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Let $A$ be an infinite cyclic group and $B$ be a cyclic group of order $n$. Suppose $$0 \to A \to G \to B \to 0$$ is a short exact sequence of abelian groups. What could $G$ be?

It is clear enough that $G = \mathbb{Z} \oplus C_m$ works, for all $m$ such that $m$ divides $n$ and $\textrm{gcd}(m, n/m) = 1$, by the Chinese remainder theorem. $A$ and $B$ are finitely generated, so $G$ is as well; thus the structure theorem for finitely-generated abelian groups tells us that this is exhaustive. However, is there a more direct approach (in the sense of not using a sledgehammer) to this easy-looking exercise? (It is exercise 6.2 in Massey's A basic course in algebraic topology.)

For example, by tensoring with $\mathbb{Q}$, we obtain a right-exact sequence $$\mathbb{Q} \to G \otimes \mathbb{Q} \to 0 \to 0$$ which tells us that the free part of $G$ has rank at most $1$, and considering $\textrm{Hom}(-, \mathbb{Z})$, we get a right-exact sequence $$0 \to \textrm{Hom}(G, \mathbb{Z}) \to \mathbb{Z} \to 0$$ which indicates the free part of $G$ has rank exactly $1$. What can we say about the torsion part?

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Actually you do not need the full structure theorem for finitely generated abelian groups, but just the fact that such a group is the direct sum of torsion and torsion-free part. Moreover tensoring with $\mathbb{Q}$ is exact, so that you directly get $\mathbb{Q}\cong G\otimes\mathbb{Q}$. –  Hagen Aug 30 '11 at 9:15
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1 Answer

Consider the short exact sequence

$1 \rightarrow A \xrightarrow{ \phi } G \xrightarrow{ \psi } B\rightarrow 1$

where $\phi(A)=\langle x|- \rangle$ be infinite cyclic group, and $B=\langle \bar{y}| \bar{y}^n\rangle$ be finite cyclic group of order $n$.

($G$ may be abelian or may not be)

Since $G \xrightarrow{ \psi } B$ is surjective, $\exists$ $y\in G$ such that $\psi(y)=\bar{y}$. Then $\psi(y^{n})=\bar{y}^n=1$, so $y^n\in ker(\psi)=\phi(A)$.

Without loss of generality, let $y^{n}=x^{i}$ for some non-negative integer $i$ . It is clear that $G$ is generated by $x,y$.

Also as $A\triangleleft G$, we have $y^{-1}xy\in \phi(A)=\langle x |-\rangle$; so let $y^{-1}xy=x^{j}$ for $j\neq 0$.

Then $y^{-2}xy^{2}=x^{j^2}$, and continuing this $y^{-n}xy^{n}=x^{j^n}$, i.e $x=x^{j^n}$, since $y^n(=x^i)$ commutes with $x$.

So $j^n=1$ in $\mathbb{Z}$. Since $n>0$.

Case1: If $n$ is odd, then $j=1$ is its only solution.

Therefore $y^{-1}xy=x$, which means $G$ is an abelian group; it is quotient of $\mathbb{Z} \oplus \mathbb{Z}$:

$G=\langle x,y | y^{-1}xy=x, y^n=x^i \rangle$

(I didn't find more simplification of this! You may be interested in structure of these groups!)

Case 2: If $n$ is even, then $j=\pm 1$. Case $j=1$ has considered above.

So let $j=-1$. As $y^{-1}xy=x^{-1}$, we have $y^{-1}x^{i}y=x^{-i}$ i.e. $x^{i}=x^{-i}$, since $x^{i}(=y^n)$ commutes with $y$. So $i=0$ i.e. $y^n(=x^i)=1$.

In this case $G$ is semidirect product of $\langle x|- \rangle \cong \mathbb{Z}$, and $\langle y|y^n\rangle \cong \mathbb{Z}/n$.

For this, we have to consider homomorphism from $\mathbb{Z}/n$ into $Aut(\mathbb{Z})\cong \mathbb{Z}_2$. Since $n$ is even, there are only two possibilities of homomorphisms -trivial and $y\mapsto \{x\mapsto x^{-1} \}$.

Therefore $G$ is one of the following:

$G=\langle x,y| y^n, y^{-1}xy=x\rangle \cong \mathbb{Z} \oplus \mathbb{Z}/n$

$G=\langle x,y| y^n, y^{-1}xy=x^{-1}\rangle \cong \mathbb{Z} \rtimes \mathbb{Z}/n$

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In case 2, your claim $x^i = x^{-i}$ is incorrect. Let us take $n = 6$, $G = \mathbb{Z} \oplus C_3$, $x = (2, 0)$, $y = (1, 1)$. Then $y^6 = (6, 0) \ne (-6, 0)$. Nonetheless, $\langle x, y \rangle / \langle x \rangle \cong C_6$. –  Zhen Lin Aug 30 '11 at 10:49
$x$ is a generator of infinite cyclic group and $y^n=x^i$ for some $i\in \mathbb{Z}^{+}\cup \{0\}$ (first relation stated after short exact sequence). Also in case 2, $y^{-1}xy=x^{j}=x^{-1}$, so taking $i^{th}$ power both side, we ger $y^{-1}x^{i}y=x^{-i}$ i.e. $y^{-1}y^{n}y=x^{-i}$ i.e. $y^{n}=x^{-i}$ i.e. $x^{i}=x^{-i}$ for $i\in \mathbb{Z}^{+}\cup \{0\}$ which means $i=0$. –  user8186 Aug 30 '11 at 10:58
In your example $xy=yx$, whereas in Case 2 considered above, we have relation $y^{-1}xy=x^{-1}$ . From this, and $y^{n}=x^{i}$ with $n>0, i\geq 0$, we have deduced that $x^{i}=x^{-i}$. –  user8186 Aug 30 '11 at 11:05
Ah, that's more convincing. You should add that to the post. –  Zhen Lin Aug 30 '11 at 11:11
Say $A=\mathbb{Z}$, $B=\mathbb{Z}/2\mathbb{Z}$, and with $y^{-1}xy=x$ as in your first part of case $2$. You seem to be saying at the end that $G$ must be the direct sum of $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$, but you can have $G=\mathbb{Z}$, and $\phi$ be the embedding that maps $A$ to $2\mathbb{Z}$. –  Arturo Magidin Aug 30 '11 at 16:08
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