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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

    mathwonk

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