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$$(\mathbb{Z}\times\mathbb{Z})/\langle (0,1)\rangle$$

How to do this coset? I know it's gonna be made of elements

$$(a,b) + \langle(0,1)\rangle$$

where $(a,b)\in \mathbb{Z}\times \mathbb{Z}$, and that $2$ elements $(a,b), (c,d)$ are in the same coset if $(a-c, b-d) = (0,k)\implies a = c, b = d+k$ for any $k$.

So for $a=0$ and $d=0$ we have $b =\cdots, -2,-1,0,1,2,\cdots$, so the elements are: $\cdots, (0,-2), (0,-1), (0,0), (0,1), (0,2),\cdots$.

For $a=1$ and $b=0$ and $d=2$ for example, we have: $c = 1, b = 2 + k =\cdots, -2,-1,0,1,2,\cdots$ too, so the elements are: $\cdots,(1,-2),(1,-1),(1,0),(1,1),(1,2),\cdots$.

In general, for any $d$, and $c$ generic, the elements on a same coset will be: $\cdots,(c,-2),(c,-1),(c,0),(c,1),(c,2),\cdots$

The set of $C$ of cosets is infinite, but they're formed by:

$$C = \{\{(a,b), b\in \mathbb{Z}\}, a\in \mathbb{Z}\}$$

My book says this quotient is isomorphic to $\mathbb{Z}$, but I quite didn't understand why.

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    $\begingroup$ $\langle(0,1)\rangle$ is the kernel of the surjective homomorphism $\Bbb Z\times \Bbb Z\to \Bbb Z$ given by $(a, b)\mapsto a$. For any homomorphism of groups, the domain divided out by the kernel is isomorphic to the image. $\endgroup$ – Arthur Apr 27 '17 at 17:24
  • $\begingroup$ Compare with this question. $\endgroup$ – Dietrich Burde Apr 27 '17 at 18:31
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First of all, it should make sense intuitively that $H = (\mathbb{Z} \times \mathbb{Z})/ \langle(0,1)\rangle$ is isomorphic to $\mathbb{Z}$. This is clear because since the cosets look like $\{(n,0), (n,1),(n,-1),(n,2),(n,-2), \ldots \}$ for $n \in \mathbb{Z}$ you can tell that there is a coset for every $n \in \mathbb{Z}$, and conversely there is an $n \in \mathbb{Z}$ for every coset.

With this in mind it should be easy to find an isomorphism. Indeed, if we denote the coset written above by $\overline{n}$, then we can define $\phi: \mathbb{Z} \to H$ such that $\phi(n)=\overline{n}$. It is easily checked that this is an isomorphism.

Or, if you like we can also apply the first isomorphism theorem. Let $\varphi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ be defined by $\varphi(m,n) = m$. Clearly, the image of $\varphi$ is all of $\mathbb{Z}$, and $\ker(\varphi)= \langle(0,1) \rangle$, so $H \cong \mathbb{Z}$ by the first isomorphism theorem.

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