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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.

Here $p$ is odd. For a finite p-group, write $N^p=⟨n^p:n\in N⟩$. A powerfully embedded subgroup satisfies $[N,G]\le N^p$; this also makes $N$ normal, since every conjugate of an element of $N$ differs from it by an element of $N$. A powerful p-group is one with $[G,G]\le G^p$, namely one powerfully embedded in itself. Use $[a,b]=a^{-1}{b}^{-1}ab$ and $a^b=b^{-1}ab$ throughout.

The lower p-series is defined byEach quotient $G_i/G_{i+1}$ is an elementary abelian p-group, and $[G_i,G]\le G_{i+1}$. These are generated subgroups, without closure bars, because the group is 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

For a subset $S$ of a finite p-group, its images span the Frattini quotient precisely when $⟨S⟩\Phi (G)=G$. By the generating property proved in part (ii), this is equivalent to $⟨S⟩=G$.

If $S$ is a minimal generating set of a group, its images must be an irredundant spanning set in the vector space $G/\Phi (G)$. A linear dependence would express one image using the others; those other images would still span, so the other elements would still generate $G$, a contradiction. Thus the images form a basis. Conversely lifts of a basis generate $G$, and deleting a lift destroys spanning and hence generation.

This proves the Burnside basis theorem, including equality of sizes for every inclusion-minimal generating set of a group:Inclusion-minimal generating sets of a group need not have equal sizes in arbitrary finite groups; the elementary-abelian Frattini quotient is the essential p-group feature.

For a finite group, define the Frattini subgroup bythe intersection is $G$ when there are no maximal subgroups, which for a finite group happens only for $G=1$. A second definition is the set of non-generators: an element $x$ such that $⟨S,x⟩=G$ always implies $⟨S⟩=G$.

If $x\in \Phi (G)$ and $H=⟨S⟩<G$, put $H$ in a maximal subgroup $M$. Then $x\in M$, so $⟨S,x⟩\le M<G$. Conversely, if $x\notin \Phi (G)$, some maximal subgroup $M$ omits it; maximality gives $⟨M,x⟩=G$, although $M\ne G$. This proves equivalence.

Every group automorphism permutes maximal subgroups, so $\Phi (G)$ is a characteristic subgroup, in particular a normal subgroup. Alsoa proper $H$ would lie in a maximal subgroup containing both factors.

Set $F=\Phi (G)$ and take a Sylow subgroup $P$ of $F$. For $g\in G$, normality of $F$ makes $P^g$ a Sylow subgroup of $F$. Conjugating inside $F$ therefore gives the Frattini argument factorization $G=F{N}_G(P)$. The preceding generating property forces $N_G(P)=G$. Thus every Sylow subgroup of $F$ is normal, and part (i) shows that $F$ is a nilpotent group.

Now suppose $G$ is a finite p-group. The normalizer condition makes every maximal subgroup $M$ normal: $M<N_G(M)$ and maximality force $N_G(M)=G$. The quotient $G/M$ has no proper nontrivial subgroup, hence is cyclic of prime order; its p-power order makes that prime $p$. Consequently every such $M$ containswhere the two factors are the commutator subgroup and the subgroup generated by pth powers. Thus $D\le \Phi (G)$.

The quotient $G/D$ is an elementary abelian group, hence a vector space over ${\mathbb{F}}_p$. If $x\notin D$, extend the nonzero vector $xD$ to a basis and choose a linear functional nonzero on it. Its kernel pulls back to a subgroup of index $p$, a maximal subgroup omitting $x$. Therefore $x\notin \Phi (G)$, provingThe two factors in the last formula are multiplied; the superscripted expression in the TeX aid is not the formula printed in the PDF.

A nilpotent group has a finite central seriesEquivalently the upper central series, defined by $Z_{j+1}/Z_j=Z(G/Z_j)$, reaches $G$.

We first prove the normalizer condition for nilpotent groups. If $H<G$, choose the least $j$ with $Z_j⊈H$. Then $Z_{j-1}\le H$, and an element $z\in Z_j\setminus H$ satisfies $[z,H]\le Z_{j-1}\le H$. Thus $z$ normalizes $H$, giving $H<N_G(H)$.

