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This question comes from a proof in Cassels and Frohlich (Chapter 4, Section 5, Proposition 5).

Let $G$ be a group, let $H$ be a subgroup and let $A$ be a $G$-module.

Claim: There exists an abelian group $X$ such that, as $H$-modules, we have $$\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}[G],A)\cong\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}[H],X).$$

The authors say this is because $\mathbb{Z}[G]$ is a free $\mathbb{Z}[H]$-module but I don't see how this follows?

I understand why $\mathbb{Z}[G]$ is free as a $\mathbb{Z}[H]$-module.

$\Big[$Proof: any set of coset representatives of $H$ in $G$ is a $\mathbb{Z}[H]$-basis.$\Big]$

But I don't see at all how to use this to prove their claim.

Help much appreciated!

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Serre (Corps Locaux) to the rescue...

We have that

$$\mathbb{Z}[G]\cong \bigoplus_{G/H}\mathbb{Z}[H]$$

as $\mathbb{Z}[H]$-modules and so, in particular, as $\mathbb{Z}$-modules. We can therefore write:

$$\mathbb{Z}[G]\cong \mathbb{Z}[H]\otimes_\mathbb{Z}\bigoplus_{G/H}\mathbb{Z}.$$

Writing $M = \bigoplus_{G/H}\mathbb{Z},$ we have that

$$\mathrm{Hom}_\mathbb{Z}(\mathbb{Z}[G],A)\cong\mathrm{Hom}_\mathbb{Z}(\mathbb{Z}[H]\otimes_\mathbb{Z}M,A)\cong\mathrm{Hom}_\mathbb{Z}(\mathbb{Z}[H],\mathrm{Hom}_\mathbb{Z}(M,A))$$ and so we can take $X=\mathrm{Hom}_\mathbb{Z}(M,A).$

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