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!


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 $X=\mathrm{Hom}_\mathbb{Z}(M,A).$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.