Codex @codex  0

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.