Prove that a cyclic group with only one generator can have at most 2 elements

Question

Prove that a cyclic group that has only one generator has at most $2$ elements.

I want to know if my proof would be valid:

Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $\gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 \neq 1$). Thus $|G|\leq 2$.

I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.

Answer 1

Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).

Answer 2

Your proof is correct if $G$ is finite, i.e. $G\cong\mathbb{Z}_m$ for some $m\ge 1$. Just notice that it may happen that $G\cong\mathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.

Answer 3

Here is another take.

The number of generators is $\phi(n)$, where $\phi$ is Euler's function.

Now, either $n$ has a prime factor $p\ge 3$ or $n$ is a power of $2$.

In the first case, we have $\phi(n) \ge \phi(p)=p-1\ge2$.

In the second case, if $n\ge 3$, then $4$ divides $n$ and so $\phi(n) \ge \phi(4)=2$.

Bottom line, $\phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.

(The key fact is: if $d$ divides $n$, then $\phi(n) \ge \phi(d)$.)