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I asked the following question on Stackexchange and got no reply so I am reposting it here. Let $G$ be a finite group. A $G$-module C is a class module if, for all subgroups $H \subset G$:

1) $H^1(H,C)=0$

2) $H^2(H,C)$ is cyclic of order $\#H$

Remark: If $G$ is cyclic then $\mathbb{Z}$ is a class module.

Question: Does every finite group admit a class module? Abelian group? If yes: is there a standard construction of such a C given G?

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The answer is yes. Let $G$ be a finite group. You can realise $G$ as a Galois group $G(L/K)$ of a finite Galois extension $L/K$ of number fields. Let $C_L$ be the idele class group of $L$. $G$ acts on $L$ and on $C_L$.

By p.196 of Tate's article on "global class fields" (in the book Algebraic Number Theory" by Cassels and Frohlich), it follows that $H^2(G,C_L)$ is cyclic of order $n=card (G)$.

On page 180 of the same article, see Theorem (9.1). Tate proves that $H^1(G,C_L)=0$.

The above two results also hold if $H\subset G$ is a subgroup, since $H=Gal (L/L^H)$.

Thus $C=C_L$ works.

I do not know if there is a standard construction of the module $C$.

[Added] These results are crucial to the definition of the Weil group of a number field as the inverse limits of certain extensions of $G(L/K)$ by $C_L$.

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  • $\begingroup$ Thanks. I did not know every finite group can be realized in this way, as an extension of some L/K number fields. Might i also find this in Cassels and Frohlich? If not, where? $\endgroup$
    – Eins Null
    Commented Mar 1, 2016 at 17:08
  • $\begingroup$ You can find the symmetric group $S_n$ as a Galois group over $\mathbb Q$; now use the fact that every finite group is embeddable in some $S_n$ $\endgroup$
    – Venkataramana
    Commented Mar 1, 2016 at 17:20

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