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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 $p$.

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.