Let $P$ be a Sylow subgroup of a finite nilpotent $G$, and write $N=N_G(P)$. If $N<G$, the normalizer condition gives some $g\in N_G(N)\setminus N$. The subgroups $P$ and $P^g$ are both Sylow subgroups of $N$, so Sylow theorem supplies $n\in N$ with $P^{gn}=P$. But then $gn\in N_G(P)=N$, forcing $g\in N$, a contradiction. Therefore every Sylow subgroup is a normal subgroup of $G$.

For normal Sylow subgroups $P,Q$ at different primes, $[P,Q]\le P\cap Q=1$. They commute, have coprime orders, and their product has order $|G|$. Multiplication consequently gives an isomorphism from their external direct product of groups onto $G$.

Conversely, a nontrivial finite p-group has nontrivial center of a group: its class equation has every noncentral conjugacy-class size divisible by $p$, so $|Z(G)|$ is a positive multiple of $p$. Induction on order, applied to $G/Z(G)$, gives a central series for $G$. Finite direct products of these central series give a central series for the product, padding shorter series by the full factor. Thus the finite nilpotent group decomposition is

For a nontrivial finite p-group $G$ of order $p^m$ with $m\ge 1$, partition into conjugacy classes. A noncentral class has size $[G:C_G(x)]$, a positive power of $p$ larger than one by Lagrange's theorem. The class equation therefore saysThe center of a group contains the identity, so $|Z(G)|\ge p$, proving its nontriviality. The positive-exponent qualification matters: the one-element group has no nonidentity central element.

If $|G|=p^2$, its center of a group has order $p$ or $p^2$. In the first case, the quotient group $G/Z(G)$ has prime order and is cyclic. Whenever a central quotient group is cyclic, writing all elements as $g^{i}z$ with $z$ central shows that they commute: $(g^{i}z)(g^{j}w)=g^{i+j}zw=(g^{j}w)(g^{i}z)$. Thus that case would already make $G$ Abelian and its center of a group all of $G$, a contradiction. Hence every group of order $p^2$ is Abelian.

If there is an element of order $p^2$, it is a generator of a group for $G$, giving $C_{p^2}$. Otherwise every nonidentity element has order $p$. Pick $a\ne 1$ and $b\notin ⟨a⟩$. Their cyclic subgroups have trivial intersection, and they commute; the $p^2$ distinct products $a^i{b}^j$ exhaust $G$. Hence the classification of groups of order p squared isBoth groups exist and are nonisomorphic, since only the first has an element of order $p^2$.

The identity belongs to the center of a group. If $z,w$ commute with every $g$, then $zw$ does too; rearranging $gz=zg$ also shows that $z^{-1}$ commutes with $g$. Hence $Z(G)$ is a subgroup. Moreover $gz{g}^{-1}=z$ for $z\in Z(G)$, so it is a normal subgroup.

Suppose $|G|=p^n>1$. The conjugation action partitions $G$ into conjugacy classes. The orbit-stabilizer theorem gives class size $[G:C_G(x)]$, where $C_G(x)$ is the centralizer of $x$. By Lagrange's theorem, this size is a power of $p$; it is one exactly when $x$ is central. Thus the class equation has the formIt follows that $p\mid |Z(G)|$. Since the identity is central, $|Z(G)|\ge 1$, and thereforeThis is the nontrivial-centre property of a finite p-group.

For a group action of a finite group $G$ on a set and a point $x$, the orbit-stabilizer theorem iswhere $G_x=\{g:gx=x\}$ is the stabilizer subgroup. The map $g{G}_x\mapsto gx$ is well defined because elements of the same coset act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their cosets agree; every orbit point has the form $gx$, giving surjectivity. This is a bijection. Lagrange's theorem, $|G|=[G:H]|H|$ for any subgroup $H$, yields the counting formula.

Apply the conjugation action of the finite p-group $G$ to its nontrivial normal subgroup $N$. Normality ensures the action stays inside $N$. The singleton orbits of a group action are exactly the elements of $N\cap Z(G)$. Every larger orbit of a group action has size a positive power of $p$, by the orbit-stabilizer theorem and Lagrange's theorem. ThereforeSince $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the central intersection property of normal subgroups of finite p-groups.

For the proper subgroup $H$, let $H$ act by left multiplication on the left cosets $G/H$. A coset $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed cosets are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton orbits of a group action have size divisible by $p$, soSince $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore the normalizer strictly contains $H$:This is the normalizer condition for finite p-groups. The proof also covers $H=\{1\}$, whose action fixes all cosets.

