Finding all abelian groups such that there exists certain short exact sequence. I have to find all abelian groups $A$ such that there exists a short exact sequence $0\rightarrow\mathbb{Z}\rightarrow A\rightarrow\mathbb{Z}\oplus\mathbb{Z}_{6}\rightarrow 0$.
I have found $A=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_{6}$, $A=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_{2}$, $A=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_{3}$, and $A=\mathbb{Z}\oplus\mathbb{Z}$. I am not able to find any other, but I am also not able to prove these are the only ones.
This exercise has come up in an Algebraic Topology course, so I would like to prove it using elementary abstract algebra and not Homological Algebra.
Any hint will be greatly appreciated.
 A: The extensions by $\def\Z{\mathbb{Z}}\Z\oplus\Z_6$ of $\Z$ are described by the group $\def\Ext{\operatorname{Ext}}\Ext(\Z\oplus\Z_6,\Z)$. By general theory, this is isomorphic to $\Ext(\Z,\Z)\oplus\Ext(\Z_6,\Z)$. As the first summand is zero, because $\Z$ is free, we need to compute the second group; this can be done by considering the exact sequence
$$
0\to\Z\xrightarrow{\cdot6}\Z\to\Z_6\to0
$$
(where $\cdot6$ denotes “multiplication by $6$) and applying to it the contravariant functor $\def\Hom{\operatorname{Hom}}\Hom(\Z,-)$:
$$
0\to\Hom(\Z_6,\Z)\to\Hom(\Z,\Z)\xrightarrow{\cdot6}\Hom(\Z,\Z)\to
\Ext(\Z_6,\Z)\to\Ext(\Z,\Z)
$$
which can be rewritten as
$$
0\to0\to\Z\xrightarrow{\cdot6}\Z\to\Ext(\Z_6,\Z)\to0
$$
which means that $\Ext(\Z_6,\Z)\cong\Z_6$ and that the isomorphism is exactly the same as the “natural” isomorphism $\Z_6\cong\Hom(\Z_6,\Z_6)$.
In particular there are six non-equivalent extensions of the kind you're looking for. How can you write them?
First, if you're given $0\to\Z\to A\to\Z\oplus\Z_6\to0$, you can say that $\Z$ is a homomorphic image of $A$ and split it off; so $A\cong B\oplus\Z$ and we have an exact sequence $0\to\Z\to B\to\Z_6\to0$. This corresponds to
$$
\Ext(\Z\oplus\Z_6,\Z)\cong\Ext(\Z,\Z)\oplus\Ext(\Z_6,\Z)=0\oplus\Ext(\Z_6,\Z).
$$
Now, let $x\in\Z_6\cong\Hom(\Z_6,\Z_6)$; this corresponds to the endomorphism $\cdot x\colon\Z_6\to\Z_6$ (the multiplication by $x$), so you can consider a pull-back diagram
$$\require{AMScd}
\begin{CD}
0 @>>> \Z @>>> B @>>> \Z_6 @>>> 0 \\
@. @| @VVV @VV{\cdot x}V @. \\
0 @>>> \Z @>{\cdot6}>> \Z @>>> \Z_6 @>>> 0
\end{CD}
$$
It can be proved that, as $x\in\Z_6$, these are the inequivalent extensions, but this would require quite a long argument, that you find in any book dealing with homological algebra.
The “split extension”, $B=\Z\oplus\Z_6$, that is the same as the one you found, corresponds to taking $x=0$.
The pullback $B$ is the subgroup of $\Z\oplus\Z_6$ consisting of the pairs $(a,b)$ such that $a+6\Z=xb$. In the case of $x=0$ it's just $6\Z\oplus\Z_6\cong\Z\oplus\Z_6$. In the case of $x=1$ it is the set of pairs $(a,b)$ such that $a+6\Z=b$. Do the same for the remaining four endomorphisms of $\Z_6$.
A: Partial Answer: Let $R$ be a commutative ring with identity and let $0 \rightarrow M \rightarrow E \rightarrow N \rightarrow 0$ be an exact sequence of $R$ modules. Then the equivalence class of extensions is given by ${\rm Ext}^1_{R}(N, M).$ (See any Homological Algebra book.)
In this case $R = \mathbb{Z}, M = \mathbb{Z}, N = \mathbb{Z} \oplus \mathbb{Z}_6.$ So need to look at the abelian group ${\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z} \oplus \mathbb{Z}_6, \mathbb{Z}).$ First of all, ${\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z} \oplus \mathbb{Z}_6, \mathbb{Z}) \cong {\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Z}) \oplus{\rm  Ext}^1_{\mathbb{Z}}(\mathbb{Z}_6, \mathbb{Z}).$ Now ${\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Z}) = 0$ and ${\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z}_6, \mathbb{Z}) \cong \mathbb{Z}_6$. (See any Homological Algebra book for this results and how to prove it.)
So we have exactly $6$ non-isomorphic choice for $A$. (But I don't know how to construct them all.)
A: Hint: For group extensions in a short exact sequence 
$$0\to A\to E\to B\to 0$$
for each of the homomorphisms $B\to{\rm Out}A$ there is an extension $E$.
Also ${\rm Out}\ \Bbb{Z}={\Bbb{Z}}_2$.
