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Groups whose orders have no common factors

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Let G and G' be finite groups whose orders have no common factor. Prove that the homomorphism [tex]\varphi[/tex] G [tex]\rightarrow[/tex] G' is the trivial one [tex]\varphi[/tex] (x) =1 for all x.



    3. The attempt at a solution

    My thoughts are that we need to use lagrange's thm. somehow. or maybe not.

    We have the order of G and G' such that. GCD( |G|, |G'|) =1.

    [tex]\varphi[/tex] G [tex]\rightarrow[/tex] G'

    Let |G| = n

    By legranges thm.

    gn [tex]\in[/tex] G = 1G

    and we know that [tex]\varphi[/tex] (gn)= 1G'

    But i dont really no what to do from here. It doesnt seem as if i am on the right track.

    Any thoughts?
     
  2. jcsd
  3. Nov 3, 2008 #2

    HallsofIvy

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    I presume you mean prove that the only homomorphism from G to G' is [itex]\phi(g)= 1_{G'}[/itex]. Let x be a member of G, not equal to 1G. Let n be the order of x: the smallest integer n such that xn= 1G. What is the order of [itex]\phi(x)[/itex]?
     
  4. Nov 3, 2008 #3
    The order of [tex]\varphi(x)[/tex] must divide n.
     
  5. Nov 3, 2008 #4

    HallsofIvy

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    But the order of any member of member of G must divide |G|. Since the order [itex]\phi(x)[/itex] divides n, it also divides |G|. And the order of [itex]\phi(x)[/itex] must, like the order of any member of G', must divide |G'|. What is the only number that divides both |G| and |G'|?
     
  6. Nov 3, 2008 #5
    Holy crap youre my hero.
     
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