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Suppose $1\ne N⊲G$ is proper. The defining property makes $N$ cyclic of order $p$. Conjugation gives a homomorphismwhose kernel is the centralizer $C_G(N)$. The automorphism group has order $p-1$, so the kernel has finite index in $G$. It contains $N$, because $N$ is abelian. If this kernel were proper, it would be an order-$p$ subgroup, and its finite index would force $G$ itself to be finite. Therefore $C_G(N)=G$, so $N$ is central.

Choose $h\notin N$. Part (i) shows that $h$ has order $p$. Its cyclic subgroup meets $N$ trivially, since two order-$p$ subgroups with nontrivial intersection are equal. Since $N$ is central,This subgroup has order $p^2$ and is proper because $G$ is infinite, contradicting the assumed order of every proper nontrivial subgroup. Thus no such $N$ exists:Together with part (i), this proves that Tarski monster groups are two-generated and simple.

The requested representation is the left coset actionIt is well defined because equal cosets remain equal after multiplication by $g$, and $\rho (g{g}^{\prime })=\rho (g)\rho (g^{\prime })$. It is transitive. For $H=1$ it is the left regular action; for general $H$ it is the associated transitive coset representation.

Its kernel is the normal core of a subgroupIndeed, $g$ fixes every $xH$ precisely when $x^{-1}gx\in H$ for every $x$. Thus $N⊲G$ and $N\le H$. If $[G:H]=i<\infty$, the image lies in the finite symmetric group $S_i$, soIn particular $N$ has finite index in $H$.

For Higman group as an amalgam of iterated HNN extensions, use the cyclic squaring presentationFirst form $L=⟨b,c\mid b^{-1}cb=c^2⟩$, an HNN extension of the infinite cyclic group $⟨c⟩$. Its base embeds, so $c$ has infinite order; $b$ has infinite order as well, since the homomorphism $L\to \mathbb{Z}$ sending $b\mapsto 1,c\mapsto 0$ is onto. Consequently $⟨b⟩$ and $⟨b^2⟩$ are isomorphic infinite cyclic subgroups. Adjoining the stable letter $a$ with $a^{-1}ba=b^2$ givesWe claim $⟨a,c⟩$ is a rank-two free group. No nonzero power of $c$ belongs to $⟨b⟩$: applying the map $b\mapsto 1,c\mapsto 0$ would make that power equal to $1$, contradicting the infinite order of $c$. Thus no nonzero power of $c$ belongs to $⟨b^2⟩$ either. Any nonempty freely reduced word in $a,c$ that contains $a$ is therefore a reduced HNN sequence: between inverse $a$ letters, a nonzero $c$ power cannot create a pinch. Britton's lemma makes it nonidentity. A word containing only $c$ is nonidentity by the base embedding. This proves the claim.

Likewiseis obtained from $⟨d,a\mid d^{-1}ad=a^2⟩$ by adjoining $c$, and its subgroup $⟨c,a⟩$ is free of rank two by the same argument. Form the amalgamated free product $P{\ast }_{⟨a,c⟩}Q$, identifying these two free subgroups generator by generator. Its presentation is exactly that of $\mathcal{H}$. The component embeddings allowed in the question show that it contains $P$, hence is infinite. It has four group generators and four relators, so it is finitely presented.

To show that this Higman group has no nontrivial finite quotient, consider any finite image and the orders of the four generator images. A relation $x^{-1}yx=y^2$ makes $y$ and $y^2$ have the same order, so every generator order is odd. Suppose one is nontrivial, and let $p$ be the smallest prime dividing any of the four orders. Choose a generator $y$ whose order is divisible by $p$, with predecessor $x$ satisfying the displayed relation. Iterating conjugation $|x|$ times givesThe multiplicative order $d$ of $2$ modulo $p$ thus divides $|x|$. Also $d\mid p-1$ and $d>1$, since $2\not\equiv 1(\mathrm{mod}p)$. Any prime divisor of $d$ is smaller than $p$ and divides the order of $x$, contradicting minimality of $p$. Hence every generator image is trivial, so every finite image is trivial. A proper finite-index subgroup would give a nontrivial transitive finite coset image, which is impossible. Therefore $\mathcal{H}$ is infinite and finitely presented, with no proper finite-index subgroups.

