# Finite $\Delta$-module of $p$-power order

I have a question concerning lemma, that I want to prove:

Let $p$ be a prime and $\Delta$ be a finite group of order prime to $p$. Let $M$ be a finite $\Delta$-module of order a power of $p$. Then there is an isomorphism of $\mathbb{F}_p[\Delta]$-modules $$_p M \cong M/pM$$ where $_p M$ denotes the kernel of the multiplication by $p$.

My problem now is that I'm able to prove the above theorem in the case of $\mathbb{Z}$-modules. But in our case - that of $\mathbb{Z}[\Delta]$-modules - I'm totally lost. In the case of $\mathbb{Z}$-module I would have just used the decomposition theorem for abelian groups. Can you help me with the case of group rings??

For example the module $\mathbb{Z}/p^2 \mathbb{Z}$ is of order $p^2$ and let the action of $\Delta$ be trivial. How do I get the above isomorphism?

Thank you for your help, Tom

• What do you mean by a finite $\Delta$-module? – Tobias Kildetoft Aug 23 '13 at 12:52
• I mean a finite $\mathbb{Z}[\Delta]$-module, where $\mathbb{Z}[\Delta]$ is the group ring of our group $\Delta$. – BIS HD Aug 23 '13 at 12:54
• This looks like an application of Maschke's Theorem or rather its proof. – Martin Brandenburg Aug 23 '13 at 16:04
• @Martin Brandenburg I already thought something like that. But, since $M$ is a $\mathbb{Z}[\Delta]$-module, Maschke's theorem doesn't really make sense, since the projection map, which is needed to construct a complement of a submodule of $M$ deals with the inverse of the order of $\Delta$ which I'm not sure about lies in $\mathbb{Z}$? Or: does Maschke's theorem hold for PID, whose characteristic isn't divisible by the group order, as well? – BIS HD Aug 25 '13 at 13:17