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I've found this exercise.

Determine the elementary divisors of the abelian group in two generators $\{a,b\}$ with the relation $3a=4b$.

I don't see how to solve this. I believe that there should be an explicit way to describe this group, but I couldn't find one. My first thought was that there should be a copy of $\mathbb{Z}$ in this group as each generator has infinite order. And the other factor could be a $\mathbb{Z}_n$.

user26857
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Johanna
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    The group is a quotient of $\mathbb{Z}\times\mathbb{Z}$ modulo the subgroup generated by $(3,4)$, identifying $a$ with the image of $(1,0)$ and $b$ with the image of $(0,1)$. When (if?) you proved the structure theorem of finitely generated abelian groups, you must have shown how to suitably express subgroups of $\mathbb{Z}\times\mathbb{Z}$ so you can read off the invariant factors/elementary divisors. – Arturo Magidin Oct 19 '21 at 20:57
  • (Actually, $b$ with the image of $(0,-1)$, or change the generator to $(3,-4)$, but it amounts to the same thing...) – Arturo Magidin Oct 19 '21 at 21:21
  • But I don't need to write it as $\frac{\mathbb{Z}}{(d_1)}\times \frac{\mathbb{Z}}{(d_2)}$? – Johanna Oct 19 '21 at 21:36
  • Yes, and if you proved the structure theorem for finitely generated abelian groups, then along the way you showed that there is a way of changing the basis for $\mathbb{Z}\times\mathbb{Z}$ so that the subgroup has a set of generators of the appropriate form that makes that computation doable. Or perhaps you learned about the Smith Normal Form, which can likewise be used here to obtain the required structure. – Arturo Magidin Oct 19 '21 at 21:39
  • For example, this post describes how to obtain the Smith Normal form and use it to describe the quotient. – Arturo Magidin Oct 19 '21 at 21:43
  • Ok, I see it. Thanks! – Johanna Oct 19 '21 at 21:57
  • Hint: Consider the map to $\Bbb Z$: $\ a\mapsto 4,\ b\mapsto 3$. – Berci Oct 19 '21 at 22:03

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