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I have an intuition why $\mathbb{Z} \times \mathbb{Z} / \langle (1,1)\rangle \cong \mathbb{Z}$ since let $(a,b) \in \mathbb{Z} \times \mathbb{Z}$ then you can fix the first entry and express the second entry as follows $(a,b) = (a,a+c) = (a,a) +(0,c)$, for some $c \in \mathbb{Z}$. Now you can let $c$ slide around the whole integers $\mathbb{Z}$ and the isomorphism is obvious. I can't find a similar argument as above, if say we divide $\mathbb{Z} \times \mathbb{Z}$ by $\langle (2,3) \rangle$. Could somebody provide an intuition and proof of why $\mathbb{Z} \times \mathbb{Z} / \langle (2,3)\rangle \cong \mathbb{Z}$ or more general $\mathbb{Z} \times \mathbb{Z} / \langle (a,b)\rangle \cong \mathbb{Z}$ if $a,b$ are relatively prime.

Martin Brandenburg
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Roger Crook
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    HINT: $1=ma+nb$ for some integers $m$ and $n$. – Ted Shifrin Dec 30 '23 at 01:13
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    I have removed the ring theory tag because it does not fit. If you are working with ideals and rings, the quotient will be very different, in fact $(R \times S)/\langle (r,s) \rangle \cong R/\langle r \rangle \times S/\langle s \rangle$ as rings. So your question is actually about abelian groups. For answers, see the linked duplicates. – Martin Brandenburg Dec 30 '23 at 01:20

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