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Proving a group is abelian

  1. Aug 26, 2009 #1
    if G is a group such that (xy)[tex]^{3}[/tex] = x[tex]^{3}[/tex]y[tex]^{3}[/tex] for all x,y in G, and if 3 does not divide the order of G, then G is abelian.

    I proved an earlier result that said if there exists an n such that
    (xy)[tex]^{n}[/tex] = x[tex]^{n}[/tex]y[tex]^{n}[/tex]
    (xy)[tex]^{n+1}[/tex] = x[tex]^{n+1}[/tex]y[tex]^{n+1}[/tex]
    (xy)[tex]^{n+2}[/tex] = x[tex]^{n+2}[/tex]y[tex]^{n+2}[/tex] for all x,y in G, then G is abelian.
    I think I'm supposed to use that but I can't quite get it. If I let n = 3, then by hypothesis 3 works and with some symbol manipulation 4 works for all groups as long as 3 does. So I'm thinking that if the hypothesis holds, I should be able to show that
    (xy)[tex]^{5}[/tex]= x[tex]^{5}[/tex]y[tex]^{5}[/tex]. I went through proving it for different orders of G, but no patterns arose. any suggestions?
     
  2. jcsd
  3. Aug 26, 2009 #2
    I'm assuming G is finite since you speak of divisibility of |G|.

    I haven't shown it using your earlier result, but I did it as follows:
    First show, using strong induction, that:
    [tex](xy)^{3k} = (yx)^{3k}= x^{3k}y^{3k}[/tex]
    [tex](xy)^{3k+1} = x^{3k+1}y^{3k+1}[/tex]
    [tex](xy)^{3k+2} = y^{3k+2}x^{3k+2}[/tex]
    Next note that every element has order 3k+1 or 3k+2 for some integer k by Lagrange's theorem and the assumption that 3 does not divide |G|. Assume x has order 3k+1, then:
    [tex]y^{3k+1} = x^{3k+1}y^{3k+1} = (xy)^{3k+1} = xy(xy)^{3k} = xyx^{3k}y^{3k}[/tex]
    From which you can show xy = yx by cancelling the rightmost y's and right-multiplying by x. Do similar manipulations for |x|=3k+2 starting with the last identity.
     
  4. Aug 26, 2009 #3
    can you be more clear on how you got this?
    the inductive step as far as a i can see would be:
    suppose (xy)[tex]^{3k}[/tex] = x[tex]^{3k}[/tex]y[tex]^{3k}[/tex].
    then (xy)[tex]^{3(k+1)}[/tex]= x[tex]^{3k}[/tex]y[tex]^{3k}[/tex]x[tex]^{3}[/tex]y[tex]^{3}[/tex]. but then what...

    edit: wait i think i got it
    x[tex]^{3k}[/tex]y[tex]^{3k}[/tex]x[tex]^{3}[/tex]y[tex]^{3}[/tex] = x[tex]^{3k}[/tex](y[tex]^{k}[/tex]x)[tex]^{3}[/tex]y[tex]^{3}[/tex]=x[tex]^{3k}[/tex]x[tex]^{3}[/tex]y[tex]^{3k}[/tex]y[tex]^{3}[/tex]=x[tex]^{3(k+1)}[/tex]y[tex]^{3(k+1)}[/tex]

    just so it's clear to anyone else reading (xy)[tex]^{3}[/tex] = x[tex]^{3}[/tex]y[tex]^{3}[/tex] for all x,y in G implies (xy)[tex]^{3}[/tex] = y[tex]^{3}[/tex]x[tex]^{3}[/tex] for all x,y in G

    also, the reason for strong induction is because for the 3k + 2 induction the base case has to start at k = 2 right?
     
    Last edited: Aug 26, 2009
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