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For a topological group module, use continuous maps with pointwise addition and the inhomogeneous group-cohomology differential. The kernel modulo the image defines continuous group cohomology as an abelian group. With general topological coefficients the quotient topology need not be Hausdorff; the cohomology here is the algebraic quotient of continuous cocycles by continuous coboundaries. With trivial discrete coefficients in a prime field, its first cohomology is the space of continuous group homomorphisms to that field's additive group.

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