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?
Best Answer
I think you are getting confused along the way.
You want to show that if $G$ is an infinite cyclic group, then it has exactly two generators. This can be done by showing two things: that there are at most two generators, and then exhibiting two generators.
Exhibiting two generators is easy: if $G=\langle a\rangle$, then $a$ and $a^{-1}$ both generate; and $a\neq a^{-1}$, since $a$ has infinite order.
The bulk of your argument is an attempt at showing the other direction, namely you are trying to show:
You analyse this correctly until the end of the paragraph that begins with a parenthetical remark. You successfully conclude $n=1$ or $n=-1$. So you are done.
But then you seem to be getting confused, and continue to argue; you are already done showing that there are at most two generators, so that's where the proof should end.
If the second part of the proof was meant to be what my first part was, then you are not clear in the first part. There should be an explicit statement where you say that your argument shows there are at most two generators. Finally, there is also the issue of noting that $a\neq a^{-1}$ (which is easy, but needs to be said).