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

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 $$\mathbb Z$$ under addition. This group has two distinct generators, therefore so does $$G$$.

Suppose finite: Then $$G$$ is isomorphic to $$\mathbb Z_n$$ under addition modulo $$n$$. We know order of $$G$$ is $$\geqslant 3$$. If order of $$G$$ is odd, both 1 and 2 are generators of $$\mathbb Z_n$$ 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?

• 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

## 3 Answers

Hint: if $a\in G$ is a generator, than also $a^{-1}$ (they have the same order).

• I think I'm being obtuse, but I don't see why they have to have the same order. I get that a^-1 can't have order greater than a (since a has order = number of elements in G), but why can't a^-1 have a smaller order than a? – anon_swe Feb 25 '14 at 15:26
• 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
• to be clear, in the first part a^n = 1 => a^-n = 1 because (a^n)(a^-n) = 1 by def, and we have a^n = 1. yes? – anon_swe Feb 25 '14 at 15:40
• Yes. You can prove it many ways - try assume by contradiction (Than it will become an obvious claim). – Astro Nauft Feb 25 '14 at 15:52

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 $m$)

It suffices to prove that $$\varphi(m)\ge2$$ for $$m\gt2$$, where $$\varphi$$ is Euler's totient function. But for $$p$$ prime, $$\varphi(p^n)=p^n-p^{n-1}$$, by counting. And by the Chinese remainder theorem, $$\varphi$$ is multiplicative. The result follows.