Why is an Abelian group finitely generated iff $A/mA$ is finite for some $m\gt 1$ and $A$ has a norm function? I know that $mx$ where $x$ is an element of $A$ is equivalent to $0$ in $A/mA$, and I tried to check if I can make the value of the norm between $0$ and $m$. But then I cannot proceed. Can someone help me please?

  • $\begingroup$ Looking at the definition of an abelian group norm on wikipedia [here](en.wikipedia.org/wiki/Norm_(abelian_group)), it seems the condition $v(mg)=|m|v(g)$ implies that $v(0)=0$. One can easily prove every element of a normed finite abelian group would have norm $0$, which contradicts the property that $g\ne0\Rightarrow v(g)\ne0$. Hence $A$ being finitely generated cannot imply it has a norm function. Are you missing a hypothesis of torsionfree? $\endgroup$ – anon Dec 26 '13 at 6:56
  • $\begingroup$ What's up with not being able to hyperlink? Didn't work in chat for me either. $\endgroup$ – anon Dec 26 '13 at 6:58
  • $\begingroup$ Also I suppose we should ask if you have the usual structure theory results for f.g. abelian groups at your disposal. $\endgroup$ – anon Dec 26 '13 at 7:07
  • $\begingroup$ Another issue: $A=\Bbb Q$ has a norm function (the absolute value) and $A/mA$ is finite for every nonzero integer $m$ (in fact it's the trivial group since $\Bbb Q$ is divisible), so the hypotheses on one side of the "iff" are satisfied, yet $\Bbb Q$ is not finitely-generated (the condition on the other side of the iff). Am I reading the problem incorrectly? $\endgroup$ – anon Dec 26 '13 at 7:28

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