# Cyclic groups generator problem!

1. Jan 30, 2012

### blahblah8724

Regarding finite cyclic groups, if a group G, has generator g, then every element $h \in G$ can be written as $h = g^k$ for some k.

But surely every element in G is a generator as for any $k$, $(g^k)^n$ eventually equals all the elements of G as $n$ in takes each integer in turn.

Thanks for any replies!

2. Jan 30, 2012

### morphism

This is not true. What if you choose a k such that g^k=1 (e.g. k=|G|)?

The problem with your reasoning is that (g^k)^n=g^(kn) won't hit every element of G as n runs, because you need all elements of the form g^m for any m. If you're only taking powers of the form kn then you won't get every integral power unless you choose a good k. It's pretty easy to determine what makes k good. You simply need to recall two facts:
(1) ord(g^k)=ord(g)/gcd(ord(g),k)=|G|/gcd(|G|,k), and
(2) h generates G iff ord(h)=|G|.
Combining these, we conclude that g^k will generate G=<g> iff gcd(|G|,k)=1, i.e., iff k and |G| are coprime.

So my example above with k=|G| (or more generally k a multiple of |G|) is an example of the worst possible k to choose.

3. Jan 30, 2012

### blahblah8724

Ah I see it now, thank you!

4. Jan 31, 2012

### mathwonk

just try adding 2 to itself repeatedly in Z/6Z. note that you only get different results for n= 1,2,3.