A maximal normal subgroup is a proper normal subgroup $M⊲G$ with no normal subgroup strictly between $M$ and $G$. To prove that finitely generated groups have maximal proper normal subgroups, fix a proper normal subgroup $K$ and order the proper normal subgroups containing $K$ by inclusion. For a chain, its union is normal and contains $K$. It is still proper: if the union were all of $G$, each element of a finite generating set would lie in some chain member, and the chain order would put all these finitely many group generators in one member. That member would equal $G$, a contradiction. The Zorn lemma therefore supplies a maximal proper normal subgroup $M$ containing $K$.

Apply this to $\mathcal{H}$ and $K=1$. By the normal-subgroup correspondence, $\mathcal{H}/M$ is nontrivial and simple. It is finitely generated as a quotient of $\mathcal{H}$. It cannot be finite, since $\mathcal{H}$ has no nontrivial finite quotients. Thus

The final assertion for arbitrary proper subgroups is false. Take $G=S_3$ and $H=⟨(12)⟩$. If a normal subgroup contained $H$, it would contain all conjugate transpositions, which generate $S_3$, so it would equal $G$. Hence this proper subgroup lies in no proper normal subgroup, and in particular in no maximal normal subgroup. The maximal-normal existence result requires the starting subgroup itself to be normal.

Let $K=\ker \theta$. The subgroup correspondence for a surjective group homomorphism sends a subgroup $B\le H$ to ${\theta }^{-1}(B)$ and a subgroup $A\le G$ containing $K$ to $\theta (A)$. Inverse images are subgroups containing $K$, and images are subgroups. Surjectivity gives $\theta ({\theta }^{-1}(B))=B$. Conversely, if $K\le A$ and $\theta (g)\in \theta (A)$, choose $a\in A$ with $\theta (g)=\theta (a)$. Then $g{a}^{-1}\in K\le A$, so $g\in A$. Thus ${\theta }^{-1}(\theta (A))=A$. These two assignments are inverse bijections.

For $C={\theta }^{-1}(B)$, the map of left coset setsis well defined and onto. If $\theta (g_1)B=\theta (g_2)B$, then $\theta (g_2^{-1}{g}_1)\in B$, so $g_2^{-1}{g}_1\in C$ and $g_{1}C=g_{2}C$. Hence it is injective, provingNo normality of $B$ is required; these are coset sets, not necessarily quotient groups.

The abelianization is $\overline{G}=G/[G,G]$, where the commutator subgroup is generated by the commutators $[x,y]=x^{-1}{y}^{-1}xy$. Conjugating a commutator gives another commutator, so this subgroup is normal. In the quotient all commutators are trivial, and therefore the quotient is abelian. If $N⊲G$ and $G/N$ is abelian, every commutator lies in $N$, so $[G,G]\le N$. Henceis well defined and satisfies $\pi =\tilde{\pi }\circ q$, where $q:G\to \overline{G}$ is the natural map. It is uniquely determined because $q$ is onto. This is the universal factorization property of abelianization.

Now write $S_i$ for the unique index-$i$ subgroup of $S$. Since conjugation preserves index, as shown in part (ii), $g{S}_i{g}^{-1}=S_i$ for every $g\in S$. Thus every $S_i$ is normal. For each prime $p$ the quotient $S/S_p$ has order $p$ and is cyclic, so there is a surjection $S\to C_p$. Being an abelian quotient, it factors through $\overline{S}$.

The abelianization is finitely generated because $S$ is. By the Fundamental theorem of finitely generated abelian groups,where $T$ is finite. If $r=0$, choose a prime $p>|T|$. The finite group $T$ could not surject onto the order-$p$ group $C_p$, contradicting the preceding surjections. Hence $r\ge 1$, and projection from a free coordinate supplies

Every subgroup of finite index in $\overline{S}$ has an inverse image of the same index in $S$, so there cannot be two different subgroups of any given finite index in $\overline{S}$. If $r\ge 2$, reduction modulo a prime $p$ in the first and second free coordinates gives two distinct surjective maps to $C_p$ with different kernels: the first basis vector lies in one kernel but not the other. Both kernels have index $p$, a contradiction. Therefore $r=1$.

If $T\ne 1$, choose a prime $p$ dividing its order. The decomposition of a finite abelian group into cyclic factors gives a surjection $\chi :T\to C_p$. In $\mathbb{Z}\oplus T$, the mapshave distinct index-$p$ kernels, again distinguished by $(1,0)$. This is another contradiction, so $T=1$. We have proved uniqueness of finite-index subgroups forces infinite cyclic abelianization:

A residually finite group is one in which every nonidentity element has nonidentity image under some homomorphism to a finite group. Equivalently, the intersection of all finite-index normal subgroups is trivial: a finite separating map has such a kernel, and conversely quotienting by such a normal subgroup gives a finite separating map.

