Question: An infinite cyclic group has exactly two generators.
Answer: Suppose $G=\langle a\rangle$ is an infinite cyclic group. If $b=a^{n}\in G$ is a generator of $G$ then as $a\in G,\ a=b^{m}={(a^{n})}^{m}=a^{nm}$ for some $m\in Z$.
$\therefore$ We have $a^{nm-1}=e.$
(We know that the cyclic group $G=\langle a\rangle$ is infinite if and only if $0$ is the only integer for which $a^{0}=e$.) So, we have, $nm-1=0\Rightarrow nm=1.$ As $n$ and $m$ are integers, we have $n=1,n=-1.$
Now, $n=1$ gives $b=a$ which is already a generator and $n=-1$ gives $H=\langle a^{-1}\rangle =\{(a^{-1})^{j}\mid j\in Z\} =\{a^{k}\mid k\in Z\}=G$ That is $a^{-1}$ is also generator of $g$
My question is that am I approaching this question correctly?