Proof for $Z^2$ not being cyclic

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

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

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 $(1,0)$?
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 $k$.
• 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 $(a,b)=k(p,q).$ May 15, 2018 at 5:44