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Why the need for Lagrange's Theorem?

  1. Sep 21, 2008 #1
    I've been working on a problem and I can't seem to figure out as to why the book wants you to use Lagrange's theorem. The question is "Prove that if k is an integer that is relatively prime to the order of a finite group, then the map x --> x^k is surjective." My idea was that since n and k are relatively prime, then ns + kt = 1 so that x = x^(ns+kt) which simplifies to x = x^(ks) so that x is the image of x^s and it is surjective. So where does Lagrange's theorem come into play?
     
    Last edited: Sep 21, 2008
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  3. Sep 21, 2008 #2

    morphism

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    How do you know that x^n=1?
     
  4. Sep 21, 2008 #3
    oh, hah, sorry. I meant to say that the order of G is n, so that makes sense now.
     
  5. Sep 21, 2008 #4

    morphism

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    Just checking: does the need for Lagrange make sense now?
     
  6. Sep 21, 2008 #5
    well, not really. I'm unsure about it. Something about the order of the subgroup generated by x^k dividing the order of group, though I'm not sure why we need that.
     
  7. Sep 21, 2008 #6

    morphism

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    OK, let me rephrase me question then: how do you know that x^n=1 without using Lagrange's theorem?
     
  8. Sep 21, 2008 #7
    well, I wasn't sure if it was something that was already proved (x^(order of G) = 1). Which is what I was thinking, but yes, I know that Lagrange's theorem is necessary for "x^n = 1", though would this be the only way that we would use Lagrange's theorem in this problem?

    ah, so it makes perfect sense now why they say use lagrange's theorem. I was definitely taking this for granted since I had been doing lots of algebra before this and it was probably something I found they had covered, since I haven't felt the need to read through the book.
     
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