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For a finitely generated pro-p group, its profinite Frattini subgroup has finite elementary abelian quotient. Continuous group homomorphisms to ${\mathbb{F}}_p$ are precisely the linear functionals on this quotient. The topological generation criterion identifies a minimal generating set with a basis, giving the displayed equality. Finite generation is essential when generator number means the least cardinality of an unrestricted dense generating family.

A topological group is a group with a topology for which $(x,y)\mapsto xy$ and $x\mapsto x^{-1}$ are continuous; we use the Hausdorff convention. Translations and inversion are then homeomorphisms.

For the final generator-extraction assertion, use the profinite Frattini subgroupFor a finitely generated pro-p group, the Burnside basis theorem says that $G/\Phi (G)$ is a finite-dimensional vector space over ${\mathbb{F}}_p$ and a subset topologically generates $G$ precisely when its image spans that vector space. In particular its dimension is the minimal number $d(G)$ of topological generators. A given $d$-element generating set shows $d(G)\le d$. The image of any other generating set spans the same finite-dimensional vector space, so it contains a basis of $d(G)$ elements. Choose their preimages in that generating set and apply the Burnside basis theorem again. Every generating set contains a generating subset of size $d(G)\le d$. This argument also applies to an infinite generating set.

A topological group is a group with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous. A profinite group has either of the following equivalent descriptions: a compact Hausdorff totally disconnected topological group, or an inverse limit of finite discrete groups. Totally disconnected means that every connected component is a singleton.

For clarity, an inverse system over a directed set $I$ consists of groups $G_i$ and maps ${\varphi }_{ij}:G_j\to G_i$ for $i\le j$, with ${\varphi }_{ii}=1$ and ${\varphi }_{ik}={\varphi }_{ij}{\varphi }_{jk}$. Its inverse limit iswith coordinatewise multiplication and the subspace topology. Equivalently, it is a group with compatible continuous projections such that every other compatible family of maps into the $G_i$ factors uniquely through it.

The displayed limit is a closed subgroup of the compact Hausdorff product of the finite groups, and is totally disconnected, since two different tuples are separated by a clopen coordinate condition. Conversely a compact Hausdorff totally disconnected group has a neighbourhood basis of open normal subgroups. One way to see the group part of this standard clopen-basis fact is to choose a clopen identity neighbourhood $U$ inside a prescribed neighbourhood. Compactness of $U$ and continuity give a symmetric identity neighbourhood $V$ with $VU\subseteq U$. The stabilizer $H=\{g:gU=U\}$ contains $V$, is open, and is contained in $U$ because $1\in U$. Its index is finite by part (ii); its normal core is the intersection of finitely many conjugates and is open and normal. The canonical mapis injective because these $N$ separate points, and surjective because compatible cosets have the finite intersection property and $G$ is compact. It is a homeomorphism by compactness and the Hausdorff property. This explains the equivalence of the two definitions.

The profinite Frattini subgroup is $\Phi (G)=\bigcap M$, where $M$ ranges over all maximal proper open subgroups; for the trivial group the empty intersection is $G$. For a pro-p group,Here powers and group commutators mean generated subgroups and the bar supplies closure. Every maximal open subgroup in a pro-p group is normal of index $p$, as follows by passing to the finite p-group quotient by its normal core. Such a subgroup therefore contains all powers and group commutators. Conversely the quotient by $D=\overline{G^p[G,G]}$ is an abelian pro-p group of exponent $p$. If an element of this quotient is nontrivial, some finite elementary abelian quotient detects it; a linear functional to ${\mathbb{F}}_p$ then detects it as well. Its group homomorphism kernel pulls back to a maximal open subgroup of $G$. Thus the intersection of these kernels is exactly $D$.

We also use the topological generation criterion: a subset generates $G$ topologically if and only if its image generates $G/\Phi (G)$ topologically. To prove the nontrivial direction, a proper closed generated subgroup $K$ lies in a proper open subgroup $KN$ for a suitably small open normal $N$, and hence in a maximal open subgroup obtained from the finite quotient $G/N$. Since that maximal subgroup contains $\Phi (G)$, the image of $K$ cannot generate the Frattini quotient.

