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The question is:

For what kind of Abelian group $A$ is there a short exact sequence: $$ 0\to\mathbb{Z}\to A\to\mathbb{Z}_{n}\to0. $$

This is an exercise from Allen Hatcher's Algebraic Topology textbook. I could name a few $A$s like $\mathbb{Z}, \mathbb{Z}\oplus\mathbb{Z}_{n},$ etc.

But I have no idea what kind of general $A$ can but put in this short exact sequence. Does anyone know if there is some topological background for this problem?

My result is: $A=Z\oplus Z_{d}$, with some d|n.

My method is to consider a commutative diagram where the upper sequence is: $0 \to Z \to Z \oplus Z \to Z \to 0 $ and the bottom is the exact sequence in the problem. We could try to construct (using the exactness of the bottom line) a surjective map from $Z\oplus Z$ to A to make the diagram commutative.

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  • $\begingroup$ it would be interesting if you can share how you get your result. $\endgroup$
    – rafforaffo
    Commented Apr 28, 2015 at 9:38

2 Answers 2

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The point of the question is that there aren't that many $A$ that fit in such a sequence. A useful lemma to prove might be the following:

Let $0 \to X \to Y \to Z\to 0$ is a short exact sequence of Abelian groups, and $X$ and $Z$ are generated by $m$ and $n$ elements, respectively. Then $Y$ has a generating set with $m+n$ or fewer elements.

In your example, this would imply that $A$ is an Abelian group with at most $2$ generators. Such groups are classified by the structure theorem, and it only remains to see which of those fit into such an exact sequence.

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HINT: Any such $A$ is finitely generated. Then we know, from the structure of a finitely generated abelian group, that $A\simeq\Bbb Z^r\times T$ with $r\geq0$ and $T$ a finite abelian group.

At this point the list of possible $A$ should be clear.

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