Apply part i toThe element $\alpha$ generates the additive cyclic group $\mathbb{Z}/p^n\mathbb{Z}$ exactly when it is coprime to $p$, which is equivalent to its reduction modulo $p$ being nonzero. Hence $\{\alpha \}$ is a topological generating set of the additive group ${\mathbb{Z}}_p$ if and only if $\alpha \not\equiv 0(\mathrm{mod}p)$.

A subset $S\subseteq G$ topologically generates $G$ exactly whenfor every $j$. In the usual presentation by surjective finite quotients this reads $⟨p_j(S)⟩=G_j$. Indeed, a subgroup is dense exactly when its image in every finite discrete quotient is the whole quotient. This is the finite-quotient criterion for topological generation.

Suppose first that $\Gamma$ is conjugacy separable. If $h\in \Gamma$ is not conjugate to $\gamma$ in $\Gamma$, some homomorphism to a finite group sends them to nonconjugate elements. This homomorphism factors through a finite quotient of $\hat{\Gamma }$, so $h$ cannot be conjugate to $\gamma$ in $\hat{\Gamma }$. Thus

Conversely, suppose this equality holds and $h$ is not conjugate to $\gamma$ in $\Gamma$. Then they are not conjugate in $\hat{\Gamma }$. By the finite-quotient criterion for conjugacy in a profinite group, their images fail to be conjugate in some finite quotient. This is precisely conjugacy separability.

The canonical image of $\Gamma$ is dense in its profinite completion $\hat{\Gamma }$. Therefore every $x\in \hat{\Gamma }$ is a limit of a net $({\delta }_{\lambda })$ in $\Gamma$. Continuity of conjugation givesso every element of ${\mathrm{Cl}}_{\hat{\Gamma }}(\gamma )$ lies in the closure of ${\mathrm{Cl}}_{\Gamma }(\gamma )$. The reverse inclusion follows because the larger conjugacy class contains the smaller one and is closed by part b(iii). Hence

If $g=xh{x}^{-1}$ in $G$, applying each projection map gives $p_j(g)=p_j(x)p_j(h)p_j(x{)}^{-1}$ in $G_j$.

Conversely, suppose $p_j(g)$ and $p_j(h)$ are conjugate for every $j$. Define the nonempty finite setEvery transition map $G_j\to G_i$ carries $X_j$ into $X_i$, so the $X_j$ form an inverse system. By the nonemptiness theorem for inverse limits of finite sets, there is a compatible tuple $(x_j)\in {\underleftarrow{\lim }}_j{X}_j\subseteq G$. Coordinatewise equality then gives $xh{x}^{-1}=g$. This proves the finite-quotient criterion for conjugacy in a profinite group.

The profinite group $G$ is compact, and the preceding part shows thatis its image. By the continuous image of a compact space theorem, the conjugacy class is compact. Since a profinite group is Hausdorff, every compact subset is closed, so ${\mathrm{Cl}}_G(h)$ is closed.

For fixed $h\in G$, the conjugation map is the compositefollowed by multiplication. Inversion, constant maps, diagonal maps, and multiplication are continuous in a topological group, so $c_h(g)=gh{g}^{-1}$ is continuous.

The product ${\prod }_j{G}_j$ of the finite groups with their discrete topology is a topological group under coordinatewise multiplication and inversion. The compatibility equations defining $G={\underleftarrow{\lim }}_j{G}_j$ are preserved by both operations, so $G$ is a subgroup. Their restrictions to the subspace topology on $G$ are continuous. Hence $G$ with its standard inverse-limit topology is a topological group.

Give each nonempty finite set $X_j$ the discrete topology. The product space $X={\prod }_j{X}_j$ is compact by the Tychonoff theorem. For each $i\le j$, the compatibility condition $f_{ij}(x_j)=x_i$ defines a closed subset $C_{ij}\subseteq X$.

These sets have the finite intersection property. Indeed, for finitely many conditions choose an index $k$ above every index occurring in them, choose any $x_k\in X_k$, and use the transition maps from $k$ to define all required coordinates; choose the remaining coordinates arbitrarily. Compactness therefore givesThis is the nonemptiness theorem for inverse limits of finite sets.

For an inverse system of sets $(X_j,f_{ij})$ indexed by a directed set $J$, the inverse limit is the set of compatible tuplesIts projection to $X_j$ sends $(x_i{)}_i$ to $x_j$.

LetThe extension $K_2/K$ is the unramified quadratic extension. Since ${\mathbb{F}}_{p^2}^{\times }$ contains all $(p^2-1)$st roots of unity, $L/K_2$ is a cyclic, tamely and totally ramified extension of degree $p^2-1$, with automorphisms $u\mapsto \zeta u$.

