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A finite generating set and finitely many relators in a free pro-p group, divided out by their closed normal closure. Relators are elements of the free pro-p group; they need not be finite ordinary words. A pro-p group with such a presentation is finitely presented in the pro-$p$ category.

Enumerate all words $w_i$ on two generators in a free pro-p group. Choose powers $p^{e_i}$ so large that ${\sum }_i{\tau }^{p^{e_i}}<2\tau -1$ for some $1/2<\tau <1$, and impose the closed normal relations $w_i^{p^{e_i}}=1$. The Golod-Shafarevich infinitude criterion for pro-p groups makes the resulting pro-$p$ group infinite. Its abstract subgroup generated by the two generator images is dense and infinite; each element is represented by an enumerated word and has finite order. The ambient finite quotients separate its elements, so it is also a residually finite group. The abstract subgroup, rather than the whole topological group, is the desired finitely generated torsion group.

Embed a free pro-p group on $x_1,\dots ,x_d$ into the units of the noncommutative power-series algebra ${\mathbb{F}}_p⟨⟨X_1,\dots ,X_d⟩⟩$ by $x_i\mapsto 1+X_i$. Its Magnus degree is the lowest total degree occurring in $\iota (w)-1$, with $D(1)=\infty$. Since $(1+v{)}^{p^e}=1+v^{p^e}$ in characteristic $p$, one has $D(w^{p^e})\ge p^{e}D(w)$. High powers therefore impose relations with small weight at any fixed parameter $0<\tau <1$.

We construct an infinite residually finite torsion group using the Golod-Shafarevich infinitude criterion for pro-p groups. Its useful form says that a pro-$p$ presentation on $d$ generators with relators $r_i$ gives an infinite pro-p group wheneverHere $D$ is the Magnus degree of a free pro-p group element: map each generator to $1+X_j$ in a noncommutative power-series algebra over ${\mathbb{F}}_p$, and take the lowest degree of $w-1$. The criterion permits infinitely many relators with a convergent weight sum. It is a consequence of the Golod-Shafarevich inequality for the completed group algebra; see Ershov's survey, Theorem 3.1.

Take $p=2$, $d=2$, and $\tau =3/4$. Let $F$ be the free pro-p group on $a,b$, and enumerate all ordinary finite words $w_1,w_2,\dots$ in these generators and their inverses. Choose integers $e_i$ sufficiently large thatForm the quotient pro-p groupwhere the bar means the closed normal closure. If the Magnus image of $w_i$ is $1+v_i$, characteristic two givesfor nonidentity $w_i$; identity relators contribute zero weight. It follows thatThus the Golod-Shafarevich infinitude criterion for pro-p groups proves that $P$ is infinite.

Let $G=⟨\bar{a},\bar{b}⟩$ be the abstract subgroup generated by the two images inside $P$. This subgroup is dense because the two images topologically generate the quotient, and it is infinite: a finite subgroup of a Hausdorff profinite group is closed, so could not be dense in infinite $P$. Every element of $G$ is the image of some ordinary word $w_i$, and the imposed relation makes its order divide $2^{e_i}$. Therefore $G$ is a two-generated torsion group. Finally $P$ is an inverse limit of finite 2-groups. Any nonidentity element of $G$ is nonidentity in $P$ and hence has a nonidentity image in one of those finite groups. Restricting that map to $G$ proves that it is a residually finite group. We have obtainedThe abstract subgroup $G$ is essential here: the argument establishes finite order for ordinary words, not for every topological limit in the ambient pro-p group.

There is no surjective homomorphism from this group to an infinite soluble group. In fact finitely generated soluble torsion groups are finite. To prove this, let $S$ be finitely generated, soluble and torsion, and induct on its derived length. In the abelian case, finitely many generators of finite orders generate a finite group. In general the abelianization $S/S^{\prime }$ is a finitely generated abelian torsion group, hence finite. Its kernel $S^{\prime }$ therefore has finite index and is finitely generated by Schreier's lemma. Explicitly, for a finite coset transversal $T$ containing $1$ and a finite symmetric generating set $X$, the finitely many elements $tx{\overline{tx}}^{-1}$ generate the finite-index subgroup, where the bar denotes the chosen representative of the coset. Now $S^{\prime }$ is torsion and has smaller derived length, so induction makes it finite. Hence $S$ is finite as well. Every quotient of $G$ is finitely generated and torsion, so any soluble quotient must be finite.

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