Cyclic groups generator problem

[blahblah8724]

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

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

Thanks for any replies!

 

[morphism]

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

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.

 

[blahblah8724]

Ah I see it now, thank you!

 

[mathwonk]

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

 

[blue_raver22]

You are correct in saying that every element in a finite cyclic group can be a generator. This is because every element in a finite cyclic group has a unique order, which is the smallest integer n such that g^n = e, where e is the identity element. This means that for any element h in G, there exists an integer k such that h = g^k.

However, the statement in the problem is referring to a specific generator g, which is an element that generates the entire group G. This means that every element in G can be written as g^k, where k is some integer. So while every element in G can be a generator, there is a specific element g that generates the entire group.

In other words, the statement is saying that g is a "universal" generator for the group G, in the sense that it can generate all the elements in the group. This is a useful property of cyclic groups and is often used in group theory and cryptography.

I hope this clarifies the concept for you. Keep exploring and asking questions, as that is the essence of science!