The Frobenius automorphism of ${\mathbb{F}}_{p^2}/{\mathbb{F}}_p$ extends by fixing $u$ and conjugates $\zeta$ to ${\zeta }^p$. Hence $L/K$ is Galois, its inertia group isand its residue-field Galois group isThe tame ramification also gives $G_1=1$.

Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. The lower ramification groups arewith $G_{-1}=G$. In particular $G_0$ is the inertia group, and $G_1$ is the wild inertia group. When the extension is totally ramified, the uniformizer criterion for lower ramification groups permits the equivalent test on one uniformizer.

The reduction of $P=(0,0)$ is a nonidentity point of the group $\tilde{E}({\mathbb{F}}_3)$ of prime order seven. Hence $7P$ reduces to the identity and lies in $E_1({\mathbb{Q}}_3)$. Part i shows that $P$ has infinite order, so $7P\ne O$. Every nonidentity point in $E_1({\mathbb{Q}}_3)$ has formal parameter $z=-x/y$ of positive valuation and thereforehas negative valuation. Thus $7P$ does not have integral coordinates.

Let $E_r({\mathbb{Q}}_2)$ be the $r$th term in the filtration of elliptic-curve points over a local field. Reduction givesof order four, whilehas order two. The formal logarithm is injective on $E_2({\mathbb{Q}}_2)$ and identifies it with an additive subgroup of ${\mathbb{Q}}_2$, so $E_2({\mathbb{Q}}_2)$ is torsion-free. A finite subgroup of $E({\mathbb{Q}}_2)$ therefore injects into $E({\mathbb{Q}}_2)/E_2({\mathbb{Q}}_2)$, whose order is $4\cdot 2=8$. Thus $|E({\mathbb{Q}}_2{)}_{\mathrm{tors}}|$ divides $8$.

For good reduction at $p$, the reduction of an elliptic curve gives a mapwhose kernel is its formal group of an elliptic curve, and the torsion in that kernel is $p$-primary. Hence the prime-to-$p$ part of any torsion subgroup injects into $\tilde{E}({\mathbb{F}}_p)$.

At $p=2$, the odd part of $E(\mathbb{Q}{)}_{\mathrm{tors}}$ must divide $4$, so it is trivial. The torsion group is therefore a $2$-group. At $p=3$, all of this group has order prime to $3$ and injects into a group of order $7$, so it too is trivial. Thus

After passing to a subsequence, weak compactness and the Rellich-Kondrachov compactness theorem giveFor fixed $w\in H^1(U)$, the Sobolev inequality gives $w\in L^6(U)$, soMeanwhile ${\partial }_j{u}_{n_k}\rightharpoonup {\partial }_{j}u$ in $L^2(U)$. Pairing this weak convergence with the strong convergence of the products, or equivalently using the weak-strong product convergence lemma, yields

Linearity in $u$ follows from linearity of the weak derivative and Lebesgue integration. By the Holder inequality, the three-dimensional Sobolev inequality, and the preceding $L^3$ interpolation estimate,Thus $\Phi$ is a linear functional and a continuous linear map on $H^1(U)$.

The Holder inequality interpolates between $L^2(U)$ and $L^6(U)$:Taking cube roots and applying the three-dimensional Sobolev inequality $H^1(U)\hookrightarrow L^6(U)$ givesThis is the H1 L3 interpolation inequality in three dimensions.

The boundedness in the Sobolev space $H^1(U)$ and the weak sequential compactness of bounded sequences in a reflexive Banach space give a subsequence converging weakly to some $u\in H^1(U)$. For each integer $m\ge 2$, the Rellich-Kondrachov compactness theorem makes $H^1(U)\hookrightarrow L^m(U)$ compact because the dimension is two. Repeated extraction followed by the diagonal argument gives one subsequence $u_{n_k}$ converging strongly to $u$ in every $L^m(U)$ with integral $m\ge 2$.

For any finite real $p\ge 1$, choose an integer $m\ge \max \{2,p\}$. Since $U$ has finite measure, the Lp inclusion on a finite measure space givesThe same subsequence therefore works for every finite $p$.

In polar coordinates, setaway from the origin, assigning any value at the origin. This is unbounded as $r=\sqrt{x^2+y^2}\downarrow 0$. It belongs to $L^2(U)$ becauseMoreoverand henceThus $u\in H^1(U)$ but $u\notin L^{\infty }(U)$, exhibiting the failure of first-order Sobolev embedding into Linfinity in two dimensions.