A subgroup $B$ of an finitely generated group $A$ such that $|A/B| = n$ 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.
 A: 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?
A: 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 $B=\langle z^n, A'\rangle$.]
