Originally Posted by

**goroner** I cant find answer to this:

If G is finite group of order n, G is cyclic iff there is element in G with order n, that is a^n = e (the identity in G for the operation in G).

This is absurd: in a group of order n we have that $\displaystyle x^n=e$ for ANY element in the group, and the

group may perfectly well not be cyclic. What you probably meant is that there's an element x

in the group s.t. $\displaystyle x^n =e$ AND $\displaystyle x^m\neq e\,\,\,\forall m<n\,,\,\,m,n\in\mathbb{N}$

And this is simply the definition of cyclic group...

Tonio

i dont know where to start, i thought of taking it this way but im afraid that's not the answer, or is it?..

if a is any element in G, and G is of order n, and you get this element a to power n for any n E Z, and if you get this a^n = e, then you can do this..

a^n = e => a^n-1 = a^-1, so the inverse of a is a^n-1, and you can go on like this and find all the inverses for a^k, for all k =1,..,n,

since you got a^n, you can find a^1, and you can do that like this:

a^n = e => a^n+1 = a^1, and you can go on like this and get all a^k for k=1,..,n so you have all the elements and their inverses in G making a the generator of G (G = <a>).

But im afraid this is not quite true cause i cant tell if a is the generator for sure , i lack of confidence to tell that, so if someone can guide me here :) thanks!!