Is $\mathbb{Z}_2 \times \mathbb{Z}$ is cyclic? True/false

Question

Is $\mathbb{Z}_2 \times \mathbb{Z}$ is cyclic? True/False

My attempt: I think NO

Element of $\mathbb{Z}_2$ is $\{0,1\}$ .Now any element of the form $(1,k)$ for any $k \in \mathbb{Z} $ can not generate an element of the form $(0,m)$ since

$$(1,k)^n=(\underbrace{1+\dots+1}_{n\text{ times}},\underbrace{k+\dots+k}_{n\text{ times}})=(n,nk)$$

for all $n \in \mathbb{Z}$

Is it true ?

Answer 1

Almost. When you wrote $a+a+\cdots+a$, you should have written $k+k+\cdots+k$. Besides,$$(1,k)^n=\begin{cases}(1,nk)&\text{ if $n$ is odd}\\(0,nk)&\text{ if $n$ is even.}\end{cases}$$So, you don't get every element of $\Bbb Z_2\times\Bbb Z$. Furthermore, $(0,k)^n=(0,nk)$ and so, again, you don't get every element of $\Bbb Z_2\times\Bbb Z$.

Answer 2

You can check that $(1,0)$ and $(0,1)$ cannot be powers/multiples of the same group element: if $(a,b)$ generates the group then there are some integers $k,n$ such that $k\cdot(a,b)=(1,0)$ and $n\cdot(a,b)=(0,1)$; then

$ka\equiv1\pmod2$, $kb=0$, $na\equiv0\pmod2$ and $nb=1$.

By the first and the last it can be seen that $k$ and $n$ cannot be zero; on the other hand $kb=0$, so at least one of them is zero.