ is a commutator of a lift of $\gamma$ with $x$. ConsequentlyConversely, in the quotient by $(\gamma -1)X$, the lift of $\gamma$ centralizes the image of $X$. Its powers are dense in the procyclic subgroup they generate. Since these powers map densely onto $\Gamma$, compactness shows that $\mathcal{G}$ is generated by $X$ and that procyclic subgroup. The quotient is therefore abelian. HenceOne can also see this by choosing a lift of $\gamma$. Since $\mathcal{G}$ is a pro-p group, the closed subgroup generated by this lift is ${\mathbb{Z}}_p$, and its map to $\Gamma$ is an isomorphism. This gives a semidirect product splitting.

The maximal abelian subextension $M/F$ is the fixed field of the closed commutator subgroup. Because $F_{\infty }/F$ is abelian, $F_{\infty }\subseteq M$. Restriction of the Galois correspondence to $X$ now gives the coinvariant formulaThe closed image above is important: the formula is a quotient of compact groups, not a quotient by an unclosed abstract subgroup. This is abelianization over a Zp-extension.

Let $F_n$ be the degree-$p^n$ layer, ${\Gamma }_n={\Gamma }^{p^n}$, andConjugation makes $Y$ a compact $\Lambda$-module. The goal is to show that it is finitely generated and has Iwasawa-module rank zero. This proof applies to every Zp-extension, not only a cyclotomic one, and makes no Leopoldt conjecture assumption.

By the ramification argument above, only primes over $p$ ramify in $F_{\infty }/F$, and at least one does. Every nonzero closed inertia group in $\Gamma \cong {\mathbb{Z}}_p$ is open. After replacing $F$ by a finite layer, every prime which ramifies is totally ramified in the remaining tower. The number $s\ge 1$ of such primes then stays constant. This replacement does not affect whether $Y$ is a torsion module: ${\mathbb{Z}}_p[[\Gamma ]]$ is finite free over the Iwasawa algebra of an open subgroup, and the two module ranks vanish together.

Let ${\mathcal{G}}_n=\mathrm{Gal}(L_{\infty }/F_n)$, and let $D_n$ be its maximal abelian quotient. The abelianization over a Zp-extension formula gives an exact sequenceChoose a lift of a generator of ${\Gamma }_n$ if a splitting is desired. We will bound the ${\mathbb{Z}}_p$-rank of the middle term independently of $n$.

At each of the $s$ ramified primes, inertia in $L_{\infty }/F_n$ maps isomorphically onto ${\Gamma }_n$: its kernel is inertia in the unramified extension $L_{\infty }/F_{\infty }$, hence trivial, and the map onto the totally ramified base inertia is surjective. Thus each image in $D_n$ is procyclic and has ${\mathbb{Z}}_p$-rank at most one. No other finite prime contributes inertia. Quotienting $D_n$ by the closed subgroup generated by these $s$ images gives an abelian extension of $F_n$ unramified at all finite primes. Its Galois group is finite by class field theory, using the ordinary or narrow ideal class group according to the treatment of real places. Any infinite-place inertia is finite and does not affect the rank bound. Therefore $D_n$ is finitely generated over ${\mathbb{Z}}_p$ andThe preceding exact sequence now yields the uniform boundIt does not assert that these coinvariant modules are always finite when several primes ramify.

At $n=0$, $Y/TY$ is finitely generated over ${\mathbb{Z}}_p$, so $Y/(p,T)Y$ is finite. Lift a finite basis of this quotient to $Y$. The Compact Nakayama lemma shows that these lifts generate $Y$ over $\Lambda$. Briefly, the quotient by their compact generated image satisfies $Q=(p,T)Q$. Every finite continuous $\Lambda$-quotient of $Q$ has nilpotent action by $(p,T)$: $p$ is nilpotent, and modulo $p$ a finite pro-$p$ action makes $\gamma -1$ nilpotent. Such a quotient must be zero. Finite quotients separate points of a compact pro-p group, so $Q=0$.

Suppose now that $Y$ had positive $\Lambda$-rank $r$. The Iwasawa module structure theorem provides a pseudo-isomorphism towith $E$ a finitely generated torsion module. Passing to ${\Gamma }_n$-coinvariant modules leaves a finite cokernel. Butis free of rank $p^n$ over ${\mathbb{Z}}_p$. Thus the coinvariant modules would have rank at least $r{p}^n$, contradicting the uniform bound $s-1$. Equivalently their ranks have asymptotic form $r{p}^n+O(1)$, with the torsion elementary divisors contributing only a bounded rank.

It follows that $r=0$. Since $\Lambda$ is an integral domain, rank zero means every element is killed by a nonzero scalar. ThereforeThis is the unramified Iwasawa torsion theorem. The crucial arithmetic inputs are finite class fields and the bounded number of ramified primes; neither vanishing of a $\mu$-invariant nor a cyclotomic main conjecture is required.

A normal subgroup $N$ is powerfully embedded in a finite p-group or pro-p group $G$ if $[N,G]\le N^p$ for odd $p$, or $[N,G]\le N^4$ for $p=2$; use closed generated subgroups in the pro-p group case. The condition implies that $N$ is a powerful p-group. For odd $p$, both $[N,G]$ and $N^p$ remain powerfully embedded.