Choose the surjection $\sigma :S\to \overline{S}\cong \mathbb{Z}$. For each positive integer $i$, the subgroup ${\sigma }^{-1}(i\mathbb{Z})$ has index $i$ and hence equals $S_i$. Its kernel $[S,S]$ is therefore contained in every finite-index subgroup. Indeed,If $S$ is residually finite, this intersection is trivial. Thus $[S,S]=1$ and $\sigma$ is an isomorphism. The residually finite group with one subgroup of each finite index is infinite cyclic conclusion isResidual finiteness is used exactly to eliminate the commutator subgroup that all finite quotients would otherwise fail to detect.

Write $p:U(H)\to PU(H)$ for the scalar quotient. Local strongly continuous sections exist: near a fixed projective unitary, fix its phase by requiring one nonzero matrix coefficient, taken relative to a fixed representative, to be positive real. Thus the pullback $E=\{(z,V):p(V)=\rho (z)\}$ is a locally trivial central extension of the circle by the circle. It is compact, since finitely many closed arcs inside such trivializing neighbourhoods cover the compact base and each pullback is a product with the compact fibre.

Any two lifts commute up to a scalar. Their commutator defines a continuous alternating bicharacter $c(z,w)$ on the base: the scalar commutator is independent of phase choices and multiplicative in each argument. If $z$ is an irrational rotation, its powers are dense and $c(z,z^m)=1$ for every integer $m$. Continuity makes $c(z,w)=1$ for every $w$. Such $z$ are themselves dense, so all the lifts commute, and $E$ is abelian.

The tautological representation $R(z,V)=V$ is strongly continuous. Choose a unit vector $\xi$ and average its rank-one orthogonal projection over Haar measure on $E$:The integrand is trace-norm continuous, so $K$ is positive compact with trace one, and commutes with $R(E)$. A positive eigenvalue has a nonzero finite-dimensional eigenspace invariant under $E$. Commuting unitary matrices on that space have a common unit eigenvector $\eta$. Therefore $R(e)\eta =\chi (e)\eta$ for a continuous character $\chi$ of $E$.

On the central fibre, $R(1,\lambda I)=\lambda I$, so $\chi (1,\lambda I)=\lambda$. Consequentlyis independent of that lift. It is a homomorphism and strongly continuous, because the continuous map on $E$ descends through its quotient to the base. It satisfies $p(U_z)=\rho (z)$. This proves circle projective representations lift to unitary representations using an actual central character, rather than assuming a global choice of phases.

First, all lifts of the real one-parameter projective group commute. Its scalar commutator $c(s,t)$ is again a continuous alternating bicharacter. For fixed nonzero $t$ and an integer $n>0$, a lift of $t$ is a phase multiple of the $n$th power of a lift of $t/n$; hence $c(t,t/n)=1$ and $c(t,mt/n)=1$ for every integer $m$. Rational multiples of $t$ are dense in $\mathbb{R}$, so continuity makes $c(t,s)=1$ for every $s$.

Choose a unitary $U$ representing $\rho (1)$ and the allowed self-adjoint logarithm $A\in U^{\prime \prime }$ with $U=e^{iA}$. Every lift of $\rho (t)$ commutes with $U$, hence with $A$ by the double commutant theorem for star-algebras condition. Thereforeis a continuous projective homomorphism, with $\tilde{\rho }(1)=1$. It is periodic of period one and descends to $\mathbb{R}/\mathbb{Z}$. Part (b) lifts it to a continuous unitary homomorphism $L$ of that circle. Each $L(t)$ is a lift of $\tilde{\rho }(t)$ and commutes with $A$, since a representative has the form of an original lift times $e^{-itA}$. Thusis a strongly continuous unitary homomorphism with projective class $\rho (t)$. This gives the required lift of the whole real group, not just unrelated unitary representatives at each time.

