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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 $p$.

The objective is a continuous faithful representation of a uniform pro-p group $G$ in some ${\mathrm{GL}}_N({\mathbb{Z}}_p)$. The proof proceeds from the intrinsic Lie lattice to a small matrix group, then extends the resulting representation to all of $G$. This last step matters: a representation defined only on an open subgroup does not yet establish linearity of the original group.

Let $L=L_G$ be the Lie algebra of a uniform pro-p group and put $\mathfrak{g}={\mathbb{Q}}_p{\otimes }_{{\mathbb{Z}}_p}L$. The uniform-group/Lie-lattice correspondence identifies the group law on $L$ with BCH. The substantial finite-dimensional Lie-theoretic input is the Ado theorem: every finite-dimensional Lie algebra over a characteristic-zero field admits a faithful finite-dimensional representation. Apply it to obtainAn Adjoint representation would not suffice here, since it kills the centre; Ado supplies injectivity even in central directions.

Choose a p-adic lattice $\Lambda$ spanning $V$. Since $L$ has finitely many lattice generators, their representing matrices have bounded denominators. There is therefore an integer $a\ge 0$ such thatOn this matrix lattice the series $\exp X={\sum }_{n\ge 0}{X}^n/n!$ and $\log (1+Y)={\sum }_{n\ge 1}(-1{)}^{n+1}{Y}^n/n$ converge and are inverse. For example $v_p(n!)\le n/(p-1)$, so the valuations of the exponential terms tend to infinity with this choice of $ϵ$; the logarithm terms do as well. They take values in $1+p^{ϵ}\mathrm{End}(\Lambda )$ and in $p^{ϵ}\mathrm{End}(\Lambda )$ respectively. The formal identity $\exp (\mathrm{BCH}(X,Y))=\exp X\exp Y$ holds for the convergent series.

Under the correspondence, $U=G^{p^a}$ has Lie lattice $p^{a}L$ and is an open characteristic subgroup. Definewhere ${\log }_G$ denotes the identification of $U$ with $p^{a}L$. A Lie homomorphism preserves BCH, so the exponential identity proves $\eta (uv)=\eta (u)\eta (v)$. If $\eta (u)=I$, the matrix logarithm gives $\rho ({\log }_{G}u)=0$; faithfulness of $\rho$ gives $u=1$. Thus $\eta$ is a continuous faithful integral representation of $U$.

To extend it, take the moduleValues on a transversal of $U\backslash G$ determine $f$, so $W$ is a free module over ${\mathbb{Z}}_p$ of rank $[G:U]\dim V$. These functions are continuous, since on each open coset they are given by the continuous representation $\eta$. Define $(T_{g}f)(x)=f(xg)$. The equivariance condition is preserved, and $T_g{T}_h=T_{gh}$. In a transversal basis each $T_g$ is a permutation of blocks followed by matrices from $\eta (U)$, so it belongs to $\mathrm{GL}(W)$ and depends continuously on $g$.

Faithfulness can be checked directly. If $g\notin U$, right multiplication moves the coset $U$ to a different coset, so $T_g$ moves a function supported on one block and cannot be the identity. If $g\in U$ and $T_g=1$, evaluation at $1$ gives $f(g)=f(1)$ for every $f$, whereas equivariance gives $f(g)=\eta (g)f(1)$. Since $f(1)$ is arbitrary, $\eta (g)=I$, and hence $g=1$. ThereforeCompactness of $G$ makes this continuous injection a homeomorphism onto its closed image. This proves the linearity of a uniform pro-p group for the full group, including its centre, and explains why taking a sufficiently small power lattice permits matrix integration before finite-index induction restores the whole group.