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Let $K/\mathbb{Q}$ be a Galois number field and $H/K$ its Hilbert class field. Then we have $\mathrm{Gal}(H/K) \cong \mathrm{CL}_K$ from class field theory, and moreover $H/\mathbb{Q}$ is Galois. We obtain an extension $$0 \to \mathrm{CL}_K \to \operatorname{Gal}(H/\mathbb{Q}) \to \operatorname{Gal}(K/\mathbb{Q}) \to 0.\tag{$*$}\label{492847_star}$$ The group $\operatorname{Gal}(K/\mathbb{Q})$ acts on $\mathrm{CL}_K$ in a natural way, which agrees with the conjugacy action induced by the extension.

My question: is the group structure on $\operatorname{Gal}(H/\mathbb{Q})$ uniquely determined by this action?

To make this more precise, recall that extensions as in \eqref{492847_star} are classified by the group cohomology $H^2(\operatorname{Gal}(K/\mathbb{Q}),\mathrm{CL}_K)$. Write $c_K \in H^2(\operatorname{Gal}(K/\mathbb{Q}),\mathrm{CL}_K)$ for the class of the extension determined by $K$. Let $L$ be another number field with the same Galois group which acts in the same way on the class group. Then are the group extensions \eqref{492847_star} the same, i.e. do we have an equality $c_K = c_L$ upon identifying the relevant groups?

This is automatic provided $H^2(\operatorname{Gal}(K/\mathbb{Q}),\mathrm{CL}_K)$ is trivial, in which case the extension splits. There are examples where the extension is non-trivial. Still it is a logical possibility that one obtains the same non-trivial extension in all cases with the same Galois group and class group action. Counter-examples could be obtained by giving two fields with the same Galois group and class group action, but where one extension splits and the other does not.

I'm also happy to consider the slightly more general and more flexible problem of considering the group extension determined by a quotient of the class group.

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    $\begingroup$ Do you want $\mathrm{Gal}(H/\mathbb{Q})$ in your exact sequence? $\endgroup$ Commented May 19 at 10:03
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    $\begingroup$ Presumably you are assuming $K/\mathbb Q$ is Galois. $\endgroup$ Commented May 19 at 10:21

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I don't think so. Take $K=\mathbb{Q}(\sqrt{-3},\sqrt{-35})$. According to LMFDB, this has class number $2$. Since $\mathbb{Q}(\sqrt{5})$ becomes unramified over $K$, we have $H=\mathbb{Q}(\sqrt{-3},\sqrt{-35},\sqrt{5})$, with Galois group $(\mathbb{Z}/2\mathbb{Z})^3$.

On the other hand, take $K=\mathbb{Q}(\sqrt{-3},\sqrt{13})$, which again has class number $2$. This time, it's easy to see that no quadratic extension of $\mathbb{Q}$ becomes unramified over $K$. So whatever $\mathrm{Gal}(H/\mathbb{Q})$ is, it can't be $(\mathbb{Z}/2\mathbb{Z})^3$ (I computed it to be $D_8$ in Magma).

In both cases, $\mathrm{Gal}(H/\mathbb{Q})$ is an extension of $(\mathbb{Z}/2\mathbb{Z})^2$ by $\mathbb{Z}/2\mathbb{Z}$, and the action is necessarily trivial.

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    $\begingroup$ Thanks these examples are great! I notice that the difference between the 2 comes from genus theory, since the class group in the first example is part of the genus group. Do you know any examples not coming from genus theory? $\endgroup$ Commented May 20 at 8:24
  • $\begingroup$ Just to follow up: In private communication, examples not coming from genus theory can be constructed by taking $K$ with Galois group $(\mathbb{Z}/2\mathbb{Z})^3$ and class number $2$. One shows they don't come from genus theory as there are examples where the Galois group of $H/\mathbb{Q}$ is non-abelian $\endgroup$ Commented May 21 at 11:49

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