A finite p-group is powerful if $[G,G]\le G^p$ for odd $p$, or $[G,G]\le G^4$ for $p=2$. Here powers denote generated subgroups. The same definition applies to pro-p groups using closed power and commutator subgroups. Its Frattini subgroup is $G^p$. For odd $p$, successive power layers admit linear correction of roots, which makes every element of $G^p$ an actual $p$th power.

A pro-p group whose closed commutator subgroup lies in its closed power subgroup $G^{p^{ϵ}}$, with $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. This is the profinite analogue of a powerful p-group. Finitely generated torsion-free examples are exactly the uniform pro-p groups. The additive group of a finite product of p-adic integers is an abelian example.

For odd $p$, set $P_0=G$ and $P_{i+1}=P_i^p$. Powerful embedding gives $[P_i,G]\le P_{i+1}$. Modulo $P_{i+2}$, commutators with $P_i$ are central and have exponent dividing $p$, so $(xy{)}^p\equiv x^p{y}^p$ when $y\in P_i$. Also the power map $P_i/P_{i+1}\to P_{i+1}/P_{i+2}$ is a surjective homomorphism. First find a root modulo $P_2$, then multiply it by successive corrections in $P_i$. The process terminates for a finite p-group, and converges for a pro-p group with separated power filtration. This proves that the generated power subgroup consists of actual powers.

A profinite group $G$ is procyclic if every quotient by an open normal subgroup is cyclic. Equivalently $G=\overline{⟨g⟩}$ for one element $g$. For the forward implication, the sets of lifts of generators of $G/N$ are nonempty clopen sets. Any finite intersection contains a lift of a generator of the cyclic quotient by the intersection of those open normal subgroups. Compactness supplies one element generating every finite quotient, and hence generating densely. The reverse implication follows by projecting that generator. For a pro-p group, two distinct maximal open subgroups would give a quotient $C_p^2$, so a nontrivial procyclic group has a unique maximal open subgroup $M$. Its profinite Frattini subgroup is $M$, and any $g\notin M$ generates topologically: otherwise the proper closed subgroup it generates lies in a maximal open subgroup. The additive group of p-adic integers is an infinite example, generated topologically by $1$.

A profinite abelian group is an inverse limit of finite abelian groups. Its topology is compact and totally disconnected. An abelian pro-p group acquires scalar multiplication by p-adic integers by taking limits of integer multiples.

The intersection of the maximal proper open subgroups of a profinite group. A subset topologically generates the profinite group exactly when its image topologically generates its quotient by this intersection: a proper closed generated subgroup is contained in a maximal open subgroup, obtained from a suitable finite quotient. For a pro-p group, $\Phi (G)=\overline{G^p[G,G]}$, where the bar denotes closure of the algebraically generated power and commutator subgroups. Closedness cannot simply be discarded in infinite rank.

The inverse limit of the quotients of an abstract group by its normal subgroups of finite $p$-power index. Maps to pro-p groups factor uniquely through the canonical dense image in this completion. Unlike profinite completion, only finite p-groups are used.

A pro-$p$ ring is a topological ring with continuous ring operations whose underlying additive group is a pro-p group. Standard local analytic examples are complete Noetherian local rings with finite residue field of characteristic $p$, such as ${\mathbb{Z}}_p$ and ${\mathbb{F}}_p[[t]]$. The topology is essential to the convergence of power series used for analytic group operations.

A closed subgroup of a profinite group that is a pro-p group with the induced topology. Its images in finite continuous quotients are p-groups.

For a pro-p ring $R$, an $R$-analytic pro-p group is a pro-p group equipped with an analytic manifold structure locally modelled on open subsets of a finite free $R$-module, such that multiplication and inversion are given locally by convergent power series. The number of local coordinates is its analytic dimension. The Nottingham group admits no such structure over any pro-$p$ ring: its hereditary just infiniteness, unbounded finite cyclic torsion and sub-base-three derived-series growth exclude both alternatives in the analytic just-infinite structure theorem.

Every abstract subgroup of finite index in a finitely generated pro-p group is open. Fixed-generator commutator collection in nilpotent groups makes the abstract commutator subgroup closed. The resulting abelian quotient is a quotient of a finite power of the p-adic integers, so its subgroup of $p$th powers is closed and of finite index. Thus the algebraic $G^p[G,G]$ is open and every abstract index-$p$ subgroup is open. An arbitrary finite quotient is a finite p-group; induction on its order, using a normal subgroup of index $p$ and the Schreier lemma, gives the assertion. The normal core reduces nonnormal finite-index subgroups to this case.

For a pro-p group $P$, vanishing of $H^{n+1}(P,{\mathbb{F}}_p)$ implies vanishing in that degree for every discrete $p$-primary module. Finite such modules have composition factors equal to the trivial module ${\mathbb{F}}_p$, so the long exact sequence gives the result by induction. General modules are unions of finite stable submodules, and continuous cohomology commutes with filtered unions. Dimension shifting gives ${\mathrm{cd}}_p(P)\le n$. The converse is immediate from the definition.

A finitely generated powerful p-group in the category of pro-p groups is uniform if all successive indices in its lower p-series are equal. Equivalently it is torsion-free. If its generator number is $d$, the successive indices are $p^d$, the power map carries $G^{p^n}$ bijectively onto $G^{p^{n+1}}$, and an ordered minimal generating set gives unique coordinates $g_1^{{\lambda }_1}\cdots g_d^{{\lambda }_d}$ with ${\lambda }_i$ in the p-adic integers. For $p=2$, powerfulness means $[G,G]\le G^4$.