Now suppose $G$ is a nontrivial procyclic pro-p group. If it had distinct maximal open subgroups $M_1,M_2$, the homomorphism to $C_p\times C_p$ induced by the two quotient maps would be onto: its projections are onto, and a proper such subgroup would be a one-dimensional graph, forcing the two kernels to coincide. Hence $G/(M_1\cap M_2)\cong C_p^2$, contradicting procyclicity. Thus there is exactly one maximal open subgroup $M$, and $\Phi (G)=M$. Any $g\notin M$ generates $G/\Phi (G)\cong C_p$, so the criterion givesFor the trivial group use its identity as generator. The additive group ${\mathbb{Z}}_p$ is an infinite example, topologically generated by $1$: its finite continuous quotients are cyclic $\mathbb{Z}/p^n\mathbb{Z}$.

For a profinite group, its profinite Frattini subgroup isUse $G$ for an empty intersection. This is a closed normal subgroup, invariant under continuous group automorphisms, since these automorphisms permute the maximal open subgroups. Let $H=\overline{⟨X⟩}$. If $H=G$, its image certainly topologically generates $G/\Phi (G)$. Conversely, suppose $H\ne G$. For $g\notin H$, the open normal subgroup basis of a profinite group gives $N{⊴}_{o}G$ with $g\notin HN$: choose a sufficiently small neighbourhood $gN$ disjoint from the closed subgroup $H$. The proper subgroup $HN/N$ of the finite group $G/N$ lies in a maximal proper subgroup. Its preimage $M$ is a maximal proper open subgroup of $G$, contains $H$, and contains $\Phi (G)$. Since $H\Phi (G)$ is compact and therefore closed, density of the image of $X$ in $G/\Phi (G)$ would force $H\Phi (G)=G$, contradicting $H\Phi (G)\le M<G$. Generation can therefore be tested modulo the profinite Frattini subgroup.

For a finite p-group, every maximal proper subgroup is normal and has index $p$. One proof uses the normalizer condition for finite p-groups: maximality forces its normalizer to be the whole group, and its simple $p$-group quotient has order $p$. Therefore every such maximal subgroup contains powers and group commutators. Conversely, the quotient by the algebraic $G^p[G,G]$ is an elementary abelian p-group; its nonzero elements are separated by linear maps to ${\mathbb{F}}_p$. Their kernels give maximal subgroups. This proves the finite Frattini subgroup formula.

Now take a pro-p group and put $D=\overline{G^p[G,G]}$, retaining the closure of the algebraically generated subgroups. A maximal open subgroup is the preimage of a maximal subgroup of a finite $p$-group quotient, so it contains $D$. If $g\notin D$, some finite quotient of the profinite group $G/D$ detects $gD$. That quotient is elementary abelian, and a linear functional nonzero at $gD$ gives a continuous group homomorphism $G\to {\mathbb{F}}_p$ whose kernel is a maximal open subgroup missing $g$. HenceThe closure bar is present in the original PDF; the converted TeX fraction is an extraction error.

For the finite-index assertion, first pass from the arbitrary abstract subgroup $K$ to its normal core of a subgroup $C$. The action on the finitely many cosets of $K$ shows $[G:C]<\infty$, without assuming that $K$ or $C$ is closed. Write $m=[G:C]=p^{r}q$ with $(p,q)=1$, and let $X_m=\{h^m:h\in G\}$. By Lagrange theorem, $X_m\subseteq C$. For $N{⊴}_{o}G$, the quotient $G/N$ is a finite p-group of order, say, $p^s$. Choose an integer $a$ with $aq\equiv 1(\mathrm{mod}{p}^s)$. Thenso $g^{p^r}\in X_{m}N$ for every such $N$. The set $X_m$ is the continuous image of the compact space $G$, hence is closed. Consequently $g^{p^r}\in {\bigcap }_N{X}_{m}N=X_m\subseteq C$. Every element of the abstract finite quotient $G/C$ thus has order dividing $p^r$. By Cauchy theorem for groups, no prime other than $p$ divides its order. Therefore $[G:C]$, and its divisor $[G:K]$, are powers of $p$. This proof does not assume continuity of the finite quotient map.