Suppose $(L,({\pi }_i))$ and $(L^{\prime },({\pi }_i^{\prime }))$ are inverse limits of the same system, in the universal-property sense. The compatible family $({\pi }_i^{\prime })$ induces a unique continuous homomorphism $u:L^{\prime }\to L$ with ${\pi }_{i}u={\pi }_i^{\prime }$; similarly there is a unique $v:L\to L^{\prime }$ with ${\pi }_i^{\prime }v={\pi }_i$. Then $uv$ and $1_L$ both induce the projections $({\pi }_i)$, so uniqueness gives $uv=1_L$. Likewise $vu=1_{L^{\prime }}$. Thus the limits are uniquely isomorphic by an isomorphism preserving every projection. In particular the inverse maps are continuous, so this is an isomorphism of topological groups, not just abstract groups.

A topological group is a group with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous. A profinite group has either of the following equivalent descriptions: a compact Hausdorff totally disconnected topological group, or an inverse limit of finite discrete groups. Totally disconnected means that every connected component is a singleton.

For clarity, an inverse system over a directed set $I$ consists of groups $G_i$ and maps ${\varphi }_{ij}:G_j\to G_i$ for $i\le j$, with ${\varphi }_{ii}=1$ and ${\varphi }_{ik}={\varphi }_{ij}{\varphi }_{jk}$. Its inverse limit iswith coordinatewise multiplication and the subspace topology. Equivalently, it is a group with compatible continuous projections such that every other compatible family of maps into the $G_i$ factors uniquely through it.

The displayed limit is a closed subgroup of the compact Hausdorff product of the finite groups, and is totally disconnected, since two different tuples are separated by a clopen coordinate condition. Conversely a compact Hausdorff totally disconnected group has a neighbourhood basis of open normal subgroups. One way to see the group part of this standard clopen-basis fact is to choose a clopen identity neighbourhood $U$ inside a prescribed neighbourhood. Compactness of $U$ and continuity give a symmetric identity neighbourhood $V$ with $VU\subseteq U$. The stabilizer $H=\{g:gU=U\}$ contains $V$, is open, and is contained in $U$ because $1\in U$. Its index is finite by part (ii); its normal core is the intersection of finitely many conjugates and is open and normal. The canonical mapis injective because these $N$ separate points, and surjective because compatible cosets have the finite intersection property and $G$ is compact. It is a homeomorphism by compactness and the Hausdorff property. This explains the equivalence of the two definitions.

The profinite Frattini subgroup is $\Phi (G)=\bigcap M$, where $M$ ranges over all maximal proper open subgroups; for the trivial group the empty intersection is $G$. For a pro-p group,Here powers and group commutators mean generated subgroups and the bar supplies closure. Every maximal open subgroup in a pro-p group is normal of index $p$, as follows by passing to the finite p-group quotient by its normal core. Such a subgroup therefore contains all powers and group commutators. Conversely the quotient by $D=\overline{G^p[G,G]}$ is an abelian pro-p group of exponent $p$. If an element of this quotient is nontrivial, some finite elementary abelian quotient detects it; a linear functional to ${\mathbb{F}}_p$ then detects it as well. Its group homomorphism kernel pulls back to a maximal open subgroup of $G$. Thus the intersection of these kernels is exactly $D$.

We also use the topological generation criterion: a subset generates $G$ topologically if and only if its image generates $G/\Phi (G)$ topologically. To prove the nontrivial direction, a proper closed generated subgroup $K$ lies in a proper open subgroup $KN$ for a suitably small open normal $N$, and hence in a maximal open subgroup obtained from the finite quotient $G/N$. Since that maximal subgroup contains $\Phi (G)$, the image of $K$ cannot generate the Frattini quotient.

Now suppose $G$ is a nontrivial procyclic pro-p group. If it had distinct maximal open subgroups $M_1,M_2$, the homomorphism to $C_p\times C_p$ induced by the two quotient maps would be onto: its projections are onto, and a proper such subgroup would be a one-dimensional graph, forcing the two kernels to coincide. Hence $G/(M_1\cap M_2)\cong C_p^2$, contradicting procyclicity. Thus there is exactly one maximal open subgroup $M$, and $\Phi (G)=M$. Any $g\notin M$ generates $G/\Phi (G)\cong C_p$, so the criterion givesFor the trivial group use its identity as generator. The additive group ${\mathbb{Z}}_p$ is an infinite example, topologically generated by $1$: its finite continuous quotients are cyclic $\mathbb{Z}/p^n\mathbb{Z}$.

