Proof. Let \(g:A\to B\) be any morphism of \(F\)-algebras. If \(e\) is an idempotent of \(A\), then \(g(e)^2=g(e^2)=g(e)\). In other words, if the image of an idempotent is nonzero, then it is an idempotent. In addition, if \(g\) is surjective, and \(e\) is central in \(A\), then \(g(e)\) has to be central in \(B\).

Now assume that \(f(e)=e'_1+e'_2\) admits a decomposition into a sum of two central idempotents in \(B\). Note that \(f^{-1}:B\to A\) is a surjective morphism of \(F\)-algebras. We have \(e=f^{-1}(f(e))=f^{-1}(e_1')+f^{-1}(e_2')\). So, the preimage of a non-primitive idempotent is also non-primitive. And we are done for the proof. ◻

Proof. Let \(e_1\) be a primitive idempotent of \(A\), and \(\{e_1,\dots,e_l\}\) be the set of primitive idempotents generated by the action of \(G\). That is, \(\{e_1,\dots,e_l\}=\{\varphi_A(g)(e_1)\mid g\in G\}\). Note that by the Lemma 1, each \(\varphi_A(g)(e_1)\) must be a primitive central idempotent of \(A\). Now define an element \(e\) by \[e:=\sum_{i=1}^l e_i.\] It is central because it is the sum of central elements. It is still an idempotent because \(e_i^2=e_i\) and \(e_ie_j=0\) for all \(1\leq i\neq j\leq l\). Moreover, it is fixed by the action of \(G\), that is, \(\varphi_A(g)(e)=e\) for all \(g\in G\). So, \(e\in A^G=F\). But \(F\) is a field, the only (central) idempotent is the identity \(1_F=1_A\). Recall that \(A\cong M_1\times\cdots\times M_n\), as an étale \(F\)-algebra. This implies that \(A\) is a semisimple \(F\)-algebra (Artinian with trivial Jacobson radical), and \(1=e_1+\cdots+e_l\) is the decomposition of \(1\) into primitive idempotents in \(A\). So, the action of \(G\) on \(e_1\) generates all the primitive idempotents. And we are done. ◻

Proof of Theorem 1. Let \(A\) be a Galois \((G,F)\)-algebra. We first explain the structure of \(A\) as an \(F\)-algebra. It is a finite dimensional étale algebra by definition, so it must be of the form \[A\cong\prod_{i=1}^l M_i\] where \(M_i/F\) is a (finite) separable extension, for each \(1\leq i\leq l\). By the Lemma 2, we know that the action of \(G\) is transitive on the idempotents, hence \(M_i\)-s are all isomorphic as fields. Let \(n:=\lvert G\rvert=[A:F]\), and \(m:=[M_i:F]\), then we have \[n=ml.\] Fix a primitive idempotent \(e\) of \(A\) and let \(G_e:=\operatorname{Stab}_G(e)\). We have \(l=\lvert G/G_e\rvert\) and \(m=\lvert G_e\rvert\). Clearly for each \(g\in G_e\), we have \(g(eA)=eA\), so there is a group homomorphism \(\varphi_{eA}:G_e\to\operatorname{Aut}_F(eA)\) defined by \(\varphi_{eA}(g):=\varphi_A(g)\vert_{eA}\).

Consider the morphism of \(F\)-algebras \(\pi_e:A\to eA\) defined by \(x\mapsto e\cdot x\). Clearly \(\pi_e\) is surjective. Let \(ex\in(eA)^{G_e}\). Define \[x':=\frac{1}{\lvert G_e\rvert}\sum_{g\in G}\varphi_A(g)(ex)\in A.\] By the Lemma 1 above, the element \(\varphi_A(g)(e)\) is some primitive idempotent. And \(\varphi_A(g)(e)=e\) if and only if \(g\in G_e\). Therefore \(ex'=ex\), that is, \(x'\) is a preimage of \(ex\) under \(\pi_e\). By the construction of \(x'\), it is fixed by the action of \(G\). So \(x'\in A^G=F\), which implies that \(ex\in\pi(F)\). Conversely, it is clear that \(\pi(F)\subseteq(eA)^{G_e}\). This shows that \((eA)^{G_e}=\pi(F)\cong F\), which already implies that \(eA/F\) is Galois, for the subgroup \(\varphi_{eA}(G_e)\) of \(\operatorname{Aut}_F(eA)\) fixes \(F\). And this also implies that \(\varphi_{eA}:G_e\to\operatorname{Gal}(eA/F)\) is surjective. Recall that \(m=\lvert G_e\rvert\), so the map \(\varphi_{eA}\) has to be injective. Therefore, the algebra \((eA/F,\varphi_{eA})\) is a Galois \(G_e\)-field extension.

We define an \(F\)-morphism \(\tilde{f}:\operatorname{Ind}^G_{G_e}eA\to A\) by the universal property of the functor \(\operatorname{Ind}^G_{G_e}\). To be precise, the identity map \(1_{eA}:eA\to eA\) is an isomorphism of Galois \(G_{e}\)-extensions over \(F\), hence also \(G_{e}\)-equivariant. Let us define \(f:eA\to A\) by composing \(1_{eA}\) and \(eA\hookrightarrow A\). Define \(i:eA\to\operatorname{Ind}^G_{G_e}eA\) by \(ex\mapsto\tau_{ex}\), where \(\tau_{ex}\in\operatorname{Map}(G,A)\) is the map defined by \[\tau_{ex}(g)=\left\{\begin{align} & gex & \text{ if }g\in G_{e} \\ & 0 & \text{ else if }g\notin G_{e}. \end{align}\right.\] It is clear that \(\tau_{ex}\in\operatorname{Ind}^{G}_{G_{e}}eA\), and \(i\) is a morphism of Galois \((H,F)\)-algebras. By the universal property of \(\operatorname{Ind}^G_{G_e}\), there is a unique \(G\)-morphism \(\tilde{f}:\operatorname{Ind}^G_{G_e}eA\to A\) such that \(\tilde{f}\circ i=f\). In particular, for each simple tensor \(g\otimes m\), we have \(\tilde{f}(g\otimes m)=\varphi_A(g)\cdot m\). The map \(\tilde{f}\) is automatically \(G\)-equivariant by its definition. And it is surjective, for \(G\cdot eA=A\) (transitivity on idempotents). The injectivity follows from the comparison of dimensions: The algebras \(A\) and \(\operatorname{Ind}^G_{G_e}eA\) both have dimension \(\lvert G\rvert\) over \(F\), so \(\tilde{f}\) has to be injective if it is surjective. So \(\tilde{f}\) gives an isomorphism between \(\operatorname{Ind}^G_{G_e}eA\) and \(A\) as Galois \((G,F)\)-algebras. And we are done for the proof. ◻

Proof. First of all, by the Lemma 2, we know that there exists some \(g\in G\) such that \(\varphi_A(g)(e)=e'\). If \(G_e\) is the stabilizer of \(e\) in \(G\), then clearly the conjugate \(G_{e'}=gG_eg^{-1}\) is the stabilizer of \(e'\). So the maps in the statement are well-defined.

