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Realize a finite p-group $P$ as the Galois group of a totally ramified extension $E/{\mathbb{F}}_p((t))$, using Artin-Schreier embedding problems over a characteristic-p field. A uniformizer $u$ of $E$ gives $E\cong {\mathbb{F}}_p((u))$: every element has a unique convergent Laurent expansion with coefficients in the residue field. An automorphism has $u\mapsto a_{1}u+a_2{u}^2+\cdots$, and $a_1$ defines a homomorphism $P\to {\mathbb{F}}_p^{\times }$. Its image is trivial because $|P|$ is a $p$-power and $p\nmid |{\mathbb{F}}_p^{\times }|$. Thus the faithful Galois action lies in the Nottingham group, embedding $P$.

An infinite profinite group is hereditarily just infinite if every open subgroup is a just-infinite profinite group. To prove this, it suffices to show that every nonidentity element of every open subgroup has open closed normal closure: any nontrivial closed normal subgroup then contains that normal closure. The Nottingham group is an example. In odd characteristic its depth-commutator formula constructs nonzero elements of each sufficiently high layer inside any such normal subgroup, and successive coefficient correction fills the entire tail.

Over $k={\mathbb{F}}_p$, the Nottingham group is $\mathcal{N}(k)=\{f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i:a_i\in k\}$ with composition and the $t$-adic topology. Equivalently it is the group of continuous $k$-automorphisms of the Laurent series field $k((t))$ that act trivially on $(t)/(t^2)$. With ordinary series composition, the isomorphism sends $f$ to the automorphism $h(t)\mapsto h(f^{-1}(t))$; this inverse avoids reversing the group product. It is a pro-p group, recovered from its finite truncations, and contains elements of arbitrarily large finite $p$-power order.

We next bound the closed derived series. For odd $p$, the depth calculation yieldsFor the inclusion from left to right, two series of depth at least $m$ commute modulo ${\mathcal{N}}_{2m+1}$: at depth $2m$ the only possible leading pair has equal depths and its coefficient vanishes. For the reverse inclusion, every depth $k\ge 2m+1$ can be written $r+s=k$ with $r,s\ge m$ and $r\not\equiv s(\mathrm{mod}p)$. Start with $r=m,s=k-m$; if their difference vanishes modulo $p$, replace them by $r=m+1,s=k-m-1$, which is allowed except at $k=2m+1$, where the original difference is already nonzero. The difference changes by $2$, nonzero for odd $p$. Commutators thus supply every high leading coefficient, and closed successive coefficient correction fills the tail.

With ${\mathcal{N}}^{(0)}={\mathcal{N}}_1$, induction on $n$ givesFor $p=2$ use the standard characteristic-two derived-series collection bound, explicitly stated as an additional result:for a fixed constant $C$. One can take a fixed enlarged constant from the tail estimate ${\mathcal{N}}^{(n)}\supseteq {\mathcal{N}}_{\lceil 4{\kappa }^n\rceil +1}$. The more involved two-coefficient collection is necessary in characteristic two; the exact odd-prime tail formula is not asserted there.

In either case there are constants $C^{\prime }>0$ and $\lambda <3$ such that ${\log }_p[\mathcal{N}:{\mathcal{N}}^{(n)}]\le C^{\prime }{\lambda }^n$. If an analytic structure existed, its dimension $d$ would be positive, since a zero-dimensional analytic pro-p group is discrete and compact, hence finite. The second alternative of the supplied theorem would requirefor all sufficiently large $n$, for some fixed $s$. Combining the bounds gives $d{3}^{-s}\le C^{\prime }(\lambda /3{)}^n$, impossible as $n\to \infty$.

The root solution proved hereditary just infiniteness, so the supplied theorem applies. Part (i) excluded its linear alternative, and this calculation excludes its growth alternative. Hence the Nottingham group is not analytic over any pro-$p$ ring.

