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Heres my proof for $Z^2$ not being a cyclic group can anyone confirm if this looks good.

Let $Z^2=\{ (a,b):a,b∈Z \}$

Since $Z^2$ is additive group, if it were cyclic there must be a fixed p,q such that $k(p,q) = (a,b)$ for any $ a,b∈Z$. So $(p,q) = (\frac{a}{k},\frac{b}{k})$. But if k doesnt divide a,b $(p,q)$ wont be integers and therefore not in the group so not cyclic.

Can anyone confirm if this is a valid proof

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    $\begingroup$ But this same idea would show $\mathbb{Z}$ is not cyclic. $\endgroup$
    – coffeemath
    May 14, 2018 at 20:31
  • $\begingroup$ K is multiplying the generator because Z is additive so it would be (p,q) + (p,q) + .... for any K $\endgroup$
    – NoteBook
    May 14, 2018 at 20:33
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    $\begingroup$ Regard $\Bbb{Z}^2$ as a two-dimensional vector space. Any cyclic subgroup is one-dimensional. $\endgroup$ May 14, 2018 at 20:36
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/2379053/…. $\endgroup$
    – lhf
    May 14, 2018 at 21:41

2 Answers 2

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Hint:

If $k(p,q)=(a,b)$ then $k$ must divide both $a,b$. What can you say about $(p,q)$ if you try to generate, say, $(0,1)$ and $(1,0)$?

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It is not correct. The assertion “if it were cyclic there must be a fixed $(p,q)$ such that $k(p,q) = (a,b)$ for any $a,b\in\mathbb Z$” is meaningless, since you say nothing about $k$.

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  • $\begingroup$ I've let (p,q) be added to itself k times because it would be the generator of the additive Z group so ive multiplied by K for any K $\endgroup$
    – NoteBook
    May 14, 2018 at 20:35
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    $\begingroup$ @NoteBook And the sentence that I quoted is still meaningless. $\endgroup$ May 14, 2018 at 20:36
  • $\begingroup$ Jose-- It would make sense if it were quantified. Say fixed $(p,q)$ such that for any $a,b \in \mathbb{Z}$ there exists $k \in \mathbb{Z}$ for which $(a,b)=k(p,q).$ $\endgroup$
    – coffeemath
    May 15, 2018 at 5:44

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