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I am reading Edwin Wiess' book called Cohomology of groups and I cant see why the following claim he makes is true in proposition 3-1-9.

He says that if $A$ is a finitely generated $G$-module for a finite group $G$ where $|G|=m$, that $A/mA$ is finite, where m is the multiplication by m map.

How does he know its finite? is this a standard result?

Thank you

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up vote 1 down vote accepted

Do you know that a f.g. abelian torsion group is finite? Well, that's precisely what $\,A/mA\,$ is: a f.g. abelian torsion group generated, as a $\,G\,$ module (and this usually means $\,A\,$ is a f.g. $\,\Bbb Z[G]-$module. Check this), by the images of a $\,G-$generator set in the quotient $\,A/mA\,$

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Thank you, I knew this for abelian groups, but since it was G-modules I was a bit unsure. –  user39947 Dec 2 '12 at 11:41
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