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

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.

  • 1
    $\begingroup$ $\langle(0,1)\rangle$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.