Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Sep 15, 2010 #1
    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?
  2. jcsd
  3. Sep 18, 2010 #2
    I guess not then :biggrin:

    I was thinking that since [tex]\phi(1)->g \neq e[/tex] 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 [tex]\phi(0) = e=g^0,g^1,...,g^{p-1}[/tex] 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 [tex]\phi (k)[/tex] divides the order of k, which is p.
  4. Sep 20, 2010 #3


    User Avatar
    Science Advisor
    Homework Helper

    i guess you are essentially using lagrange for cyclic groups, which is all you need here.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook