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I've been asked to show that a finitely generated abelian group G is finite iff $G/pG = \{0\}$ for some prime number $p$, and to find a group such that that is true for all prime $p$. Not really sure where to start with this, I've tried finding a homomorphism with kernel $pG$ unsuccessfully. Hints or help are appreciated.

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By the classification theorem, any FG abelian group is of the form $$\mathbb{Z}^d \times \mathbb{Z}_{d_1} \times \dots \times \mathbb{Z}_{d_i}$$ where $d_1 \mid \dots \mid d_i$.

Hint: what does $G/pG$ look like, if $d=0$ and if $d>0$ (that is, if $G$ is finite or infinite respectively)?

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  • $\begingroup$ If it is finite, do you just use a prime factor of $d_1$ as your p? $\endgroup$ – Simon Means Apr 22 '16 at 12:50
  • $\begingroup$ That works, yes. (I think. I'm quite tired at the moment.) $\endgroup$ – Patrick Stevens Apr 22 '16 at 13:00
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For your second question: we know that multiplication by $p$ is an isomorphism for all $p$. So every element $g \in G$ must have some element $h$ such that $h + \dotsb + h = g$. So every element is infinitely divisible. Therefore look for groups where the ability to divide is of the essence.

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  • $\begingroup$ So Q is an example of this? $\endgroup$ – Simon Means Apr 22 '16 at 13:49
  • $\begingroup$ @SimonMeans Yes. $\endgroup$ – Eric Auld Apr 22 '16 at 16:44

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