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 objective is a continuous faithful representation of a uniform pro-p group $G$ in some ${\mathrm{GL}}_N({\mathbb{Z}}_p)$. The proof proceeds from the intrinsic Lie lattice to a small matrix group, then extends the resulting representation to all of $G$. This last step matters: a representation defined only on an open subgroup does not yet establish linearity of the original group.

Let $L=L_G$ be the Lie algebra of a uniform pro-p group and put $\mathfrak{g}={\mathbb{Q}}_p{\otimes }_{{\mathbb{Z}}_p}L$. The uniform-group/Lie-lattice correspondence identifies the group law on $L$ with BCH. The substantial finite-dimensional Lie-theoretic input is the Ado theorem: every finite-dimensional Lie algebra over a characteristic-zero field admits a faithful finite-dimensional representation. Apply it to obtainAn Adjoint representation would not suffice here, since it kills the centre; Ado supplies injectivity even in central directions.

Choose a p-adic lattice $\Lambda$ spanning $V$. Since $L$ has finitely many lattice generators, their representing matrices have bounded denominators. There is therefore an integer $a\ge 0$ such thatOn this matrix lattice the series $\exp X={\sum }_{n\ge 0}{X}^n/n!$ and $\log (1+Y)={\sum }_{n\ge 1}(-1{)}^{n+1}{Y}^n/n$ converge and are inverse. For example $v_p(n!)\le n/(p-1)$, so the valuations of the exponential terms tend to infinity with this choice of $ϵ$; the logarithm terms do as well. They take values in $1+p^{ϵ}\mathrm{End}(\Lambda )$ and in $p^{ϵ}\mathrm{End}(\Lambda )$ respectively. The formal identity $\exp (\mathrm{BCH}(X,Y))=\exp X\exp Y$ holds for the convergent series.

Under the correspondence, $U=G^{p^a}$ has Lie lattice $p^{a}L$ and is an open characteristic subgroup. Definewhere ${\log }_G$ denotes the identification of $U$ with $p^{a}L$. A Lie homomorphism preserves BCH, so the exponential identity proves $\eta (uv)=\eta (u)\eta (v)$. If $\eta (u)=I$, the matrix logarithm gives $\rho ({\log }_{G}u)=0$; faithfulness of $\rho$ gives $u=1$. Thus $\eta$ is a continuous faithful integral representation of $U$.

To extend it, take the moduleValues on a transversal of $U\backslash G$ determine $f$, so $W$ is a free module over ${\mathbb{Z}}_p$ of rank $[G:U]\dim V$. These functions are continuous, since on each open coset they are given by the continuous representation $\eta$. Define $(T_{g}f)(x)=f(xg)$. The equivariance condition is preserved, and $T_g{T}_h=T_{gh}$. In a transversal basis each $T_g$ is a permutation of blocks followed by matrices from $\eta (U)$, so it belongs to $\mathrm{GL}(W)$ and depends continuously on $g$.

Faithfulness can be checked directly. If $g\notin U$, right multiplication moves the coset $U$ to a different coset, so $T_g$ moves a function supported on one block and cannot be the identity. If $g\in U$ and $T_g=1$, evaluation at $1$ gives $f(g)=f(1)$ for every $f$, whereas equivariance gives $f(g)=\eta (g)f(1)$. Since $f(1)$ is arbitrary, $\eta (g)=I$, and hence $g=1$. ThereforeCompactness of $G$ makes this continuous injection a homeomorphism onto its closed image. This proves the linearity of a uniform pro-p group for the full group, including its centre, and explains why taking a sufficiently small power lattice permits matrix integration before finite-index induction restores the whole group.

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 $p$.

The Chinese remainder theorem says that for pairwise coprime positive moduli $m_i$, the map $\mathbb{Z}/(\prod m_i)\to {\prod }_i\mathbb{Z}/m_i$ is an isomorphism; in particular prescribed residues have a unique solution modulo the product.

Write $n=p^{e}m$, with $e\ge 2$ and $(p,m)=1$. Choose $z\equiv 1+p^{e-1}(\mathrm{mod}{p}^e)$ and $z\equiv 1(\mathrm{mod}m)$. The Chinese remainder theorem supplies $z$. The binomial theorem gives $z^p\equiv 1(\mathrm{mod}{p}^e)$, since every nonconstant term is divisible by $p^e$; also $z^p\equiv 1(\mathrm{mod}m)$. Thus $z^p\equiv 1(\mathrm{mod}n)$, whereas $z\not\equiv 1(\mathrm{mod}n)$. Its multiplicative order is exactly $p$.

