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The displayed descending filtration combines powers and group commutators. In a pro-p group, take the closure of the generated product at each step. For a powerful p-group these layers are $P_i(G)=G^{p^{i-1}}$, with iterated closed power subgroups in the infinite case. The successive quotients are elementary abelian p-groups.

For an odd prime number $p$, a finite p-group that is a product of $d(G)$ cyclic subgroups is a powerful p-group. Indeed, $H=G/G^p$ has exponent of a finite group $p$ and the same minimal generator number $d$. Its cyclic factorization gives $|H|\le p^d$, while its Frattini quotient has order $p^d$. Thus $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. The condition on the number of factors is essential: the nonabelian Heisenberg group over ${\mathbb{F}}_p$ has exponent $p$ and is a product of three cyclic subgroups, although its minimal generator number is two.

First, the images of $a_1^p,\dots ,a_n^p$ span $G_2/G_3$ by the surjective group homomorphism in (ii). Part (i) identifies $G_3$ with $\Phi (G_2)$, so the Burnside basis theorem givesThis proves generation by the indicated powers; it does not yet prove that an arbitrary product of powers is itself a power.

For that stronger assertion, let $y\in P_1=G^p$. Part (ii) supplies $x$ with $x^p\equiv y(\mathrm{mod}{P}_2)$. Suppose at some stage $x^p\equiv y(\mathrm{mod}{P}_{j+1})$, where $j\ge 1$. Apply (ii) to the powerful p-group $P_j$: its first two power layers are $P_{j+1},P_{j+2}$. We can therefore choose $u\in P_j$ withThe preliminary power congruence gives $(xu{)}^p\equiv y(\mathrm{mod}{P}_{j+2})$. Replacing $x$ by $xu$ improves the root by one layer. The lower p-series terminates, so finitely many corrections yield $x^p=y$ exactly. Every element of $G^p$ is a $p$-th power in $G$.

Apply (iii) successively to the powerful p-groups $G_i$. Starting with $G_1=⟨a_1,\dots ,a_n⟩$, if $G_i=⟨a_1^{p^{i-1}},\dots ,a_n^{p^{i-1}}⟩$, thenThus

Throughout this question $p$ is odd, $A^p$ means the subgroup generated by the $p$-th powers of elements of $A$, and the commutator of group elements convention is $[x,y]=x^{-1}{y}^{-1}xy$. A subgroup $N$ is powerfully embedded in $G$ when $[N,G]\subseteq N^p$. In particular $N$ is normal, since $n^g=n[n,g]\in N$. A finite p-group $G$ is a powerful p-group when $[G,G]\subseteq G^p$.

First put $M=N^p$. It is normal in $G$. By the reduction permitted in the question, it suffices to prove $[M,G]\subseteq M^p$ after factoring out $[M,G,G]$. In that quotient, $[M,G]$ is central. For $n\in N$ and $g\in G$, put $c=[n,g]\in M$. Then $[c,n]$ is central, and the elementary identity $[ab,g]=[a,g{]}^b[b,g]$ gives, by induction,Both $c$ and $[c,n]$ belong to $M$, and $p$ divides $p(p-1)/2$ because $p$ is odd. Hence $[n^p,g]\in M^p$. The subgroup $M^p$ is normal in $G$, so the commutator identity extends this inclusion from the generating elements $n^p$ to every element of $M$. Applying the allowed reduction proves $N^p$ is powerfully embedded in $G$.

For $H=⟨N,x⟩$, normality of $N$ gives $H=N⟨x⟩$. Its quotient by $N$ is cyclic. Modulo $[N,H]$, the image of $N$ is central and the remaining generator is $x$, so that quotient is abelian. ThereforeThus $⟨N,x⟩$ is powerful.

