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Question: Determine the order of $(\mathbb{Z} \times \mathbb{Z})/ \left<(4,2)\right>$. Is the group cyclic?

I want to first apologize for the way this post is written. I'm on the road and have no access to a computer; this is typed with a mobile.

The order of the factor group is infinite due to the fact that the order of the external dirext product of the two group of integers is infinite. From here I am able to determine the order of the factor group via Lagrange theorem.

The next part is giving me a bit of a problem in the sense that I am unable to determine the order of the element of the permutation $(4,2)$.

Thanks in advance.

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    $\begingroup$ Two comments: 1. You cannot apply Lagrange's Theorem to infinite groups. 2. $(4,2)$ is not a permutation, but an element of ${\mathbb Z} \times {\mathbb Z}$. $\endgroup$ – Derek Holt Jul 10 '16 at 13:38
  • $\begingroup$ Thank you. Poor memory I have. Thanks for the heads up $\endgroup$ – Mathematicing Jul 10 '16 at 16:43
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$\mathbb{Z}\times\mathbb{Z}/\langle(4,2)\rangle$ is an infinite group since the order of $(1,0)+\langle(4,2)\rangle$ is infinite, namely $S=\{(n,0)+\langle(4,2)\rangle\mid n\in\mathbb{Z}\}$ is infinite.

The group is not cyclic since $(2,1)+\langle(4,2)\rangle$ has order $2$ and any infinite cyclic group is isomorphic to $\mathbb{Z}$ which does not have any element of finite order.

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Hint. Think of $\mathbb{Z} \times \mathbb{Z}$ as the set of lattice points in the plane. When do two lattice points differ by an integer multiple of the vector $(4,2)$?

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