Prove that any finite cyclic group with more than two elements has at least two different generators

Prove that any finite cyclic group with more than two elements has at least two different generators.

Attempt: Let $G$ be a finite cyclic group with 3 elements, say $e,g,g^{-1}\in G$ such that $e=$identity element. If $G$ is a cyclic group then it must have a generator, say $g$. If $g$ generates $G$ then so does $g^{-1}$ since $o(g)=n$ implies $g^n=e$, and $g^{-n}=(g^{-1})^n=e.$ Therefore, $G$ has at least two generators; $g$ and $g^{-1}.$

(I know this is a possible duplicate but I am asking a question specifically about my proof. Is it correct?)

• I think that you should say that $G$ is a finite cyclic group with at least three elements, then label three of them as $e$, $g$, and $g^-1$. The rest of the proof looks fine to me. – Andrew Stelzer Nov 5 '16 at 18:09
• Or you could say, for n>2, there are two prime numbers less than n. – jnyan Nov 5 '16 at 18:13
• @jnyan I cannot see the relevance of prime numbers less than $n$, and it in any case it is not true for $n=3$. – Derek Holt Nov 5 '16 at 18:57
• Cyclic group is isomorphic to Zn. So one generator is fixed. For given Zn, prime numbers will be generator. For this case consider 1 as prime, since it will also be generator. – jnyan Nov 5 '16 at 19:15
• The title is a duplicate, but what you are asking about your proof is OK. However, a much better idea would be to see in general that a cyclic group of order $n$ has $\phi(n)$ different generators. And $\phi(n)\ge 2$ for $n\ge 2$. – Dietrich Burde Nov 5 '16 at 19:27

As mentioned in the comments $g^{-n} = e$ does not imply that $g^{-1}$ has order $n$, only that the order of $g^{-1}$ is a divisor of $n$. You should base your proof that the number of elements of order $n$ is $\Phi(n)$ (the Euler totient function).