For odd $n=\prod p_i^{e_i}$ define the Jacobi symbol $(a/n)={\prod }_i(a/p_i{)}^{e_i}$, using the Legendre symbol, with $(a/1)=1$. It is zero if $a$ shares a prime factor with $n$. For example $(2/9)=(2/3{)}^2=1$, but the square residues modulo nine are $0,1,4,7$, so two is not a square.

For positive odd coprime $u,v$, Jacobi reciprocity isFactor $u=\prod p_i^{a_i}$ and $v=\prod q_j^{b_j}$ and multiply quadratic reciprocity for each prime pair. Its sign exponent is ${\sum }_{i,j}{a}_i{b}_j(p_i-1)(q_j-1)/4$. For odd numbers, $(rs-1)/2\equiv (r-1)/2+(s-1)/2(\mathrm{mod}2)$, so this exponent has the same parity as $(u-1)(v-1)/4$. Multiplicativity of the Legendre symbol gives the two Jacobi products, proving the formula. If $u,v$ are not coprime, both sides of the displayed reciprocity equality are zero.

For the previously constructed $z$, every prime divisor of $n$ sees residue one, so $(z/n)=1$. But $p\nmid (n-1)/2$, because $p\mid n$ and $p$ is odd. The order-$p$ unit therefore has $z^{(n-1)/2}\not\equiv 1(\mathrm{mod}n)$. This supplies a unit failing the proposed test.

Finally the passing units form the kernel of the homomorphismMultiplicativity makes this a homomorphism, and the failing unit shows its kernel is proper. A proper subgroup has index at least two. Hence the test fails on at least half the units modulo $n$.

An action assigns each $g\in G$ a homeomorphism of $X$, with $e\cdot x=x$ and $g\cdot (h\cdot x)=(gh)\cdot x$. For the covering assertion use a properly discontinuous free action: every $x$ has an open neighbourhood $U$ such that $gU\cap U=\varnothing$ for every $g\ne e$. Then the translates $gU$ are pairwise disjoint. The quotient map $p$ is open because $p^{-1}(p(U))={\bigcup }_{g}gU$ is open. Each restriction $p:gU\to p(U)$ is a continuous open bijection and hence a homeomorphism. Thus $p(U)$ is evenly covered and $p:X\to X/G$ is a covering map.

For the surjective homomorphism one additionally needs $X$ path connected; choose $x_0$ and its image $y_0$. Lift a based loop at $y_0$ from $x_0$. Its endpoint has a unique form $g{x}_0$. Homotopy lifting makes this assignment well-defined on the fundamental group. With concatenation meaning first the first loop and then the second, the second lift from $g{x}_0$ is $g$ times the lift from $x_0$, so the endpoint is $gh{x}_0$. This proves the homomorphism property. For each $g$, a path in $X$ from $x_0$ to $g{x}_0$ projects to a loop realizing $g$. HencePointwise freeness alone is insufficient for a covering: an irrational rotation action of $\mathbb{Z}$ on the circle is free but its orbits are dense. Path connectedness is also essential for the final surjectivity claim as written: the discrete space $X=G$ with left translation is a free properly discontinuous action whose quotient is a point, and its trivial fundamental group cannot surject onto nontrivial $G$. The proof supplies the standard connected covering-space interpretation rather than hiding these hypotheses.

For a plane crystallographic group, the translation subgroup consists of translations by vectors in a rank-two lattice $\Lambda$; the lattice is this additive group of translation vectors. The point group is the image under the linear-part homomorphism $G\to O(2)$, equivalently $G/\Lambda$. Conjugating a translation by an isometry with linear part $A$ gives translation by $Av$, so every point-group element preserves $\Lambda$.

Choose a lattice basis. A point-group rotation is represented in that basis by an integral matrix, because it maps lattice vectors to lattice vectors, and its inverse does likewise. The trace is therefore an integer. In an orthonormal basis the same rotation has trace $2\cos \theta$, which lies between minus two and two. Hence $2\cos \theta \in \{-2,-1,0,1,2\}$. The value two gives the identity, while the other values give angles $\pi$, $\pm 2\pi /3$, $\pm \pi /2$, or $\pm \pi /3$. Thus every nonidentity rotation has orderThis proves the crystallographic restriction theorem from the lattice action rather than assuming the classification of wallpaper groups.