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For a profinite group $G={\underleftarrow{\lim }}_j{G}_j$, a subset $S\subseteq G$ is a topological generating set exactly when $⟨p_j(S)⟩=p_j(G)$ in every finite quotient. Indeed, a subgroup of a profinite group is dense exactly when its image in every finite continuous quotient is surjective.

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.

Choose $\hat{\zeta }\in H^2(\hat{\Gamma },M)$ with ${\iota }^{\ast }(\hat{\zeta })=\zeta$, and letbe its extension. By the given fact, $E$ is a profinite group, hence is residually finite. The extension $E_{\zeta }^{\prime }$ over $\Gamma$ is the pullback of $E$ along the injective map $\iota$. Part 5(a)(iii) embeds $E_{\zeta }^{\prime }$ into $E$. Since every subgroup of a residually finite group is residually finite, so is $E_{\zeta }^{\prime }$.

Let $I(L/K)$ be the inertia group and let $\phi$ denote Frobenius on the residue-field extension. The Relative Weil group isIts Weil-group topology makes $I(L/K)$ an open profinite group with its usual topology and gives $W(L/K)/I(L/K)$ the discrete topology. Thus every inertia coset is an open copy of $I(L/K)$.

Take $L=K^{\mathrm{nr}}$, the maximal unramified extension of $K$. ThenThe subgroup $\{0\}\subset \mathbb{Z}$ is open in the discrete Weil-group topology, but it is not open in the profinite topology inherited from $\hat{\mathbb{Z}}$.

Every homomorphism from a group $G$ to a profinite group $P$ extends uniquely to a continuous homomorphism $\hat{G}\to P$. Construct the extension on every finite quotient of $P$ and invoke the universal property of an inverse limit; uniqueness follows from the density of $G$ in $\hat{G}$.