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The Lazard isomorphism compares continuous group cohomology of suitable $p$-adic analytic groups with Lie algebra cohomology of their associated Lie algebras. It converts a cohomological question about a nonlinear multiplication law into a calculation with alternating multilinear maps. The hypotheses matter: one cannot assert this comparison for every topological group.

One precise integral version takes a uniform pro-p group $G$ and a finite free ${\mathbb{Z}}_p$-module $M$. For odd $p$, require the continuous action to have image in $1+p\mathrm{End}(M)$; for $p=2$, use $1+4\mathrm{End}(M)$. With $\mathfrak{g}$ the associated integral Lazard Lie algebra,This is an integral refinement of the rational comparison.^[ref]

For the additive group $G={\mathbb{Z}}_p$ with trivial coefficients $M={\mathbb{Z}}_p$, continuous one-cocycles are $x\mapsto ax$. The associated rank-one abelian Lie algebra has $H^1\cong {\mathrm{Hom}}_{{\mathbb{Z}}_p}({\mathbb{Z}}_p,{\mathbb{Z}}_p)\cong {\mathbb{Z}}_p$, giving the same answer. Rationalizing coefficients alone would not establish an integral comparison in general.

In the setting of a uniform pro-p group $G$, its integral Lazard Lie algebra is a finite free ${\mathbb{Z}}_p$-module whose addition and Lie bracket recover infinitesimal versions of group multiplication and the group commutator. It carries more integral information than the associated ${\mathbb{Q}}_p$-Lie algebra. The construction requires hypotheses on the group and its filtration; it is not available unchanged for every profinite group.^[ref]

For odd $p$, a concrete example is $G=1+p{M}_n({\mathbb{Z}}_p)$. The convergent matrix logarithm and matrix exponential identify its Lie algebra with $p{M}_n({\mathbb{Z}}_p)$, using ordinary matrix addition and bracket $[A,B]=AB-BA$. Multiplication corresponds to the Baker--Campbell--Hausdorff formula, beginningFor $p=2$, the analogous safe neighborhood is $1+4M_n({\mathbb{Z}}_2)$ with Lie algebra $4M_n({\mathbb{Z}}_2)$. The smaller neighborhood $1+2{\mathbb{Z}}_2$ contains $-1$, so torsion already prevents it from being uniform.^[ref]

For the additive group ${\mathbb{Z}}_p$, the associated Lie algebra is rank one with zero bracket. Under suitable coefficient hypotheses, the Lazard isomorphism compares its Lie algebra cohomology with continuous group cohomology.