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I am looking at this released exam question number two which states:

Let $G$ be the group $\mathbb{Z} \times \mathbb{Z}$, let $a = (6,9) \in G$. Prove that $G/\langle a \rangle$ has an element of order 3.

I am not sure what is meant by the group $\mathbb{Z} \times \mathbb{Z}$, is it implied to be under addition or multiplication?

I have a feeling that the answer is to choose:

$$(2,3)$$ $$(2,3)^2=(4,6)$$ $$(2,3)^3=(6,9) \stackrel{?}{=} (0,0)$$

But:

  1. that would be assuming under addition, however I cannot think of a situation where multiplication would work.
  2. Doesn't $G/ \langle a \rangle$ mean that there is no $(6,9)$ in this set and $(2,3)^3$ would be undefined?
walkar
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Dair
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  • I also wanted to comment on your second bullet there. The quotient $G/\left\langle a\right\rangle$ doesn't mean that anything related to $a$ is undefined, but that we "collapse" all group elements related by some factor of $a$ into the same point, i.e. $n \sim m$ iff $n = m+ka$ for some $k$, and we partition the group into classes based on that relation. – walkar Feb 13 '15 at 04:00
  • Related: http://math.stackexchange.com/questions/419777/mathbbz-times-mathbbz-langle-a-b-rangle-is-isomorphic-to?rq=1 – Watson Aug 22 '16 at 14:41

1 Answers1

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The Cartesian product of groups follows the natural definition. In this case, we have $(z_1,z_2) + (y_1,y_2) = (z_1+y_1,z_2+y_2)$. Multiplication here is not a group operation as we are dealing with $\mathbb{Z}$, which has no multiplicative inverses for any elements that are not 1 or -1.

Recall $\left\langle a\right\rangle = \{ ...(-6,-9),(0,0),(6,9),(12,18),(18,27),... \} = \{(z6,z9) | z \in \mathbb{Z}\} $, i.e. integer multiples of $(6,9)$. (Here, multiplication stands in only for repeated addition and notional convenience.)

The quotient group $G/\left\langle a\right\rangle$ follows the normal definition for quotient groups, i.e. we associate the element $(1,2)$ with $(7,11)$, $(13,20)$, $(-5,-7)$... i.e., every element $(z_1,z_2)$ represents the equivalence class $(z_1+k6,z_2+k9)$ for any $k \in \mathbb{Z}$. In this case, $\mathbb{Z}\times \mathbb{Z} / \left\langle (6,9)\right\rangle = \mathbb{Z}_6 \times \mathbb{Z}_9$, so we must show some element of $\mathbb{Z}_6 \times \mathbb{Z}_9$ has an element of order 3. You are right in your choice of $(2,3)$, as $\overline{(2,3)} + \overline{(2,3)} + \overline{(2,3)} = \overline{(6,9)} = \overline{(0,0)}$, so $\overline{(2,3)}$ has order 3.

walkar
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  • Oh, I was thinking the quotient of two sets, so we would remove them completely, I guess group notation is slightly different then. Thank you for also clarifying there is no multiplication. :) – Dair Feb 13 '15 at 03:52
  • @Bair No problem! The notation is the only difficult bit for you, it seems. We usually use exponentiation for repeated operation for groups with multiplicative notation but multiplication for groups with additive notation, in this case. So $3\overline{(2,3)} = \overline{(6,9)} = \overline{(0,0)}$ is the more standard way to write it. – walkar Feb 13 '15 at 03:54