It suffices to verify that \((\varphi_A(g),c_g)\) is really an isomorphism of the Galois \(F\)-algebras. Since \(\varphi_A(g)\) is an element of \(\operatorname{Aut}_F(A)\), it is clear that \(\varphi_A(g)\) restricted to \(eA\) induces an isomorphism of field extensions \(eA\cong e'A\) over \(F\). It is also clear that \(c_g:G_e\to G_{e'}\) is a group isomorphism. Now let \(\sigma\) be an element of \(G_e\), and \(x\in eA\). We have \[\begin{align} \varphi_A(g)\bigl(\varphi_{eA}(\sigma)(x)\bigr)=&\varphi_A(g\sigma)(x) \\ =&\varphi_A(g\sigma g^{-1}g)(x) \\ =&\varphi_{e'A}\bigl(c_g(\sigma)\bigr)\bigl(\varphi_A(g)(x)\bigr). \end{align}\] In short, we’ve shown that \[\varphi_A(g)(\sigma\cdot x)=c_g(\sigma)\cdot\varphi_A(x).\] This shows that \(\varphi_A(g)\) and \(c_g\) are compatible. So the pair \((\varphi_A(g),c_g)\) is indeed an isomorphism of the Galois \(F\)-algebras, and we are done. ◻

Proof. First assume that \((A,\varphi_A)\cong(B,\varphi_B)\) as Galois \((G,F)\)-algebras. By definition, there exists a morphism \(f:A\to B\) of \(F\)-algebras, such that \(f\) is bijective and \(G\)-equivariant. Let \(e\) be any primitive idempotent of \(A\). By the Lemma 1, we know that \(e':=f(e)\) is a primitive idempotent of \(B\). Since \(f\) is \(G\)-equivariant, we see that \(H:=G_e=G_{e'}\). Then we see that \((f\vert_{eA}:eA\to e'B,1_{H}:H\to H)\) is automatically an isomorphism of the Galois \(F\)-algebras.

For the opposite direction, let \((eA,G_e)\) and \((e'B,G_{e'})\) and \((f:eA\to e'B,c_g:G_e\to G_{e'})\) be as in the statement. By the Lemma 3, we see that \[(\varphi_B(g^{-1}),c_{g^{-1}}):(e'B,G_{e'})\to((g^{-1}e')B,G_{g^{-1}e'})\] is an isomorphism of Galois \(F\)-algebras. Note also that \(H:=G_{g^{-1}e}=G_e\). This implies that \((\varphi_B(g^{-1})\circ f,1_{H}):(eA,H)\to(g^{-1}e'B,H)\) is an isomorphism of the Galois \(F\)-algebras. Since the morphism of groups is just the identity map of \(H\), it reduces to an isomorphism of Galois \((H,F)\)-field extensions: \[\varphi_B(g^{-1})\circ f:(eA,\varphi_{eA})\to(g^{-1}e'B,\varphi_{g^{-1}e'B}).\] In particular, the map \(f':=\varphi_B(g^{-1})\circ f\) is \(H\)-equivariant. It induces an isomorphism \[\operatorname{Ind}^G_H f':\operatorname{Ind}^G_H eA\to\operatorname{Ind}^G_H g^{-1}e'B\] of Galois \((G,F)\)-algebras. Because \(\operatorname{Ind}^G_H\) is a functor, and the resulting map \(\operatorname{Ind}^G_H f'\) is a morphism of \(F\)-algebras that is \(G\)-equivariant with an inverse \(\operatorname{Ind}^G_H (f')^{-1}\). In the Theorem 1 we have shown that \(\operatorname{Ind}^G_H eA\cong A\) as Galois \((G,F)\)-algebras, and similarly \(\operatorname{Ind}^G_H g^{-1}e'B\cong B\). Therefore, \(A\) and \(B\) are isomorphic as Galois \((G,F)\)-algebras. ◻

Proof. Given \(g\in G\) such that \(\varphi_2=\varphi_1\circ c_g\). Let \(\sigma:=\varphi_1^{-1}(g)\) and one can check that \(\sigma:(K,\varphi_1)\to(K,\varphi_2)\) is an isomorphism of Galois algebras. Conversely, if \(\sigma\in\operatorname{Gal}(K/F)\) is an isomorphism of Galois algebras, then let \(g:=\varphi_1(\sigma)\) and one can check that \(\varphi_2=\varphi_1\circ c_g\). ◻

Proof. Let \(\bar{F}\) be the algebraic closure of \(F\), and \(\rho:G_F\to G\) be a continuous homomorphism with image \(H\). Then \(K:=\bar{F}^{\ker\rho}\) is a Galois \(H\)-extension over \(F\), where the map \(\varphi_K:H\to\operatorname{Aut}_F(K)\) is induced by \(\rho\). And \(A:=\operatorname{Ind}^G_H K\) is naturally a Galois \((G,F)\)-algebra.

If \(\chi:G_F\to G\) is another continuous homomorphism such that there exists some \(g\in G\) for all \(\sigma\in G_F\) we have \(\chi(\sigma)=g\rho(\sigma)g^{-1}\), then \(K=\bar{F}^{\ker\chi}\) is a Galois \(gHg^{-1}\)-extension over \(F\). This implies that \((K,H)\) and \((K,gHg^{-1})\) are isomorphic as Galois \(F\)-algebras by \((1_K,c_g)\). Let \(B:=\operatorname{Ind}^G_{gHg^{-1}}K\). By the Proposition 4, we see that \(A\) and \(B\) are isomorphic as Galois \((G,F)\)-algebras.

Therefore, we have a well-defined map \[\Phi:\operatorname{Hom}(G_F,G)/\sim\to\{\text{Galois }(G,F)\text{-algebras}\}/\cong,\quad\varphi\mapsto\operatorname{Ind}^G_{\varphi(G_F)}\bar{F}^{\ker\varphi}.\] Let \(A\) be a Galois \((G,F)\)-algebra. By the Theorem 1, if \(e\) is a primitive idempotent, then \(eA\) is a Galois \((G_e,F)\)-field extension. As an algebraic extension over \(F\), we could choose an embedding \(eA\hookrightarrow\bar{F}\). Then we have a surjective continuous homomorphism \(\rho_{eA}:G_F\to\operatorname{Aut}_F(eA)\), and the isomorphism \(\varphi:G_e\to\operatorname{Aut}_F(eA)\) induces a continuous homomorphism \(\rho_A:=\varphi_A^{-1}\circ\rho_{eA}:G_F\to G\) with image \(G_e\). By the Corollary 1, we see that different embeddings \(eA\hookrightarrow\bar{F}\) induces conjugate surjective homomorphisms \(G_F\to\operatorname{Aut}_F(eA)\). Therefore the map \(A\mapsto\rho_A\) is defined up to conjugation. That is, the algebra \(A\) corresponds to a class of homomorphisms.

If \(B\) is another Galois \((G,F)\)-algebra that is isomorphic to \(A\), then by the Proposition 4, we may assume without loss of generality that \(e\), resp. \(e'\), is a primitive idempotent of \(A\), resp. of \(B\), such that \(G_e=G_{e'}\) and \(f:(eA,\varphi_{A})\to(e'B,\varphi_{B})\) is an isomorphism of Galois \((G_e,F)\)-algebras. Given any embedding \(\sigma:e'B\hookrightarrow\bar{F}\), we see that \(\sigma\circ f:eA\hookrightarrow\bar{F}\) is an embedding such that the induced maps \(\rho_A=\varphi_A^{-1}\circ\rho_{eA}\) and \(\rho_B=\varphi_B^{-1}\circ\rho_{eB}\) coincide. This implies that \(A\) and \(B\) correspond to the same class of continuous homomorphisms. So, we have a well-defined map \[\Psi:\{\text{Galois }(G,F)\text{-algebras}\}/\cong\to\operatorname{Hom}(G_F,G)/\sim,\quad(A,\varphi_A)\mapsto\rho_A.\] Then it suffices to check that \(\Phi\) and \(\Psi\) are inverse to each other. Let \(\rho:G_F\to G\) be a continuous homomorphism with image \(H\) and kernel \(G_K\). Up to the equivalence relations, we have \(\Phi(\rho)=\operatorname{Ind}^G_H K\), denoted by \(A:=\Phi(\rho)\). The Galois \((G,F)\)-algebra \(A\) has a component \(K\) which is a Galois \(H\)-extension over \(F\). It is already a subfield of \(\bar{F}\) by its construction, so \(\varphi_K:H\to\operatorname{Aut}_F(K)\) induces the map \(\rho_A:G_F\to G\) by composing with the restriction of Galois actions \(G_F\to\operatorname{Aut}_F(K)\). But this is exactly the original map \(\rho\), for the map \(\varphi_K:H\to\operatorname{Aut}_F(K)\) is induced by \(\rho\). That is, the action of \(H\) on the field extension \(K/F\) is given by \(\rho:G_F\to G\). So, we see that \(\Psi\circ\Phi\) is the identity map.