To prove openness under finite generation, first establish commutator collection with fixed generators in nilpotent groups. If a nilpotent group $P$ is generated by $a_1,\dots ,a_d$, every element of $P\prime$ isInduct on the nilpotency class $c$. The assertion is trivial for $c=1$. Apply the induction hypothesis in $P/{\gamma }_c(P)$; the remaining error lies in the central subgroup ${\gamma }_c(P)=[{\gamma }_{c-1}(P),P]$. Since commutators in this last layer are central, collecting against words in the $a_i$ expresses the error as ${\prod }_i[y_i,a_i]$ with $y_i\in {\gamma }_{c-1}(P)$. Also $[P\prime ,{\gamma }_{c-1}(P)]\le {\gamma }_{c+1}(P)=1$. Thus replacing each $x_i$ by $x_i{y}_i$ multiplies the desired product by precisely that central error, proving the formula.

Let $a_1,\dots ,a_d$ now topologically generate $G$. Every finite continuous quotient is a nilpotent group, so the formula holds there with these fixed generators. For $z\in \overline{[G,G]}$, the sets of tuples $(x_1,\dots ,x_d)\in G^d$ solving $z\equiv {\prod }_i[x_i,a_i](\mathrm{mod}N)$ are nonempty closed subsets of a compact space. They have the finite intersection property, by replacing finitely many $N$ by their intersection. A common tuple solves the equality in $G$. It follows that the algebraic commutator subgroup is the compact image of $G^d$ under this product map, and is closed.

Its quotient $A=G/[G,G]$ is a topologically finitely generated abelian pro-p group. The continuous map ${\mathbb{Z}}_p^d\to A$ taking $({\lambda }_i)$ to ${\prod }_i{\bar{a}}_i^{{\lambda }_i}$ is onto, because its compact image contains a dense generating subgroup. Therefore $A/pA$ is generated as an elementary abelian group by at most $d$ elements, and $pA$ is compact. The preimage $G^p[G,G]$ of $pA$ is closed of finite index, hence open: its complement is a finite union of closed cosets. The finite-index result and the maximal-subgroup property of finite p-groups show that any abstract index-$p$ subgroup of $G$ is normal. It contains this open subgroup, so is itself open.

Finally, a finite-index normal subgroup $C$ has finite $p$-group quotient. If $[G:C]>1$, choose a normal index-$p$ subgroup in that quotient and call its preimage $H$. The preceding argument makes $H$ open. An open subgroup of a finitely generated pro-p group is finitely generated: intersect a finitely generated dense abstract subgroup with $H$, use the Schreier lemma, and observe that the intersection is dense in the open subgroup $H$. Induction on $[G:C]$ now makes $C$ open in $H$, hence in $G$. Returning to the normal core of a subgroup, $K$ is a union of open cosets of $C$. Every finite-index subgroup of a finitely generated pro-p group is open.

A profinite group $G$ is procyclic if every quotient by an open normal subgroup is cyclic. Equivalently $G=\overline{⟨g⟩}$ for one element $g$. For the forward implication, the sets of lifts of generators of $G/N$ are nonempty clopen sets. Any finite intersection contains a lift of a generator of the cyclic quotient by the intersection of those open normal subgroups. Compactness supplies one element generating every finite quotient, and hence generating densely. The reverse implication follows by projecting that generator. For a pro-p group, two distinct maximal open subgroups would give a quotient $C_p^2$, so a nontrivial procyclic group has a unique maximal open subgroup $M$. Its profinite Frattini subgroup is $M$, and any $g\notin M$ generates topologically: otherwise the proper closed subgroup it generates lies in a maximal open subgroup. The additive group of p-adic integers is an infinite example, generated topologically by $1$.