The lower p-series is $G_1=G$, $G_{i+1}=G_i^p[G_i,G]$. For the later calculations write $P_j=G_{j+1}$, so $P_0=G$. Part (i) will show $P_{j+1}=P_j^p$ and $[P_j,G]\subseteq P_{j+1}$. A useful power lifting in a powerful p-group congruence isIndeed, modulo $P_{j+2}$ the commutators $[u,x]$ are central, lie in $P_{j+1}$ and have exponent dividing $p$. Collecting the $p$ copies of $xu$ therefore gives only the correction $[u,x{]}^{\left(\genfrac{}{}{0ex}{}{p}{2}\right)}$, which is trivial. This argument also covers $j=0$. The congruence is the reason roots and coordinates can be corrected one power layer at a time.

For odd $p$, a subgroup $N$ is powerfully embedded in $G$ if $[N,G]\le N^p$, where $N^p=⟨n^p:n\in N⟩$. The condition implies normality since $n^g=n[n,g]\in N$. The group $G$ is a powerful p-group if $[G,G]\le G^p$, equivalently if $G$ itself is powerfully embedded. For a finite p-group, the Burnside basis theorem gives $d(Q)={\dim }_{{\mathbb{F}}_p}Q/\Phi (Q)$, and $\Phi (Q)=Q^p[Q,Q]$. Thus for powerful $Q$, $d(Q)=\dim Q/Q^p$.

We prove $d(H)\le d(G)$ by induction on $|G|$, with the trivial group immediate. The supplied embedding result applied to $N=G$ makes $G^p$ powerfully embedded, and hence powerful. It is a proper subgroup when $G\ne 1$, because $G^p=\Phi (G)$ is proper in a nontrivial finite p-group. SetAll the displayed quotients are elementary abelian. The supplied onto homomorphism $\theta :V\to W$, $x{G}^p\mapsto x^p{G}^{p^2}$, is therefore linear. Write $v=\dim V=d(G)$, $w=\dim W=d(G^p)$, $r=\dim A$, and $t=\dim \theta (A)$.

Since $H/K$ embeds in $V$, $\Phi (H)\le K$. AlsoHere $(G^p{)}^p=G^{p^2}$ is the iterated-power convention, valid for powerful groups. Hence the natural map $K/\Phi (K)\to W$ is defined. The image under this map of the subspace $\Phi (H)/\Phi (K)$ contains $\theta (A)$, because for every $h\in H$ its power $h^p$ lies in $\Phi (H)\cap K$. It follows thatThe exact sequence $1\to K/\Phi (H)\to H/\Phi (H)\to H/K\to 1$ consists of elementary abelian groups, and $\dim H/K=r$. By induction inside $G^p$, $d(K)\le d(G^p)=w$. ThereforeThis proves $d(H)\le d(G)$, including subgroups $H$ that are not themselves powerful.

The rank of a profinite group is $\mathrm{rk}(P)=\sup \{d(K):K\le P\text{ closed}\}$, where $d(K)$ is its least topological generator number. If $P$ has finite rank then every closed subgroup, including the open subgroup $H$, has no larger rank. Conversely suppose $H$ has rank $r<\infty$. Its normal core $C={\bigcap }_{g\in P}{H}^g$ is an open normal subgroup: there are finitely many conjugates because $H$ has finite index. It has rank at most $r$.

For every closed $K\le P$, the subgroup $K\cap C$ is normal in $K$ and has at most $r$ topological generators. The quotient $K/(K\cap C)$ embeds in the finite group $P/C$. A finite group of order at most $M=[P:C]$ has at most $\lfloor {\log }_{2}M\rfloor$ generators: in a minimal sequential generating list every strict enlargement at least doubles the subgroup order. Lift such quotient generators and adjoin generators of $K\cap C$. Their closed generated subgroup contains the group homomorphism kernel and maps onto the quotient, so equals $K$. ConsequentlyThus an open subgroup has finite rank exactly when the whole profinite group does.

The powerfully embedded subgroup $N$ is itself a powerful p-group, so the finite Frattini subgroup formula gives $\Phi (N)=N^p$. Conjugation by $G$ acts trivially on $N/N^p$, because $[N,G]\le N^p$. Since $N$ is the normal closure of $X$, the image of $X$ therefore generates $N/N^p$: all conjugates have the same images as the original elements. The Burnside basis theorem now gives $N=⟨X⟩$. Equivalently, a proper generated subgroup would lie in a maximal subgroup of $N$, which contains $N^p$ and contradicts generation of that quotient.