Now let \((A,\varphi_A)\) be a Galois \((G,F)\)-algebra, \(e\) be a primitive idempotent, and \(\sigma:eA\hookrightarrow\bar{F}\) be a choice of embedding with the induced surjective map \(\rho_{eA}:G_F\to\operatorname{Aut}_F(eA)\). By definition, \(\rho_A=\Psi(A)=\varphi_A^{-1}\circ\rho_{eA}\). Clearly, \(K:=\bar{F}^{\ker\rho_A}=\sigma(eA)\), and \(K\) is a Galois \((G_e,F)\)-field by the map \(\rho_A:G_F\to G\). So, we see that \(\sigma:eA\to K\) is an isomorphism of Galois \(G_e\)-field extensions over \(F\), and \(\Phi(\rho_A)=\operatorname{Ind}^G_{G_e}K\) is isomorphic to \(A\) as Galois \((G,F)\)-algebras by the Proposition 4. This implies that \(\Phi\circ\Psi\) is also the identity (of the opposite direction). And we are done. ◻

Proof. The case when \(\mathfrak{p}\nmid\lvert G\rvert\infty\) follows from the fact that if \(L_{\mathfrak{p}}\) and \(\rho_{\mathfrak{p}}\) correspond to each other, then the inertia generator \(y_{\mathfrak{p}}\) of \(L_{\mathfrak{p}}\) is exactly the image of \(y_{\mathfrak{p}}^t\) (up to conjugate).

For each \(\mathfrak{p}\mid\lvert G\rvert\infty\) of \(K\), we could define \[C'_{\mathfrak{p}}(\rho_{\mathfrak{p}}):=C_{\mathfrak{p}}(L_{\mathfrak{p}})\] whenever \(\rho_{\mathfrak{p}}\) is the continuous homomorphism corresponding to \(L_{\mathfrak{p}}\). If a Galois \((G,K_{\mathfrak{p}})\)-algebra \(\Sigma_{\mathfrak{p}}\) (up to isomorphism) never shows up as a local specification of the global field extension \(L/K\), then we could simply assign the value \(1\) to it. That is, if the isomorphism class \(\Sigma_{\mathfrak{p}}\) of the Galois \((G,K_{\mathfrak{p}})\)-algebras corresponds to \(\rho_{\mathfrak{p}}\), and if for each Galois \(G\)-extension \(L/K\) the specification \(L_{\mathfrak{p}}\) at \(\mathfrak{p}\) is not isomorphic to \(\Sigma_{\mathfrak{p}}\), then define \[C'_{\mathfrak{p}}(\rho_{\mathfrak{p}})=1.\] So, by combining these two cases, for each \(\mathfrak{p}\) of \(K\), we have obtained a local generalized discriminant \(C'_{\mathfrak{p}}\) for the equivalence classes of the continuous homomorphisms \(\operatorname{Hom}(G_{K_{\mathfrak{p}}},G)\) such that \[C_{\mathfrak{p}}(L_{\mathfrak{p}})=C'_{\mathfrak{p}}(\rho_{\mathfrak{p}}),\] whenever \(L_{\mathfrak{p}}\) and \(\rho_{\mathfrak{p}}\) corresponds to each other. ◻

Proof. It suffices to show that \(\#\mathcal{C}(\Sigma)\leq\lvert\operatorname{Hom}(\operatorname{Cl}^{S}_{k},G)\rvert\). Assume without loss of generality that \(K/k\) is a solution to \(\mathcal{C}^{\Sigma}\). The statement follows from the short exact sequence \[1\to J_k^{S}/k^{S}\to\operatorname{C}_k\to\operatorname{Cl}_k^{S}\to1.\] Apply the functor \(\operatorname{Hom}(\mathrm{-},G)\) and we have a long exact sequence \[1\to\operatorname{Hom}(\operatorname{Cl}_k^{S},G)\to\operatorname{Hom}(\operatorname{C}_k,G)\to\operatorname{Hom}(J_k^{S}/k^{S},G)\to\cdots.\] Since \(K/k\) is clearly a solution to \((G,\Sigma)\), let \(\chi_K:\operatorname{C}_k\to G\) be the surjective continuous homomorphism that corresponds to \(K\). Denote the restriction of \(\chi_K\) to \(J_k^S/k^S\) by \(\chi_K^S\).

Claim: For each solution \(L\in\mathcal{C}(\Sigma)\), the restriction of \(\chi_L\) to \(J_k^S/k^S\) is the same as \(\chi_K^S\). By Wood [9], there is a one-to-one correspondence between the set of the isomorphism classes of \(G\)-structured \(k_{\mathfrak{p}}\)-algebras and \(\operatorname{Hom}(k_{\mathfrak{p}}^*,G)\). For each \(\mathfrak{p}\in S\), the induced local map \(\chi_{L,\mathfrak{p}}:k_{\mathfrak{p}}^*\to G\) corresponds to \(K_{\mathfrak{p}}\), so it coincides with \(\chi_{K,\mathfrak{p}}\). And for each \(\mathfrak{p}\notin S\), the algebra \(L_{\mathfrak{p}}/k_{\mathfrak{p}}\) is unramified, which implies that the induced map \(\chi_{L,\mathfrak{p}}:U_{\mathfrak{p}}\to G\) must be trivial. Since we can represent the map \(\chi_K^S\) by a continuous homomorphism \(\prod_{\mathfrak{p}}\chi_{K,\mathfrak{p}}:J_k^S\to G\) that is trivial on \(k^S\), that is, the map \(\chi_K^S\) is determined by the local data. We see that \(\chi_L^S\) must be the same as \(\chi_K^S\).