Write $k={\mathbb{F}}_p$. The Nottingham group $\mathcal{N}$ is the set of formal series $f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i$ over $k$, with ordinary composition $f\circ g$ as multiplication and the $t$-adic topology. The identity is $t$. A compositional inverse exists uniquely by solving its coefficients successively, since the linear coefficient is $1$.

Equivalently $\mathcal{N}$ is the group of continuous $k$-automorphisms of $k((t))$ whose action on $(t)/(t^2)$ is the identity. Such an automorphism is determined by the image of $t$, which is $t+O(t^2)$. To identify this with ordinary series composition without reversing multiplication, send $f$ to the automorphism $h(t)\mapsto h(f^{-1}(t))$. Its composite with the automorphism for $g$ is the one for $f\circ g$.

The depth filtration isEach ${\mathcal{N}}_m$ is open and normal, and the coefficient of $t^{m+1}$ identifies ${\mathcal{N}}_m/{\mathcal{N}}_{m+1}$ with $(k,+)$. Thus $[\mathcal{N}:{\mathcal{N}}_m]=p^{m-1}$, and $\mathcal{N}$ is the inverse limit of its finite p-group truncations.

Here is a local-field proof that every finite p-group embeds in $\mathcal{N}$. We first establish the realization step rather than assume the Nottingham embedding itself. Put $F=k((t))$ and $\Gamma =\mathrm{Gal}(F^{\mathrm{sep}}/F)$, and let $I⊴\Gamma$ be the inertia subgroup, the group homomorphism kernel of the action on the algebraic closure of the residue field. We use the standard additive normal basis theorem consequence $H^j(\Gamma ,F^{\mathrm{sep}})=0$ for $j>0$. It follows by taking direct limits of the finite-Galois normal-basis modules, whose additive cohomology vanishes. The Artin-Schreier sequencethen givesThe classes $t^{-m}$ for positive $m$ prime to $p$ are linearly independent: a nonzero finite linear combination has a most negative valuation prime to $p$, whereas a negative valuation of $z^p-z$ is divisible by $p$. Thus there are infinitely many independent continuous characters $\Gamma \to C_p$.

Induct on the order of a finite p-group $E$. Choose a central subgroup $C\cong C_p$ and put $Q=E/C$. By induction there is an onto map $\varphi :\Gamma \to Q$ with $\varphi (I)=Q$, corresponding to a totally ramified $Q$-extension. The central extension $1\to C\to E\to Q\to 1$ is specified by a two-cocycle. Its pullback along $\varphi$ is a coboundary because $H^2(\Gamma ,{\mathbb{F}}_p)=0$. Correcting a set-theoretic section by that coboundary gives a continuous homomorphism ${\rho }_0:\Gamma \to E$ lifting $\varphi$.

We must make the lift onto on inertia, not merely onto on the whole Galois group. Let $A=I\cap \ker \varphi$. On $A$, the lift ${\rho }_0$ takes values in $C$. Characters vanishing on $A$ form a finite-dimensional space: $\Gamma /A$ has a finite subgroup $I/A\cong Q$ and procyclic quotient $\Gamma /I\cong \hat{\mathbb{Z}}$, so a character is determined by finitely many values on generators of these two pieces. The infinite-dimensional character space therefore has infinitely many distinct restrictions to $A$. Choose a character $\chi :\Gamma \to C$ whose restriction to $A$ does not cancel ${\rho }_0{|}_A$, and set $\rho (\gamma )=\chi (\gamma ){\rho }_0(\gamma )$. Centrality of $C$ makes this another homomorphism lifting $\varphi$. Its image on $A$ is nontrivial and hence all of $C$, while its image on $I$ maps onto $Q$. Therefore $\rho (I)=E$. The fixed field of $\ker \rho$ is the required totally ramified finite Galois extension $K/F$ with group $E$. This proves the realization step for every finite p-group.