For a powerful p-group, $G_2=G^p$, and $G_3=(G^p{)}^p[G^p,G]$. In $G/G_3$, the image of $G_2$ is central and has exponent dividing $p$, while the commutator subgroup is contained in that image. The quotient thus has nilpotency class at most two. Direct collection in such a group givesbecause $p$ is odd. If $z\in G_2$, the same calculation gives $(xz{)}^p\equiv x^p$ modulo $G_3$, since $z^p\in G_3$. Consequently the power map is well defined on $G/G_2$ and is a group homomorphism to $G_2/G_3$. Its image contains all the power generators of that quotient. It is therefore surjective.

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.

Set $P_0=G$ and $P_{i+1}=P_i^p$. The permitted power-embedding result, applied repeatedly, gives $[P_i,G]\le P_{i+1}$, and each $P_i$ is a powerful p-group. These subgroups eventually reach $1$: if $P_i\ne 1$, its Frattini subgroup $P_i^p$ is proper. In fact these are the lower p-series layers, with their indices shifted by one.

For $x\in G$ and $y\in P_i$, the commutator $[y,x]$ lies in $P_{i+1}$, is central modulo $P_{i+2}$, and has exponent dividing $p$ there. The class-two collection calculation from part (iv) therefore givesApplying part (iv) inside $P_i$, whose third lower-series term is contained in $P_{i+2}$, shows that the map $y\mapsto y^p$ is onto $P_{i+1}/P_{i+2}$.

Given $a\in P_1$, first choose $x$ with $x^p\equiv a(\mathrm{mod}{P}_2)$. Suppose $x^p\equiv a(\mathrm{mod}{P}_{i+1})$. Choose $y\in P_i$ with $y^p\equiv x^{-p}a(\mathrm{mod}{P}_{i+2})$. The displayed collection congruence then gives $(xy{)}^p\equiv a(\mathrm{mod}{P}_{i+2})$. Replacing $x$ by $xy$ improves the approximation by one layer. Since the filtration on a group terminates, a finite sequence of corrections gives $x^p=a$ exactly. The reverse inclusion is the definition of $G^p$. Hence

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 pro-p group whose closed commutator subgroup lies in its closed power subgroup $G^{p^{ϵ}}$, with $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. This is the profinite analogue of a powerful p-group. Finitely generated torsion-free examples are exactly the uniform pro-p groups. The additive group of a finite product of p-adic integers is an abelian example.

For odd $p$, a subgroup $H$ of a finite powerful p-group $G$ satisfies $d(H)\le d(G)$. Here $d$ is the dimension of the Frattini quotient. Set $K=H\cap G^p$, $V=G/G^p$, $W=G^p/G^{p^2}$, and $A=H{G}^p/G^p$. The onto power map $\theta :V\to W$ sends $A$ into the image of $H^p$ in $K/\Phi (K)$. Since $\Phi (K)\le G^{p^2}$, this image has dimension at least $\dim \theta (A)$. Also $\Phi (K)\le \Phi (H)\le K$. Induction on $|G|$, applied to the powerful subgroup $G^p$, gives $d(K)\le \dim W$. ConsequentlyThe last inequality is rank-nullity theorem.

A finitely generated powerful p-group in the category of pro-p groups is uniform if all successive indices in its lower p-series are equal. Equivalently it is torsion-free. If its generator number is $d$, the successive indices are $p^d$, the power map carries $G^{p^n}$ bijectively onto $G^{p^{n+1}}$, and an ordered minimal generating set gives unique coordinates $g_1^{{\lambda }_1}\cdots g_d^{{\lambda }_d}$ with ${\lambda }_i$ in the p-adic integers. For $p=2$, powerfulness means $[G,G]\le G^4$.