# Prove: Every cyclic group with order > 2 has at least 2 distinct generators

Prove: Every cyclic group with order > 2 has at least 2 distinct generators

Here's what I've got so far:

Either order of our group, $G$, is finite or infinite:

Suppose infinite: then our group is isomorphic to $Z$ under addition. This group has two distinct generators, therefore so does G.

Suppose finite: Then $G$ is isomorphic to $Z$n under addition modulo $n$. We know order of $G$ is >= 3. If order of G is odd, both 1 and 2 are generators of $Z$n under addition modulo $n$, so $G$ has at least two generators.

If order of $G$ is even, then both 1 and some odd number between 1 and $n$ are generators. How do we determine this odd number?

Does this look like I'm on the right track?

Thanks for the help, Mariogs

• If $G$ is a group, how does the order of an element $g\in G$ relate to the order of its inverse $g^{-1}$? – froggie Feb 24 '14 at 20:21

Hint: if $a\in G$ is a generator, than also $a^{-1}$ (they have the same order).
• Denote $ord(a)=n$, than $a^n=1$. so $a^{-n} =1$, therefore $(a^{-1} )^n=1$. If $ord(a^{-1} )=m<n$ than $a^{-m} =1$ so $a^m=1$, which is a contradiction (n is minimal) – Astro Nauft Feb 25 '14 at 15:28
Let $g$ be a generator of $G$ then $|g|=|G|=m$. Take a number $k$ relatively prime to $m$ then $a^k$ is also a generator of $G$. So it has exactly $\phi (m)$ generators. (the number of numbers relatively prime to $m$ smaller than $m$)