The claim shows that \(\mathcal{C}(\Sigma)\) is a subset of the preimage of \(\chi_{K}^S\). The size of the preimage is \(\lvert\operatorname{Hom}(\operatorname{Cl}_k^S,G)\rvert\) by the exactness at \(\operatorname{Hom}(\operatorname{C}_k,G)\). Therefore, we have the estimate \[\#\mathcal{C}(\Sigma)\leq\lvert\operatorname{Hom}(\operatorname{Cl}^{S}_{k},G)\rvert.\] ◻

Proof. (1): Let \(\rho:G_{k}\to G\) be the surjective group homomorphism that corresponds to \(K\). Clearly, the \(G\)-specification \(K_{\mathfrak{p}}\) at \(\mathfrak{p}\) corresponds to the continuous map \(\rho_{\mathfrak{p}}:G_{k_{\mathfrak{p}}}\to G\). Since \(N\) is a normal subgroup, we have induced maps \[\bar{\rho}:G_{k}\to G/N\quad\text{and}\quad\bar{\rho}_{\mathfrak{p}}:G_{k_{\mathfrak{p}}}\to G/N.\] Clearly the map \(\bar{\rho}\) corresponds to the Galois \(G/N\)-extension \(K^N/k\). So the map \(\bar{\rho}_{\mathfrak{p}}\) corresponds to the \(G\)-specification \((K^{N})_{\mathfrak{p}}\) by the Proposition 5. If \(L\) is a Galois \(G\)-extension over \(k\) with the corresponding continuous homomorphism \(\chi:G_{k}\to G\) such that \(L_{\mathfrak{p}}\cong K_{\mathfrak{p}}\), then by the Proposition 5 we know that \(\chi_{\mathfrak{p}}\) is conjugate to \(\rho_{\mathfrak{p}}\) in the sense that there exists some \(g\in G\) such that for all \(\sigma\in G_{k_{\mathfrak{p}}}\) we have \[\chi_{\mathfrak{p}}(\sigma)=g\rho_{\mathfrak{p}}(\sigma)g^{-1}.\] This implies immediately that for each \(\sigma\in G_{k_{\mathfrak{p}}}\), we have \[\bar{\chi}_{\mathfrak{p}}(\sigma)=\bar{g}\bar{\rho}_{\mathfrak{p}}(\sigma)\bar{g}^{-1}\] where \(\bar{g}\) is the image of \(g\) in \(G/N\). By the Proposition 5, we know that \((L^{N})_{\mathfrak{p}}\cong(K^{N})_{\mathfrak{p}}\) as Galois \((G/N,k_{\mathfrak{p}})\)-algebras.

(2): This follows easily from the structure of \(K_{v}\), which is isomorphic to \[\prod_{w\mid v}K_{w}.\] And \(K_{v}\cong\prod_{w\mid v}K_{w}\).

(3): This follows from the fact that the discriminant could be computed locally, that is, \[\mathfrak{Nd}_{K/k}=\prod_{\mathfrak{p}\in\mathcal{P}_k}\prod_{\mathfrak{P}\mid\mathfrak{p}}\mathfrak{Nd}_{K_{\mathfrak{P}}/k_{\mathfrak{p}}}.\] For each \(\mathfrak{p}\in S\), the data of the local specification \(K_{\mathfrak{p}}/k_{\mathfrak{p}}\) determines the set of all the places \(\mathfrak{P}\) above \(\mathfrak{p}\) with the corresponding the local extensions \(K_{\mathfrak{P}}/k_{\mathfrak{p}}\). Therefore the local relative discriminant are fixed, hence the relative discriminant. ◻

Proof. (1): For finite cyclic \(p\)-groups, we see that \[\operatorname{Hom}(\mathbb{Z}/p^a\mathbb{Z},\mathbb{Z}/p^b\mathbb{Z})\cong\mathbb{Z}/p^{\min\{a,b\}}\mathbb{Z}.\] In general, write \(A\cong\prod_{i=1}^{\operatorname{rk}_p A}\mathbb{Z}/p^{a_i}\mathbb{Z}\) and \(B\cong\prod_{i=1}^{\operatorname{rk}_p B}\mathbb{Z}/p^{b_i}\mathbb{Z}\), we have \[\operatorname{Hom}(A,B)\cong\prod_{i=1}^{\operatorname{rk}_p A}\prod_{j=1}^{\operatorname{rk}_p B}\mathbb{Z}/p^{\min\{a_i,b_j\}}\mathbb{Z}.\] This implies that \[\lvert\operatorname{Hom}(A,B)\rvert\leq\prod_{j=1}^{\operatorname{rk}_p B}\prod_{i=1}^{\operatorname{rk}_p A}p^{a_i}=\lvert A\rvert^{\operatorname{rk}_p B}.\] And we are done for (1).

(2): By (1), we have \[\begin{align} \lvert\operatorname{Hom}(\operatorname{Cl}_K,A)\rvert=&\prod_{p}\lvert\operatorname{Hom}(\operatorname{Cl}_K[p^\infty],A[p^\infty])\rvert \\ \leq&\prod_p\lvert\operatorname{Cl}_K[p^\infty]\rvert^{\operatorname{rk}_p A} \leq\lvert\operatorname{Cl}_K\rvert^{r_A}, \end{align}\] where \(r_A=\sup_p\operatorname{rk}_pA\). By Minkowski bound of the size of the class group, for each \(\varepsilon>0\), there exists a constant \(c=c(n,\varepsilon)\) such that \[\lvert\operatorname{Cl}_K\rvert^{r_A}\leq c d_K^{\frac{1+\varepsilon}{2}r_A},\] where \(d_K\) is the absolute discriminant of \(K/\mathbb{Q}\). And this shows the existence of \(c\) and \(N\). ◻

Proof. First of all, the expression (2 ) is well-defined in the sense that it is independent of the choice of \(K\in\mathcal{C}(\Sigma)\). Because for each \(K'\in\mathcal{C}(\Sigma)\), the extension \(K'/k\) is unramified outside \(S(k)\), and its local specification at \(S(k)\) is given by \(\Sigma\). So, we have \(d_{K'}=d_{K}\). Moreover, Lemma 5(2) shows that \[\mathfrak{Nd}_{K/K_i}=\mathfrak{Nd}_{K'/K'_i},\] because as \(G_i\)-extensions, they have exactly the same local specification at \(S(K_i)\). Since \(d_{K}=\mathfrak{Nd}_{K/K_i}\cdot d_{K_i}^{\lvert G_i\rvert}\), we see that \(d_{K_i}=d_{K'_i}\). And the formula does not depend on the choice of \(K\in\mathcal{C}(\Sigma)\).

We prove the statement by induction on \(n\), the length of the tower of subgroups of the solvable group \(G\). If \(G\) is abelian, then this reduces to the case of abelian extensions, and the upper bound simply comes from the Minkowski bound for the size of the class group. In other words, the statement is true when \(n=1\).

Now assume that the statement is true for \(1,\dots,n-1\), where \(n\geq2\). The subgroup \(G_1\) is normal in \(G\), and it is also solvable with the tower \[\{e_G\}=G_{l}\triangleleft G_{l-1}\triangleleft\cdots\triangleleft G_1.\] By Lemma 5, for each \(K\in\mathcal{C}(\Sigma)\), the field extension \(K_1/k\) has the same local specification at \(S(k)\), and \(K/K_1\) has the same local specification at \(S(K_1)\). So, we could count the number \(\#\mathcal{C}(\Sigma)\) by the following method. Define \(\mathcal{C}(G/G_1,\Sigma)\) to be the set of solutions to the \(G/G_1\)-extensions \(K_1'/k\) unramified outside \(S(k)\) such that \(K'_{1,\mathfrak{p}}\cong K_{1,\mathfrak{p}}\) for each \(\mathfrak{p}\in S(k)\). And for each \(K'_1\in\mathcal{C}(G/G_1,\Sigma)\), define \(\mathcal{C}(G_1,K'_1,\Sigma)\) to be the set of solutions to the \(G_1\)-extensions \(K'/K'_1\) unramified outside \(S(K_1')\) such that \(K'_{\mathfrak{P}}\cong_{G_1} K_{\mathfrak{P}}\) for each \(\mathfrak{P}\in S(K_1')\). Consider the sum \[\sum_{K'_1\in\mathcal{C}(G/G_1,\Sigma)}\#\mathcal{C}(G_1,K_1',\Sigma).\] By induction assumption, we see that \(\#\mathcal{C}(G_1,K_1',\Sigma)\leq\prod_{i=1}^{n-1}c_id_{K'_i}^{N_i}\). It is independent of the choice of \(K'_1\) in the sense that \(d_{K'_i}=d_{K_i}\) for all \(K'\in\mathcal{C}({\Sigma})\). The number of \(K'_1\in\mathcal{C}(G/G_1,{\Sigma})\) is reduced to the number of abelian extensions over \(k\) unramified outside \(S\) with a fixed specification at \(S(k)\), the case when \(n=1\). So, we have \[\#\mathcal{C}({\Sigma})\leq c_0d_{K_0}^{N_0}\prod_{i=1}^{n-1}c_id_{K_i}^{N_i}=\prod_{i=0}^{n-1}c_id_{K_i}^{N_i}.\] ◻

