# Thread: Cyclic groups of order greater then 2 have more then one generator

1. ## Cyclic groups of order greater then 2 have more then one generator

My problem is to prove that cyclic groups of order 3 or greater must have at least 2 generators.

I think I found an answer but I'm worried that it's wrong.

Here's my proof.

In a group of order $n$ generated by $a$ there exists a "last element" $a^{n-1}$. This element can be shown to generate every inverse element and therefore every element in the group.

For example $a^{n-1}a^{n-1}=a^{2n}a^{-2}=ea^{-2}=a^{-2}$
$a^{n-1}a^{n-1}a^{n-1}=a^{3n}a^{-3}=ea^{-3}=a^{-3}$

and so on until $a^{n-1}a^{n-1}\cdots(n times)\cdots a^{n-1}=a^{n^2}a^{n}=e$

Thus it is shown that in every cyclic group with $n>2$ the element $a^{n-1}$ generates the group as well as $a$.

Is this valid?

2. That's right--the inverse of a generator is a generator, and a generator is never its own inverse unless it's the only nonidentity element.

3. Right, except I'm not sure if you're allowed to assume the cyclic group is finite. (I'm not sure if saying $|G|>2$ is supposed to mean that $G$ is finite or not.) In this situation, though, just replace $a^{n-1}$ with $a^{-1}$ (which, in the case of a finite cyclic group, are the same thing).