Solution

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 $p$.