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Determine isomorphism type of quotient group $$\mathbb{Z} \times \mathbb{Z} / \langle(1,1)\rangle $$ using Fundamental Theorem Finite Generated Abelian Groups

after looking at the factor group, it seems everything in $$\mathbb{Z} \times \mathbb{Z}$$ is generated except $(0,0)$. All in all, I'm just having a very difficult time even starting this problem.

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    $\begingroup$ Do you know the First Isomorphism Theorem? $\endgroup$ – paw88789 May 12 '15 at 19:23
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The group generated by $(1,1)$ is $\mathbb{Z}(1,1)$, that is, $\{(m,m)\}$ where $m$ runs over $\mathbb{Z}$. So for instance $(1,0)$ is not equivalent to zero in the quotient and generates a subgroup $\{[(m,0)]\}$ isomorphic to $\mathbb{Z}$. Now any equivalence class has a unique representative of the form $(m,0)$, since $(p,q)$ is equivalent to $(p-q,0)$ and $(m,0)$ is not equivalent to $(n,0)$ if $m\neq n$. So, what can we conclude?

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We claim that the group is isomorphic to $\mathbb Z$. Consider the homomorphism $\mathbb Z\times\mathbb Z\to\mathbb Z$ given by $(a, b)\mapsto a-b$. The kernel is the set of pairs whose coordinates are equal, which is exactly $\langle (1, 1)\rangle$. The map is surjective, since $(n, 0)\mapsto n$ for each $n\in\mathbb Z$. Thus, by the First Isomorphism Theorem, $\mathbb Z\times\mathbb Z/\langle(1, 1)\rangle\cong\mathbb Z$.

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It's hard to find something better than answer of Nishant, but let me present one variant more, just for variety. Let $G=\mathbb Z\times\mathbb Z$, $e_1=(1,0),e_2=(0,1)$, $D=\{(a,a):a\in\mathbb Z\}$. Then $e_1,e_2$ - $\mathbb Z$-basis of $G$. Let $f_1=e_1,f_2=e_1+e_2$, then $f_1,f_2$ - $\mathbb Z$-basis of $G$ too, since each of these two systems can be expressed from others with integer coefficients. Hence $G=\mathbb{Z}f_1\oplus\mathbb{Z}f_2=\mathbb{Z}f_1\oplus D$. Therefore $G/D\cong (\mathbb{Z}f_1\oplus D)/D\cong \mathbb{Z}f_1\cong\mathbb Z$.

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