Proof. Let \(N\) be any Sylow \(p\)-subgroup. The image of \(N\) in \(G/G_{l-1}\) must be a \(p\)-subgroup of the quotient group. But according to our condition, the size of \(G/G_{l-1}\) is coprime to \(p\). This shows that \(N\subseteq G_{l-1}\). Since \(G_{l-1}\) is a subgroup of \(G\), this means that \(N\) must be a Sylow \(p\)-subgroup of \(G_{l-1}\). But according to the tower of \(G\), the subgroup \(G_{l-1}\) is abelian. In other words, we have shown that any Sylow \(p\)-subgroup of \(G\) must be included in \(G_{l-1}\) and equal to the unique Sylow \(p\)-subgroup \(N\) of \(G_{l-1}\). The rest of the statement just follows, and we are done for the proof. ◻

Proof. Note that \(G\) has the property that \(\operatorname{Stab}_{G}(1)\) is trivial. So a \(G\)-extension \(K/k\) is Galois with Galois group isomorphic to \(G\), and \(\mathcal{C}(G,k)\) is the set of Galois \(G\)-extensions. Then the statement is just a translation of the definition, so we omit the details. ◻

Proof. The proof of (1) and (2) follows the similar method, and (1) is on the field-counting while (2) is counting a specific subfamily. But first, the condition that \(\Omega\subseteq G_{l-1}\) puts more restrictions on the structure of \(G\), that is, \(\Omega\) generates an abelian normal subgroup of \(G\) included in \(G_{l-1}\). So, we may assume without loss of generality that \(G_{l-1}=\langle\Omega\rangle\).

As in the Theorem 7, for each \(0\leq i\leq l-1\), let \(c_i\) and \(N_i\) be the constants from Lemma 6 depending only on \(n_{i}:=\lvert G/G_i\rvert\) (the base field \(k\) is \(\mathbb{Q}\) in this case) and \(G_{i}/G_{i+1}\) such that for each number field \([L:\mathbb{Q}]=n_{i}\), we have \[\lvert\operatorname{Hom}(\operatorname{Cl}_{L},G_{i}/G_{i+1})\rvert\leq c_{i}d_{L}^{N_{i}}.\] Note that \(d_{K_{i}}\mid d_{K_{i+1}}\) for each integer \(0\leq i\leq l-1\). Let \(c=\prod_{i=0}^{l-1} c_{i}\) and \(N\) be a constant such that \[\label{eqn:32moment32for32solvable32extensions321} \prod_{i=0}^{l-1} c_id_{K_i}^{N_i}\leq c(\sqrt{d_{K_{l-1}}})^N,\tag{4}\] where \(\sqrt{d_{K}}\) is the radical of the absolute discriminant. In particular, for a subset \(S\) of \(\mathcal{P}\) containing the ones at \(\lvert G\rvert\infty\), for each \(G\)-specification \(\Sigma\) at \(S\), we have \[\#\mathcal{C}(\Sigma)\leq c(\sqrt{d_{\Sigma_{l-1}}})^{N},\] where \(d_{\Sigma_{l-1}}=\prod_{p}d_{\Sigma_{l-1,p}}\), the product of the local discriminant. Let \(M:=\max_{g\in G}c_{G}(g)\), and \(a\) be a positive integer such that for each \(p\in\mathcal{P}\), we have \[\#\{\Sigma_{\mathfrak{p}}\text{ is a }G\text{-specification at }p\}\leq a.\] Claim: under the condition that for all \(g\in G\) if \(c_{G}(g)>m\) then \(c_{G}(g)-N>m\), the series \[D(s):=c\prod_{p}(1+ap^{-ms}+ap^{N}\sum_{i=N+m+1}^{M}p^{-is})=\sum_{n=1}^{\infty}a_{n}n^{-s}\] satisfies the property that for each \(X>0\), we have \[N_{\mathcal{C},C}(X)\leq\sum_{n<X}a_{n}.\] If the Claim is true, then by the author [16], we know immediately that \[N_{\mathcal{C},C}(X)\leq\sum_{n<X}a_{n}\ll X^{1/m}(\log X)^{a-1}.\] Let us prove the Claim by comparing the coefficient directly. Let \(d\) be square-free, and \(f\) be a positive integer such that if a prime \(p\mid f\) then \(p^{N+m+1}\mid f\). Clearly for each \(K/\mathbb{Q}\in\mathcal{C}\), its corresponding value \(C(K)\) under the counting function is of the form \(C(K)=d^{m}f\). This shows that if \(a_{n}=0\), then \(\#\mathcal{C}(n)=0\), because every positive integer \(n\) of the form \(df\) corresponds to a nonzero coefficient \(a_{n}\). Conversely, if \(a_{n}>0\), then there exists some \(K/\mathbb{Q}\) such that \(C(K)=n\). In this case, we see that if \(p\mid d_{K_{n-1}}\), then \(p^{N+m+1}\mid C(K)\), hence \(\sqrt{d_{K_{n-1}}}\mid\sqrt{f}\) where \(\sqrt{}\) means the radical. This implies that \[\begin{align} a_{n}=&a^{\omega(n)}\cdot c(\sqrt{f})^{N} \\ \geq &\sum_{\Sigma:C(\Sigma)=n}c(\sqrt{f})^{N} \\ \geq &\sum_{\Sigma:C(\Sigma)=n}c(\sqrt{d_{\Sigma_{n-1}}})^{N} \\ \geq &\sum_{\Sigma:C(\Sigma)=n}\#\mathcal{C}(\Sigma), \end{align}\] where \(\omega(n)\) is the number of distinct prime factors of \(n\). So, for each \(n\in\mathbb{Z}_{+}\), we have that \[a_{n}\geq\#\mathcal{C}(n),\] hence the claim. For general case when \(c_{G}(g)-N\leq m\) if \(g\notin\Omega\cup\{e_{G}\}\), let us adopt the similar method. Let \(r\) be a large enough positive integer such that \(rc_{G}(g)-N>rm\) for all \(g\notin\Omega\cup\{e_{G}\}\). Then the generalized discriminant \(C^{r}(K)=C(K)^{r}\) satisfies the condition the Claim. So, there exists some positive integer \(a\) such that \[\begin{align} N_{\mathcal{C},C^{r}}(X)\ll& X^{\frac{1}{mr}}(\log X)^{a-1} \\ \Rightarrow N_{\mathcal{C},C}(X)=N_{\mathcal{C},C^{r}}(X^{r})\ll&X^{r\frac{1}{mr}}(\log X^{r})^{a-1}\ll X^{1/m}(\log X)^{a-1}. \end{align}\] Finally we move on to the proof of (3), which is the estimate for \(N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)\). Let \(T:=T_{\mathbb{Q}}(\lvert G\rvert)\), that is, \(T=\{p\in\mathcal{P}: p\nmid\lvert G\rvert\infty\}\).

