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Let $A$ be a finitely generated, infinite abelian group, and $n$ a positive integer. Show there exists a subgroup $B$ such that $|A/B| = n$.

I'm trying to use induction on the number of generators of $A$, but I don't know if this is a good approach. Any hints are greatly appreciated.

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  • $\begingroup$ @Conifold This is for an arbitrary group $A$, not just an example of a group. $\endgroup$ – gravitybeatle Dec 18 '19 at 4:44
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By the fundamental theorem of finite abelian groups, $A$ is isomorphic to a group $\mathbb{Z}\times A'$, where $A'$ is some abelian group. Hence, $A$ surjects onto $\mathbb{Z}$. As $\mathbb{Z}$ surjects onto $\mathbb{Z}_n$, the cyclic group of order $n$, it follows that $A$ also surjects onto $\mathbb{Z}_n$. The result then follows by the first isomorphism theorem.

You can actually find the subgroup $B$ if you want, although the question doesn't ask for this: Using the integers in the obvious way to denote the elements of the $\mathbb{Z}$ factor in $\mathbb{Z}\times A'$, so $A=\langle 1, A\rangle$, then $B=\langle n, A'\rangle$.


[The last paragraph is potentially ambiguous as there may be lots of $\mathbb{Z}$-factors of $A$. I wrote it like this because I wanted to use additive notation. Multiplicatively, and unambiguously, let $z$ be the generator of the $\mathbb{Z}$-factor in $\mathbb{Z}\times A'$, so $A=\langle z, A\rangle$, and then $B=\langle z^n, A'\rangle$.]

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Hint: The $k = 1$ case is when $A \cong \mathbb{Z}$. Can you describe the desired subgroup $B$ in this case? Can you modify your argument for the $k$ generator case?

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  • $\begingroup$ The subgroup $B$ is certainly $n\mathbb Z$, but for a general $A = gen(a_1, a_2, \ldots, a_k)$, I don't think $B = gen(na_1, na_2, \ldots, na_k)$ would work, right? Unless I'm missing something. $\endgroup$ – gravitybeatle Dec 18 '19 at 4:57
  • $\begingroup$ @gravitybeatle: That wouldn’t work; it would give you $(\mathbb{Z}/n\mathbb{Z})^k$. But you are close... $\endgroup$ – Arturo Magidin Dec 18 '19 at 6:24
  • $\begingroup$ I've changed the $n$ to a $k$ here, as $n$ has a different meaning in the question. $\endgroup$ – user1729 Dec 18 '19 at 9:34

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