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Suppose we have an exact sequence of abelian groups:

$0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z} \rightarrow H \rightarrow 0$, where the map $\mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}$ is given by $t \rightarrow (2t, -2t).$

What is the solution for group $H$?

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    I'd say something like $\;\Bbb Z/2\Bbb Z\oplus\Bbb Z/2\Bbb Z\;$ ... – 2017-02-01
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    what will be quotient map then? why (2,0) will go to (0,0), when (2,0) is not in the image of the first map? – 2017-02-01
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    @DomAntonio That would decrease the rank by 2, which would require a kernel of rank 2. – 2017-02-01
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    So it should be $\Bbb Z / 2\Bbb Z \oplus \Bbb Z$, with a map such as $(u,v) \mapsto (\bar u, u + v)$ – 2017-02-02
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    @DonAntonio Certainly not that, $H$ has non-torsion elements, such as the image of $(1,0)$. Alternatively, if your guess were correct, then right exactness of the tensor product applied to this sequence would have $\mathbb Q\to \mathbb Q^2\to 0\to 0$ being exact. So we know $H\otimes \mathbb Q\cong \mathbb Q$. – 2017-02-02

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Hint: Take $u=(1,1),v=(1,0)\in\mathbb Z\oplus\mathbb Z$. These generate the group, and their images generate $H$. What is the order of each element in $H$?