Now let $n=m=p$ be prime and put $\alpha =\sqrt[p]{p}$ and $A=\mathbb{Z}[\alpha ]$. The Eisenstein criterion applied at $p$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the discriminant of elements of a number field of its power basis iswhere $f(X)=X^p-p$. By the discriminant-index formula for an integral lattice, every prime dividing $[{\mathcal{O}}_L:A]$ must therefore be $p$.

Suppose $A\ne {\mathcal{O}}_L$. The finite abelian group ${\mathcal{O}}_L/A$ is then a nontrivial finite p-group, so it contains an element of order $p$. Consequently there is an $x\in {\mathcal{O}}_L\setminus A$ such thatwith some $c_k$ not divisible by $p$.

The polynomial $X^p-p$ is Eisenstein, so total ramification from an Eisenstein polynomial gives a unique prime ideal $P$ above $p$ with normalized discrete valuationLet $k_0$ be the least index for which $p\nmid c_{k_0}$. The term $c_{k_0}{\alpha }^{k_0}$ has valuation $k_0<p$; every earlier term has valuation at least $p$, and every later term with coefficient prime to $p$ has a distinct, larger valuation. The non-Archimedean valuation therefore givesThis contradicts $x\in {\mathcal{O}}_L$, since an algebraic integer has nonnegative valuation at every prime ideal. Thus $A={\mathcal{O}}_L$, and

The group $G$ is a nontrivial finite p-group, so the nontrivial center of a finite p-group gives $Z(G)\ne \{e\}$. If $Z(G)\ne G$, then $Z(G)$ is already a nontrivial proper normal subgroup, so $G$ is not a simple group. If $Z(G)=G$, then $G$ is abelian. By Cauchy's theorem for finite groups, it has a subgroup of order $p$; every subgroup of an abelian group is normal, and this subgroup is nontrivial and proper because $|G|=p^4$. Thus in every case

Using the commutation relations, every element has a normal formOn squaring, equal generators cancel and every interchange contributes only the central element $Z$, so $g^2\in ⟨Z⟩$. There are at most $2^5=32$ normal forms, and every element has order dividing four. Hence $G$ is a finite p-group with $p=2$, of order at most $32$.

A normal subgroup $N$ is powerfully embedded in a finite p-group or pro-p group $G$ if $[N,G]\le N^p$ for odd $p$, or $[N,G]\le N^4$ for $p=2$; use closed generated subgroups in the pro-p group case. The condition implies that $N$ is a powerful p-group. For odd $p$, both $[N,G]$ and $N^p$ remain powerfully embedded.

A finite p-group is powerful if $[G,G]\le G^p$ for odd $p$, or $[G,G]\le G^4$ for $p=2$. Here powers denote generated subgroups. The same definition applies to pro-p groups using closed power and commutator subgroups. Its Frattini subgroup is $G^p$. For odd $p$, successive power layers admit linear correction of roots, which makes every element of $G^p$ an actual $p$th power.

For odd $p$, set $P_0=G$ and $P_{i+1}=P_i^p$. Powerful embedding gives $[P_i,G]\le P_{i+1}$. Modulo $P_{i+2}$, commutators with $P_i$ are central and have exponent dividing $p$, so $(xy{)}^p\equiv x^p{y}^p$ when $y\in P_i$. Also the power map $P_i/P_{i+1}\to P_{i+1}/P_{i+2}$ is a surjective homomorphism. First find a root modulo $P_2$, then multiply it by successive corrections in $P_i$. The process terminates for a finite p-group, and converges for a pro-p group with separated power filtration. This proves that the generated power subgroup consists of actual powers.

The inverse limit of the quotients of an abstract group by its normal subgroups of finite $p$-power index. Maps to pro-p groups factor uniquely through the canonical dense image in this completion. Unlike profinite completion, only finite p-groups are used.

A p-subgroup is a subgroup which is a finite p-group, for a specified prime $p$. Every p-subgroup lies in a Sylow subgroup.

Every abstract subgroup of finite index in a finitely generated pro-p group is open. Fixed-generator commutator collection in nilpotent groups makes the abstract commutator subgroup closed. The resulting abelian quotient is a quotient of a finite power of the p-adic integers, so its subgroup of $p$th powers is closed and of finite index. Thus the algebraic $G^p[G,G]$ is open and every abstract index-$p$ subgroup is open. An arbitrary finite quotient is a finite p-group; induction on its order, using a normal subgroup of index $p$ and the Schreier lemma, gives the assertion. The normal core reduces nonnormal finite-index subgroups to this case.