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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

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    $\begingroup$ If $G$ is a group, how does the order of an element $g\in G$ relate to the order of its inverse $g^{-1}$? $\endgroup$ – froggie Feb 24 '14 at 20:21
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Hint: if $a\in G$ is a generator, than also $a^{-1}$ (they have the same order).

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  • $\begingroup$ 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? $\endgroup$ – bclayman Feb 25 '14 at 15:26
  • $\begingroup$ 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) $\endgroup$ – Astro Nauft Feb 25 '14 at 15:28
  • $\begingroup$ 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? $\endgroup$ – bclayman Feb 25 '14 at 15:40
  • $\begingroup$ Yes. You can prove it many ways - try assume by contradiction (Than it will become an obvious claim). $\endgroup$ – Astro Nauft Feb 25 '14 at 15:52
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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 $m$)

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