Claim: under the condition that for all \(g\in G\) if \(c_{G}(g)>m\) then \(c_{G}(g)-N>m\), the series \[\begin{align} E_{\gamma}(s):=& c\Bigl(\sum_{\Sigma_{\infty}}1\Bigr) \prod_{p\mid\lvert G\rvert}\Bigl(\sum_{\Sigma_{p}}C_{p}(\Sigma_{p})^{-s}\Bigr) \prod_{p\in T}(1+ap^{N}\sum_{i=N+m+1}^{M}p^{-is}) \Bigl(\sum_{ \substack{ d\in I_{T}^{+,\mu} \\ \omega(d)=\gamma } }a^{\gamma}d^{-ms}\Bigr) \\ =& \sum_{n=1}^{\infty}b_{\gamma,n}n^{-s} \end{align}\] satisfies the property that \[N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)\leq\sum_{n<X}b_{\gamma,n}.\] If the Claim is true, then by combining the author [16], when \(\gamma\geq1\), we have \[N_{\mathcal{C}_{\Omega}^{0},C}(X)\ll X^{1/(m'-N)}\quad\text{and}\quad N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)\ll\frac{X^{1/m}}{\log X}(\log\log X)^{\gamma-1}.\] Then we use the same trick as above for the general case and also the statement for \(\gamma=0\). To be precise, for each \(0<\epsilon<1\), let \(r\) be a large enough integer such that \(rc_{G}(g)-N>rm\) and \(N/r<\epsilon\). Then for \(\gamma\geq1\), we have \[\begin{align} N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)=& N_{\mathcal{C}_{\Omega}^{\gamma},C^{r}}(X^{r}) \\ \ll & \frac{X^{r/rm}}{\log X}(\log\log X)^{\gamma-1} \\ =& \frac{X^{1/m}}{\log X}(\log\log X)^{\gamma-1}. \end{align}\] And for \(\gamma=0\), we have \[N_{\mathcal{C}_{\Omega}^{0},C}(X)=N_{\mathcal{C}_{\Omega}^{0},C^{r}}(X^{r})\ll X^{\frac{r}{rm'-N}}=X^{\frac{1}{m'-N/r}}\leq X^{\frac{1}{m'-\epsilon}}.\] It then suffices to show the Claim. We consider the similar estimation. For each positive integer \(n\), the set \(\mathcal{C}_{\Omega}^{\gamma}(n)\) is nonempty only if \(\#\{p\in T:p^{m}\| n\}=\gamma\) and for each \(p\mid n\) we have \(p^{m}\mid n\). Let us write \(\prod_{p\in T:p^{m}\| n}p=d\). Under such condition, we have \[\begin{align} \#\mathcal{C}_{\Omega}^{\gamma}(n)\leq &\sum_{\Sigma:C(\Sigma)=n} c(\sqrt{d_{\Sigma_{l-1}}})^{N} \\ \leq & \sum_{\Sigma:C(\Sigma)=n}c(\sqrt{n/d^{m}})^{N} \\ \leq & c(\sqrt{n/d^{m}})^{N}\#\{\Sigma_{\infty}\}\prod_{p\mid\lvert G\rvert,p\mid n}\#\{\Sigma_{p}: C(\Sigma_{p})\|n\}\prod_{p\in T,p\mid n}\#\{\Sigma_{p}\} \\ \leq & c(\sqrt{n/d^{m}})^{N}\#\{\Sigma_{\infty}\}\prod_{p\mid\lvert G\rvert,p\mid n}\#\{\Sigma_{p}: C(\Sigma_{p})\|n\}\prod_{p\in T,p\mid n}a \\ =& b_{\gamma,n}. \end{align}\] So the Claim is true, hence the proof of (2). ◻

Proof. By the author [16], for each \(\gamma\in\mathbb{Z}_{+}\), we have \[N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)\gg \frac{{X}^{1/m}}{\log X}(\log\log X)^{\gamma-1}.\] Apply the Proposition 9 here, we immediately obtain the desired result. ◻

Proof. This is a direct corollary of [16]. For each rational prime \(q\), let \(e(q,L/\mathbb{Q})\) be its ramification index in the Galois extension \(L/\mathbb{Q}\in\mathcal{C}\). According to the author [16], we have \[\begin{align} \operatorname{rk}_{p}\operatorname{Cl}_{L}\geq&\#\{q\in\mathcal{P}\mid e(q,L/\mathbb{Q})\equiv 0 \bmod{p}\}-[L:\mathbb{Q}]^{2} \\ \geq&\#\{q\nmid\lvert G\rvert\infty:\rho_{q}(y_{q}^{t})\in\Omega\}-[L:\mathbb{Q}]^{2}. \end{align}\] where \(\rho_{q}:G_{\mathbb{Q}_{q}}\to G\) is the local map corresponding to \(L_{p}\). Let \(T:=T_{\mathbb{Q}}(\lvert G\rvert)=\{p\in\mathcal{P}:p\nmid\lvert G\rvert\infty\}\). We then see that \(((\mathcal{C},C),\Omega,T)\) satisfies the condition of [16]. That is, there exists some constant \(c\geq0\), for each \(L\in\mathcal{C}\), we have \[\operatorname{rk}_{p}\operatorname{Cl}_{L}\geq\#\{q\in T:\rho_{q}(y_{q}^{t})\in\Omega\}-c,\] and for each non-negative integer \(\gamma\), we have that \[N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)=o(N_{\mathcal{C}_{\Omega}^{\gamma+1},C}(X)).\] Then [16] implies immediately the zero-probability and the infinite \(\mathbb{Z}/p\mathbb{Z}\)-moment. ◻

Proof. By Neukirch [17], if \(\mathfrak{p}\) is a finite prime of \(K\) that is tamely ramified in \(K/\mathbb{Q}\), then we have \[\mathfrak{D}_{\mathfrak{p}}=\mathfrak{D}(K_{\mathfrak{p}}/\mathbb{Q}_{p})=\mathfrak{p}^{e-1},\] where \(\mathfrak{D}\) is the different and \(e\) is the ramification index of \(\mathfrak{p}\) which is the same as the ramification index of \(p\). The basic relation between the different and the discriminant is given by \[d_{\mathfrak{p}}=d(K_{\mathfrak{p}}/\mathbb{Q}_{p})=\mathfrak{ND}_{\mathfrak{p}},\] where \(\mathfrak{N}=\operatorname{Nm}_{K/\mathbb{Q}}\). See Neukirch [17] for example. Therefore, if \(p\) has the splitting \(p\mathcal{O}_{K}=(\mathfrak{p}_{1}\cdots\mathfrak{p}_{l})^{e}\) in \(K/\mathbb{Q}\), then we have \[\begin{align} d(K_{p}/\mathbb{Q}_{p})=&\prod_{i=1}^{l}d_{\mathfrak{p}_{i}}\\ =&\prod_{i=1}^{l}\mathfrak{ND}_{\mathfrak{p}_{i}}(K/\mathbb{Q})\\ =&(p^{f})^{(e-1)l}. \end{align}\] Recall the fundamental identity: \(efl=\lvert\Gamma\rvert\). So, we have \[fl(e-1)=\lvert\Gamma\rvert(1-e^{-1}),\] hence the statement. ◻

Proof. It suffices to check that the absolute discriminant \(d\) satisfies the condition of this section (15). Since \(p\) is the smallest prime that divides \(\lvert G\rvert\), we see that if \(g\in N\) has order \(p\), then \[c_{G}(g)=\lvert G\rvert(1-p^{-1}),\] where \(c_{G}\) is the weight that corresponds to the absolute discriminant. Clearly, for all nontrivial elements \(g\in G\), we have \[c_{G}(g)=\lvert G\rvert(1-r_{g}^{-1})\geq\lvert G\rvert(1-p^{-1}).\] So it is true that \(m=\lvert G\rvert(1-p^{-1})\), and only the elements of order \(p\) could reach this minimum \(m\). According to our assumption on the group structure, this means that \(\Omega=c_{G}^{-1}(m)\subseteq N\). So, in this case, all the conditions of the Theorem 10 and 11 are satisfied. And we are done. ◻

