Exercise 2.3.12 (Examples of groups which are not cyclic )

Proof.

(a)

The orders of the elements of $Z_2\times Z_2$ are $1$ or $2$, so there is not elements of order $4$. If $Z_2\times Z_2$ was cyclic, then $Z_2\times Z_2=⟨x⟩$, where $|x|=4$. This contradiction proves that $Z_2\times Z_2$ is not cyclic.

(b)

Assume for the sake of contradiction that $Z_2\times \mathbb{Z}$ is cyclic. Then every element $y\in Z_2\times \mathbb{Z}$ satisfies $|y|=1$ or $|y|=\infty$. But the order of $(\bar{1},0)$ is $2$. This contradiction shows that $Z_2\times \mathbb{Z}$ is not cyclic.

(c)

Assume for the sake of contradiction that $\mathbb{Z}\times \mathbb{Z}$ is cyclic, so that $\mathbb{Z}\times \mathbb{Z}=⟨(a,b)⟩$ for some $(a,b)\in \mathbb{Z}\times \mathbb{Z}$. In particular $(1,1)=n(a,b)$ for some $n\in \mathbb{Z}$, i.e., $1=\mathit{na},1=\mathit{nb}$. Therefore $a\ne 0,b\ne 0$. But $(1,0)\in ⟨(a,b)⟩$, thus there is some integer $m$ such that $(1,0)=m(a,b)$. Therefore $0=\mathit{mb}$, where $b\ne 0$, so $m=0$. This gives $1=0\cdot a=0$. This contradiction shows that $\mathbb{Z}\times \mathbb{Z}$ is not cyclic.