Groups of prime order are cyclic. (without Lagrange?)

[TwilightTulip]

I know full well the proof using Lagrange's thm. But is there a direct way to do this without using the fact that the order of an element divides the order of the group?

I was thinking there might be a way to set up an isomorphism directly between G and Z/pZ.

Clearly all non-zero elements of Z/pZ are generators, so sending any non-identity element of G to any equivalence class of 1,...,p-1 should induce an isomorphism... but how to prove it?

 

[TwilightTulip]

I guess not then

I was thinking that since \phi(1)->g \neq e works, there must be a way to construct such an ismorphism that makes it easy to "play" with the properties of a group to get the isom to work out.

I suppose simply showing that \phi(0) = e=g^0,g^1,...,g^{p-1} are all unique would be enough. But this would e hard without using the fact that the order of an elt must divide p, unless of course this is provable without using lagrange?

I;ve come up with a way, but it assumes we know the properties of Z/pZ and a certain property of homomorphisms: that the order of \phi (k) divides the order of k, which is p.

 

[mathwonk]

i guess you are essentially using lagrange for cyclic groups, which is all you need here.