# 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.

• @Conifold This is for an arbitrary group $A$, not just an example of a group. – gravitybeatle Dec 18 '19 at 4:44

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 $$B=\langle z^n, A'\rangle$$.]
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?
• 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. – gravitybeatle Dec 18 '19 at 4:57
• @gravitybeatle: That wouldn’t work; it would give you $(\mathbb{Z}/n\mathbb{Z})^k$. But you are close... – Arturo Magidin Dec 18 '19 at 6:24
• I've changed the $n$ to a $k$ here, as $n$ has a different meaning in the question. – user1729 Dec 18 '19 at 9:34