Cyclic groups of order greater then 2 have more then one generator

**magus** · November 12th 2010, 01:55 PM

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?

**Tinyboss** · November 12th 2010, 05:22 PM

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.

**topspin1617** · November 12th 2010, 07:12 PM

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).

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