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In a problem sheet, it has asked the following

Which of the following groups are cyclic? Either find a generator or show that no generator exists. For the cyclic groups, determine how many different generators there are.

One of the groups it has listed is $\mathbb{Z}^2$, but I'm unsure as to what groups this is. The convention of the course is $m\mathbb{Z} = \left\{mn : n \in \mathbb{Z} \right\}$ under addition, so it isn't this group. It also isn't the second cyclic group, as the convention there is to use $C_2$. Is it the square numbers? The problem here is, I don't see how the square numbers form a group since no inverses exist (under either addition or multiplication). The other possibility is $\mathbb{Z} \times \mathbb{Z}$ (in fact, it's looking like the only possibility). Thanks for any clarification!

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    $\begingroup$ Assuming usual notational conventions, it does indeed mean $\mathbb{Z}\times\mathbb{Z}$. $\endgroup$ – mdp Mar 15 '13 at 11:56
  • $\begingroup$ @MattPressland Thank you! $\endgroup$ – Noble. Mar 15 '13 at 11:57
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    $\begingroup$ $\Bbb{Z} \times \Bbb{Z}$, definitely. Now, quick, show that it is not cyclic, before somebody else does it for you! $\endgroup$ – Andreas Caranti Mar 15 '13 at 11:59
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Take an element $(n,m)\in\mathbb{Z}^2=\mathbb{Z}\times\mathbb{Z}$. If it is a generator, then $(1,0)$ and $(0,1)$ are in $\mathbb{Z}\cdot(n,m)$ the group generated by $(n,m)$. Write down what this actually means. Can you reach a contradiction?

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  • $\begingroup$ Does it imply $n = m = 0$ which is clearly not a generator for $\mathbb{Z}^2$? $\endgroup$ – Noble. Mar 15 '13 at 12:19
  • $\begingroup$ @Noble. Exactly. $\endgroup$ – Julien Mar 15 '13 at 12:20
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As per comments, the group in question is $\mathbb{Z} \times \mathbb{Z}$.

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  • $\begingroup$ I posted this so this doesn't end up as a "unanswered" question... $\endgroup$ – apnorton Mar 15 '13 at 12:04

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