Proof. This just follows from the computation. Fix a field \(L\in\mathcal{C}\) with \(K=L^{N}\). For each prime \(p\), let \(\rho_{p}:G_{\mathbb{Q}_{p}}\to G\) be the continuous map corresponding to \(L_{p}\). The quotient map \(G\to H\) naturally induces a continuous map \(\bar{\rho}_{p}:G_{\mathbb{Q}_{p}}\to H\) that corresponds to \(K_{p}\). For a prime \(p\nmid\lvert G\rvert\infty\), we have that \(p\mid C(L)\iff \rho_{p}(y_{p}^{t})\) is nontrivial. Therefore, either \(\bar{\rho}_{p}(y_{p}^{t})\) is nontrivial, or \(\rho_{p}(y_{p}^{t})\in N\backslash\{e_{G}\}\). Let \(\chi:\operatorname{C}_{K}\to N\) be the surjective continuous \(H\)-morphism that corresponds to \(L/K\). Using these basic observations, we have \[\begin{align} C(L)=& \prod_{p\mid\lvert G\rvert}d_{L_{p}}\cdot\prod_{p\nmid\lvert G\rvert\infty}C_{p}(L_{p}) \\ =& \prod_{p\mid\lvert G\rvert}d_{K_{p}}^{[L:K]}\prod_{\mathfrak{p}\mid p}\mathfrak{Nd}_{L_{\mathfrak{p}}/K_{\mathfrak{p}}} \cdot\prod_{p\nmid\lvert G\rvert\infty}p^{c_{G}(\rho_{p}(y_{p}^{t}))} \\ =& \prod_{p\mid\lvert G\rvert}d_{K_{p}}^{[L:K]}\prod_{\mathfrak{p}\mid p}\mathfrak{Nd}_{L_{\mathfrak{p}}/K_{\mathfrak{p}}} \cdot\prod_{ \substack{ p\nmid\lvert G\rvert\infty \\ \bar{\rho}_{p}(y_{p}^{t})\neq e_{H} } }p \cdot\prod_{ \substack{ p\nmid\lvert G\rvert\infty \\ \rho_{p}(y_{p}^{t})\in N\backslash\{e_{G}\} } }p \\ =& \Bigl(\prod_{p\mid\lvert G\rvert}d_{K_{p}}^{[L:K]}\prod_{p\nmid\lvert G\rvert\infty}\bar{C}_{p}(\bar{\rho}_{p})\Bigr) \cdot \Bigl(\prod_{\mathfrak{p}\mid\lvert G\rvert}\mathfrak{Nd}_{L_{\mathfrak{p}}/K_{\mathfrak{p}}}\prod_{\mathfrak{p}\nmid\lvert G\rvert\infty}C_{\mathfrak{p}}(\chi_{\mathfrak{p}})\Bigr) \\ =& \bar{C}(K)\cdot C'(L/K) \end{align}\] ◻

Proof. Let \(a\) be a positive integer such that for each rational prime \(p\) we have \[\#\{\Sigma_{p}\text{ is a }G-\text{specification at }p\}\leq a.\] Define \(T:=T_{\mathbb{Q}}(\lvert G\rvert d_{K})\) and recall \(I_{T}^{+,\mu}\) from Definition 14. For each \(\gamma>0\), define \[D_{\gamma}(s):= a^{\omega(\bar{C}(K))} \prod_{\mathfrak{p}\mid\lvert G\rvert\infty}\Bigl(\sum_{\chi_{\mathfrak{p}}:K_{\mathfrak{p}}^{*}\to N}\mathfrak{Nd}_{\chi_{\mathfrak{p}}}^{-s}\Bigr) \cdot\sum_{ \substack{ d\in I_{T}^{+,\mu}\\ \omega(d)=\gamma } }\prod_{p\mid d}ap^{-s}=\sum_{n=1}^{\infty}a_{\gamma,n}n^{-s}.\] Recall that \(C(L)=\bar{C}(K)C'(L/K)\). This implies that for each positive \(n\), the set \(\{L\in\mathcal{C}_{\Omega}^{\gamma}\mid K\subseteq L,\, C'(L/K)=n\}\) is nonempty only if \(n\) satisfies the following two conditions simultaneously:

1. for each \(p\in T\), if \(p\mid\bar{C}(K)\) then \(p\nmid n\);

2. there exists \(d\in I_{T}^{+,\mu}\) with \(\omega(d)=\gamma\) such that \[d\mid n.\]

Let \(n\) be an integer with the above property. Apply the Lemma 4 for the \(H\)-morphisms \(\operatorname{C}_{K}\to N\), we have \[\begin{align} & \#\{L\in\mathcal{C}_{\Omega}^{\gamma}\mid K\subseteq L,\, C'(L/K)=n\}\\ \leq & \lvert\operatorname{Hom}_{H}(\operatorname{Cl}_{K},N)\rvert \cdot\prod_{\mathfrak{p}\mid\gcd(n,\lvert G\rvert)}\lvert\operatorname{Hom}(K_{\mathfrak{p}}^{*},N)\rvert \\ &\cdot\prod_{p\mid n\bar{C}(K),p\in T}\#\{\Sigma_{p}\text{ is a }G\text{-specification at }p\}\\ \leq & \lvert\operatorname{Hom}_{H}(\operatorname{Cl}_{K},N)\rvert a_{\gamma,n}. \end{align}\] By the author [16], there exists some constant \(c_{\gamma}\) such that for all \(X>0\), we have \[\sum_{n<X}a_{\gamma,n}\leq c_{\gamma}\max\{1,\frac{X}{\log X}(\log\log X)^{\gamma-1}\}.\] So, for each \(K\in\mathcal{D}\), we have \[\begin{align} & \#\{L\in\mathcal{C}_{\Omega}^{\gamma}\mid K\subseteq L,\, C'(L/K)<X\} \\ \leq & \lvert\operatorname{Hom}_{H}(\operatorname{Cl}_{K},N)\rvert\sum_{n<X}a_{\gamma,n} \\ \leq & c_{\gamma}\lvert\operatorname{Hom}_{H}(\operatorname{Cl}_{K},N)\rvert\max\{1,\frac{X}{\log X}(\log\log X)^{\gamma-1}\}. \end{align}\] ◻