Since the residue field of $K$ is still $k$, choosing a uniformizer $u$ identifies $K$ with $k((u))$. Indeed successive subtraction of its residue coefficient and division by $u$ gives each element of its valuation ring a unique convergent expansion ${\sum }_{i\ge 0}{b}_i{u}^i$ with $b_i\in k$, and fractions give the Laurent expansions. Each element of $E$ preserves the valuation and fixes $k$, so sends $u$ to $a_{1}u+a_2{u}^2+\cdots$. The linear coefficient is a homomorphism $E\to k^{\times }$. Since $E$ is a $p$-group and $|k^{\times }|=p-1$, that homomorphism is trivial. Thus the faithful Galois action lies in $\mathcal{N}(k)$, completing the embedding proof.

An infinite profinite group is hereditarily just infinite when every open subgroup is infinite and every nontrivial closed normal subgroup of each such open subgroup is open. We sketch the normal-subgroup argument, giving the depth calculation in odd characteristic explicitly. For series of depths $r,s$,where $[f,g]=f^{-1}\circ g^{-1}\circ f\circ g$. The coefficient follows by comparing $f\circ g$ and $g\circ f$ in degree $r+s+1$: their cross terms are $ab(r+1)$ and $ab(s+1)$.

Let $H$ be open and $1\ne K⊴H$ closed. Choose $m$ with ${\mathcal{N}}_m\le H$ and choose $f\in K$ of depth $r$. For every $j\ge m$ with $j\not\equiv r(\mathrm{mod}p)$, commutation with $t+b{t}^{j+1}$ supplies an element of $K$ of depth $r+j$, with any prescribed nonzero leading coefficient by varying $b$. Thus every sufficiently high layer except possibly depths congruent to $2r$ modulo $p$ occurs in $K$.

When $p$ is odd, choose one available depth $u\ge r+m$ with $u\not\equiv 2r,r(\mathrm{mod}p)$. Such a residue exists even for $p=3$. An element $h\in K$ of that depth can be commuted with $t+b{t}^{k-u+1}$ for every sufficiently large missing depth $k\equiv 2r(\mathrm{mod}p)$. The new leading coefficient contains $2u-k\equiv 2(u-r)\ne 0$, so those missing layers occur too. Hence $K$ contains an element with any chosen coefficient in every depth $k\ge M$, for some $M$.

Successive coefficient correction now proves ${\mathcal{N}}_M\le K$: match the coefficient of an arbitrary target in depth $M$ by an element of $K$, correct the error in depth $M+1$, and continue. The partial products converge $t$-adically to the target and remain in $K$, which is closed. Thus $K$ is open. This proves hereditary just infiniteness for odd $p$.

For characteristic two, the leading coefficient alone misses even depths. The standard characteristic-two normal-closure collection lemma supplies the needed extra input: for every $m\ge 1$ and every nonidentity $f\in \mathcal{N}({\mathbb{F}}_2)$, the closed subgroup generated by the conjugates $f^g$ with $g\in {\mathcal{N}}_m$ contains ${\mathcal{N}}_M$ for some $M$. Its proof retains the next coefficients in pairs of group commutators to fill the parity gaps, then uses the same successive coefficient correction. Apply this lemma to $f\in K$ and ${\mathcal{N}}_m\le H$: every such conjugate lies in $K$, so $K$ contains the tail and is open. This states the additional collection result used in the characteristic-two sketch; the odd-prime leading-term argument is not being applied there. Since every open subgroup contains a tail and is infinite, the Nottingham group is hereditarily just infinite for every prime $p$.

For a pro-p ring $R$, an $R$-analytic pro-p group is a pro-p group equipped with an analytic manifold structure locally modelled on open subsets of a finite free $R$-module, such that multiplication and inversion are given locally by convergent power series. The number of local coordinates is its analytic dimension. The Nottingham group admits no such structure over any pro-$p$ ring: its hereditary just infiniteness, unbounded finite cyclic torsion and sub-base-three derived-series growth exclude both alternatives in the analytic just-infinite structure theorem.