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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.

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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 $2$).

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  • $\begingroup$ Beautiful! Thank you so much for this! $\endgroup$ – Pablo Feb 20 at 7:49
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    $\begingroup$ When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy. $\endgroup$ – Marc van Leeuwen Feb 20 at 13:03
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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.

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  • $\begingroup$ That's a very good observation, thanks! $\endgroup$ – Pablo Feb 20 at 7:49
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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 $\phi(n) \ge \phi(d)$.)

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  • $\begingroup$ @darij, thanks. $\endgroup$ – lhf Feb 20 at 10:48

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