Proof. We are going to do the computation for \(N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)\). Recall that there exists some non-negative integer \(b\) such that \[h(X):=\sum_{ \substack{ K\in\mathcal{D}\\ \bar{C}(K)<X } }\lvert\operatorname{Hom}_{H}(\operatorname{Cl}_{K},N)\rvert\ll X(\log X)^{b}\] as \(X\to\infty\). For simplicity, define \[f_{\gamma}(x):=\left\{\begin{align} &1& &\text{ if }& x\in(0,e] \\ &\max\{1,\frac{x}{\log x}(\log\log x)^{\gamma-1}\}& &\text{ else if }& x>e, \end{align}\right.\] and \[g(x):=\left\{\begin{align} &1& &\text{ if }& x\in(0,1] \\ &\max\{1,x(\log x)^{b}\}& &\text{ else if }& x>1. \end{align}\right.\] We first have the following estimate \[\begin{align} N_{\mathcal{C}_{\Omega}^{\gamma},C}(X) =&\sum_{ \substack{ K\in\mathcal{D}\\ \bar{C}(K)<X } }\#\{L\in\mathcal{C}_{\Omega}^{\gamma}\mid K\subseteq L,\, C'(L/K)<\frac{X}{\bar{C}(K)}\} \\ \leq & \sum_{ \substack{ K\in\mathcal{D}\\ \bar{C}(K)<X } }c_{\gamma}\lvert\operatorname{Hom}_{H}(\operatorname{Cl}_{K},N)\rvert f_{\gamma}(X/\bar{C}(K)) \\ =&c_{\gamma}\int_{1}^{X}f_{\gamma}(X/t)\operatorname{d}h(t) \end{align}\] where the integral of the last row is the Riemann-Stieltjes integral. It suffices to consider the integral. Since \(f_{\gamma}(x)\) is a continuous function, and \(h(x)\) is a step function, we know that the integral-by-parts holds in this case. To be precise, by Hugh L. Montgomery and Robert C. Vaughan [19], we first know that \(\int_{1}^{X}f_{\gamma}(X/t)\operatorname{d}h(t)\) exists, hence \[\begin{align} I:=\int_{1}^{X}f_{\gamma}(X/t)\operatorname{d}h(t) =& f_{\gamma}(X/t)h(t)\Big\vert_{1}^{X}-\int_{1}^{X}h(t)\operatorname{d}f_{\gamma}(X/t) \\ (u=X/t)\Rightarrow=& h(X)-f_{\gamma}(X)h(1)+\int_{1}^{X}h(X/u)\operatorname{d}f_{\gamma}(u) \end{align}\] Note that for all positive integer \(\gamma\), the function \(f_{\gamma}(x)\) is increasing when \(x\) is large enough. So, as \(X\to\infty\), we have \[\begin{align} I\ll & g(X)+\int_{1}^{X}g(X/u)\operatorname{d}f_{\gamma}(u) \\ \leq & g(X)+\int_{1}^{X}\frac{X}{u}(\log X)^{b}\operatorname{d}f_{\gamma}(u) \\ = & X(\log X)^{b}+\frac{X(\log X)^{b}}{u}f_{\gamma}(u)\Big\vert_{1}^{X}-(\log X)^{b}\int_{1}^{X}f_{\gamma}(u)\operatorname{d}\frac{X}{u} \\ \ll & X(\log X)^{b}+X(\log X)^{b-1}(\log\log X)^{\gamma-1}+X(\log X)^{b}\int_{e}^{X}\frac{(\log\log u)^{\gamma-1}}{u\log u}\operatorname{d}u \\ =& X(\log X)^{b-1}(\log\log X)^{\gamma-1}+X(\log X)^{b}+\frac{X(\log X)^{b}}{\gamma}(\log\log u)^{\gamma}\Big\vert_{e}^{X} \\ =& (\gamma^{-1}+o(1))X(\log X)^{b}(\log\log X)^{\gamma}. \end{align}\] Therefore, for each positive integer \(\gamma\), we have shown that \[N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)\ll X(\log X)^{b}(\log\log X)^{\gamma}.\] ◻

Proof. Claim: for each non-negative integer \(\gamma\), we have \[N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)=o(N_{\mathcal{C},C}(X)).\] If the Claim is true, then the statement is just a direct corollary of the author [16].

By our condition and Theorem 15, this is true for each positive integer \(\gamma\). So, it suffices to prove the case when \(\gamma=0\). For each \(K\in\mathcal{D}\), recall that the surjective continuous homomorphisms \(\operatorname{C}_{K}\to N\) that is \(H\)-equivariant correspond to the Galois \(N\)-extensions \(L/K\) such that \(L/\mathbb{Q}\) is a Galois \(G\)-field. This implies that there is a one-to-one correspondence between \(\mathcal{C}_{\Omega}^{0}\) and \(\bigcup_{K\in\mathcal{D}}\operatorname{Sur}_{H}(\operatorname{Cl}_{K},N)\). Therefore, we have \[\begin{align} N_{\mathcal{C}_{\Omega}^{0},C}(X) =& \sum_{ \substack{ K\in\mathcal{D} \\ \bar{C}(K)<X } }\#\{L\in\mathcal{C}_{\Omega}^{0}\mid K\subseteq L,\, C(L)<X\} \\ \leq & \sum_{ \substack{ K\in\mathcal{D} \\ \bar{C}(K)<X } }\lvert\operatorname{Sur}_{H}(\operatorname{Cl}_{K},N)\rvert \\ \leq & \sum_{ \substack{ K\in\mathcal{D} \\ \bar{C}(K)<X } } \lvert\operatorname{Hom}_{H}(\operatorname{Cl}_{K},N)\rvert\ll X(\log X)^{b}. \end{align}\] This proves that the Claim is true when \(\gamma=0\). So the Claim is true, hence also the theorem. ◻

Proof of Theorem 3. Note that \(S_{3}\) admits the short exact sequence \[1\to\mathbb{Z}/3\mathbb{Z}\to S_{3}\to\mathbb{Z}/2\mathbb{Z}\to1\] So, let \(\Omega:=\{(123),(132)\}\). The set \(\mathcal{C}\) of non-Galois cubic fields and the set of Galois \(S_{3}\)-fields are in the one-to-one correspondence to each other. In particular, we could define \(\mathcal{C}_{\Omega}^{\gamma}\) via this correspondence. And it admits a simple interpretation: a non-Galois cubic field \(K\in\mathcal{C}_{\Omega}^{\gamma}\) if and only if it admits exactly \(\gamma\) totally ramified primes other than \(2\) and \(3\). By Roquette and Zassenhaus [24], for a cubic field \(K\) with its Galois closure \(L\), we have that \[\begin{align} \operatorname{rk}_{3}\operatorname{Cl}_{L}\geq & \operatorname{rk}_{3}\operatorname{Cl}_{K} \\ \geq & \#\{p\nmid\infty:p\text{ is totally ramified in }K/\mathbb{Q}\} \\ \geq & \#\{p\nmid 6\infty:\rho_{p}(y_{p}^{t})\in\Omega\}-4, \end{align}\] where \(\rho_{p}:G_{\mathbb{Q}_{p}}\to S_{3}\) is the map corresponding to \(L_{p}\). We may also order \(\mathcal{C}\) by the product of ramified primes \(C\) of the Galois \(S_{3}\)-fields (the Definition 16), that is, \(C(K):=C(L)\) for a non-Galois cubic field \(K\) with its Galois closure \(L\). By Shankar and Thorne [13], we know that \[N_{\mathcal{C},\sqrt{d_{K}}}(X)\asymp X\log X.\] Though our definition of \(C(K)\) is different from the radical of the discriminant \(\sqrt{d_{K}}\), their result allows a finite collection of local specifications (see [13]). For example, counting real cubic fields unramified at \(2\) and \(3\) by the radical of the discriminant has the main term \(X\log X\). In this case, the radical of the discriminant is literally the same as the product of ramified primes defined in the Definition 16. This implies that \[N_{\mathcal{C},C}(X)\gg X\log X.\] This shows that the set \(\mathcal{C}\) of non-Galois cubic fields ordered by the product of ramified primes \(C\) satisfies the condition of the Theorem 16. So the Claim in the proof is true, that is, for each non-negative integer \(\gamma\), we have that \[N_{\mathcal{C}_{\Omega}^{\gamma},C}(X)\ll N_{\mathcal{C},C}(X).\] Moreover, let \(T:=T(6)=\{p\in\mathcal{P}\mid p\nmid 6\infty\}\). Then the tuple \(((\mathcal{C},C),\Omega,T)\) satisfies the condition of the author [16]. So, the theorem is true. ◻