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I am new to quotient spaces and I came across this problem and it confused me:

Given the group $G=\mathbf{Z}\times\mathbf{Z}$ and some normal subgroup $\langle (1,0)\rangle$.. The quotient group $G/N$ is isomorphic to $\mathbf{Z}$.

How is this the case? I understand that $\mathbf{Z}\times\mathbf{Z} = \{ (a,b) \mid a,b \in \mathbf{Z} \} $ and that the normal subgroup $N = \langle (1,0)\rangle$ is just $ \{ (a,0) \mid a \in \mathbf{Z} \} $. So the quotient group $G/N$ is the group of left cosets of $N$ in $G$. Here is where I have a little bit of trouble. So to my understanding a coset is $ aN = \{ an \mid n \in N \} $ where $a \in G$. So in this case, what exactly would be a coset? Like is $\{ (a,1) \mid a \in \mathbf{Z} \} $ one coset? Then how actually is the set of all cosets isomorphic to $\mathbf{Z}$? I am familiar with the first isomorphism theorem, but I'm getting confused with the cosets / group of cosets / quotient group.

Edit: Is it that anything in N will be mapped to 0 (given the homomorphism from $G$ to $G/N$), therefore what we are really looking at is just the second element??

Any help would be appreciated.

Shaun
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Rummannkurz
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    I suggest that since the group structure on $G = \mathbb Z \times \mathbb Z$ is usually written as vector addition, you use the $+$ sign for your operations on this group. So the coset of $a \in G$ is $a+N = {a + n \mid n \in N}$. Perhaps you can take it from here? – Lee Mosher Apr 11 '24 at 19:51
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    Also, I'd suggest that you don't use the same variable $a$ to denote both integers and pairs of integers. – jjagmath Apr 11 '24 at 19:54
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    Well, a minor issue indeed, but I still have issues with your title. If I am understanding correctly, $\mathbb{Z} \times \mathbb{Z}$ is already Abelian, and so every subgroup is normal. So I take issue with "its normal subgroup $\langle (1,0) \rangle$" because it may confuse some readers that this is the only nontrivial normal subgroup [or close to something like that]...or that you mean something else than usual with $\mathbb{Z} \times \mathbb{Z}$. – Mike Apr 11 '24 at 20:08

2 Answers2

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The group operation on $\mathbb Z \times \mathbb Z$ is given by $(a,b) \circ (c,d) = (a+c, b+d)$. Therefore the cosets are given by $(b,c) N =\{(b,c) \circ (a,0): (a,0) \in N\} = \{(b+a, c+0): a \in \mathbb Z\}$. The last set is the same as $\{(d, c): d \in \mathbb Z\}$ so the map $$\mathbb Z \to G/N, \quad c \mapsto \{(d, c): d \in \mathbb Z\}$$ gives an isomorphism of groups.

Lukas
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    Oh, so then we have that ℤ×ℤ $\to$ ℤ , { $ (a,b):a,b∈ℤ $} $\to$ $b \in$ ℤ , thus the subgroup N is the kernel of this surjective group homomorphism and thus $G/N$ is isomorphic to ℤ? – Rummannkurz Apr 11 '24 at 19:58
  • Yes exactly. That's a good observation. – Lukas Apr 11 '24 at 20:03
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Hint: $G/N$ is isomorphic to $\Bbb{Z}$ iff there is a surjective homomorphism of $G$ onto $\Bbb{Z}$ whose kernel is $N$. This is a consequence of (what is usually called) the first isomorphism theorem (but depending on your textbook, YMMV about the numbering of the three isomorphism theorems). The map from $G = \Bbb{Z}\times \Bbb{Z}$ to $\Bbb{Z}$ that maps $(m, n)$ to $n$ looks like an excellent candidate for a surjection with $(1, 0)$ in the kernel. Can you see how to fill in the details?

Rob Arthan
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