Is $\mathbb{Z}^2$ cyclic? Is $\mathbb{Z}^2$ cyclic? What does it mean for a group to be cyclic? Is it just that it has one generator?
Thanks
 A: In general one has the following for finite groups. Let $G$ and $H$ be cyclic groups then $G \times H$ is cyclic if and only if $\gcd(|G|,|H|)=1$. If $G$ is an infinite group and $H$ is any non-trivial group then $G \times H$ is never cyclic. 
We call a group $G$ (written multiplicatively) cyclic if there exists $g \in G$ such that $\{g^n : n \in \mathbb Z\}=G$. Or rather that $G$ is generated by a single element. As I've noted in my comment, this almost never means there is only one generator. 
A: $\mathrm{End}(\mathbb{Z}^2) = M_2(\mathbb{Z})$ is not commutative, hence $\mathbb{Z}^2$ is not cyclic.
A: Notice that, if $G$ is a cyclic group, for any $g,h \in G$, there exist $m,n \in \mathbb{Z}_{\neq 0}$ such that $g^m=h^n$. But in $\mathbb{Z}^2$, $h \cdot (1,0)= (h,0) \neq (0,k)=k \cdot (0,1)$ for all $k,h \neq 0$.
A: No, it isn't.
For all $a,b\in\mathbb{Z}$ it holds that $\langle (a,b)\rangle = \{(ka,kb) \mid k\in\mathbb{Z}\} \neq \mathbb{Z}^2$. So $\mathbb{Z}^2$ is not generated by a single generator and hence not cyclic.
EDIT
To address the comments "There is no argument, you just rephrase the claim." and "Nothing is wrong, but in fact this is because nothing is there." by Martin Brandenburg, I add details. By the way, I don't agree with these comments, since point 1. below is basic knowledge and 2. is quite obvious.
1. If $(G,\cdot)$ is a group and $g\in G$, then $\langle g\rangle = \{g^k \mid k\in \mathbb Z\}$.
By definition, $\langle g\rangle$ is the intersection of all subgroups of $G$ containing $g$. It is straightforward to check that $\{g^k \mid k\in \mathbb Z\}$ is a subgroup of $G$, showing "$\subseteq$". As the intersection of subgroups, $\langle g\rangle$ is a subgroup. Since any subgroup containing $g$ must contain all the $g^k$, too, we get "$\supseteq$".
Application to the additively written group $\mathbb Z^2$ yields $\langle (a,b)\rangle = \{k\cdot (a,b) \mid k\in\mathbb Z\} = \{(ka,kb) \mid k\in\mathbb Z\}$.
2. $\{(ka,kb) \mid k\in\mathbb Z\} \neq\mathbb Z^2$
If $a = 0$, then $(1,0)\notin \{(ka,kb) \mid k\in\mathbb Z\}$ and otherwise $(a,b+1)\notin \{(ka,kb) \mid k\in\mathbb Z\}$.
A: There are only two kinds of cyclic groups: $\mathbb{Z}$ and $\mathbb{Z}/\left(n\mathbb{Z}\right)$.  This is easy to see.  If $G$ is an infinite cyclic group generated by $x$, then $G=\{x^m:m \in \mathbb{Z}\}$, which suggests the isomorphism $x^m\mapsto m$.  The same argument works for $\mathbb{Z}/\left(n\mathbb{Z}\right)$.
Since $\mathbb{Z}^2$ is infinite, it would have to be isomorphic to $\mathbb{Z}$, which is easily shown to be impossible.

 Letting $\phi:\mathbb{Z}\rightarrow \mathbb{Z}^2$ and $\phi(1)=(x,y)$, we have by the homomorphic property that $\phi(m)=(mx,my)$ for any $m\in \mathbb{Z}$.  However, then there's no $z\in \mathbb{Z}$ for which $\phi(z)=(x,y+1)$.

A: Suppose $m,n\in\mathbb{Z}$ are such that $\mathbb{Z}\times\mathbb{Z}=⟨(m,n)⟩$,
 then there exists $x\in\mathbb{Z}$ s.t.
$x(m,n)=(1,1)\implies xm=1=xn\implies x=\frac{1}{m}=\frac{1}{n} \implies m=n=1$
Well, now it must also be true that there exists $y \in \mathbb{Z}$ s.t.∶
$y(m,n)=(0,1)\implies ym=0,yn=1
\implies m=\frac{0}{y}=0 ,n=\frac{1}{y}\implies y=1,
m=0 ,n=1$     Contradiction...
