What is $\mathbb{Z}^2/\left( (m,n)\mathbb{Z} \right)$ where $m, n$ are bigger than one and co-prime? Since it is abelian it must be $\mathbb{Z}^r$ plus some torsion but I can't figure out what it is precisely.
Asked
Active
Viewed 1,251 times
0
-
2Other hint: What's the order of the group? What's the order of $(1, 1)$? – Connor Harris Apr 28 '17 at 14:42
-
@ConnorHarris well I can show that there are no elements of finite order. Probably I should make the question more precise. I cant determine whether it is $\mathbb{Z}$ or $\mathbb{Z}^2$? – J.E.M.S Apr 28 '17 at 14:46
-
1I’m guessing that “$(m,n)\Bbb Z$” is the infinite cyclic group generated by the element $(m,n)$ of $\Bbb Z\times\Bbb Z$. Is that right? – Lubin Apr 28 '17 at 14:46
-
@Lubin Yes it is. – J.E.M.S Apr 28 '17 at 14:46
-
1Did you see this question, or this one, etc.? – Dietrich Burde Apr 28 '17 at 14:59
-
@DietrichBurde No, I was searching but found some less helpful answers. Thank you very much for pointing them. – J.E.M.S Apr 28 '17 at 15:02
-
Ah, I thought your notation meant $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$. – Connor Harris Apr 28 '17 at 15:10
1 Answers
3
The clue to the solution is that $\begin{pmatrix}m\\n\end{pmatrix}$ may be extended to a basis of $\Bbb Z^2$. Indeed, if $Am+Bn=1$, then $$ \left\lbrace \begin{pmatrix}m\\n\end{pmatrix}, \begin{pmatrix}-B\\A\end{pmatrix} \right\rbrace\,, $$ is a basis, as you easily see. Thus the quotient by your subgroup is isomorphic to $\Bbb Z$.
Lubin
- 65,209