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Orders of products of group elements.

  1. Nov 1, 2008 #1
    I'm looking to show that:
    Order (ab(a^-1)) = Order b
    So far...
    Let x be order of ab(a^-1), so we have:
    e = (ab(a^-1)^x)
    = (a^x)(b^x)(a^(x-1))
    = (a^x)(b^x)((a^-1)^x) so by associativity we have...
    = (a^x)((a^-1)^x)(b^x)
    = (((a)(a^-1))^x)(b^x)
    = (e^x)(b^x)
    = b^x
    Hence x is the order of b aswell.
    Im really not sure of how indices work with group elements, is it the same as with actual numbers? ie does (ab)^x = (a^x)(b^x) in groups? Im not sure it does!
    Any help greatly appreciated!
  2. jcsd
  3. Nov 1, 2008 #2


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    Exercise: Suppose G is a group with the property that, for all a and b, [itex](ab)^2 = a^2 b^2[/itex]. Show that G is Abelian.
  4. Nov 1, 2008 #3


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    Let n be the order of [itex]a b (a^-1)[/itex]. Then
    [tex](a b a^{-1})^n = (a b a^{-1}) (a b a^{-1}) \cdots (a b a^{-1}) = e[/tex]

    (Hint: use that the group multiplication is associative -- i.e. you can put the brackets differently!)
  5. Nov 2, 2008 #4
    Let [tex]o(aba^{-1})=m, o(b)=n[/tex] So it follows that :[tex] (aba^{-1})^m=e, b^n=e[/tex]

    now lets consider the following

    [tex] [aba^{-1}]^n=aa^{-1}(aba^{-1})*(aba^{-1})*(aba^{-1})*(aba^{-1})...*(aba^{-1})=a(a^{-1}a)*b*(a^{-1}a)*b*(a^{-1}a)*b*(a^{-1}a)*.....*b*a^{-1}=[/tex]

    [tex]ab^na^{-1}=a*e*a^{-1}=e=> m|n[/tex]


    [tex] b^m=e*b*e*b*e*b*e*......*b*e=(a^{-1}a)*b*(a^{-1}a)*b*(a^{-1}a)*b*......*(a^{-1}a)*b*(a^{-1}a)=[/tex]


    Now since n|m and m|n => n=m what we had to prove.

    P.S. This was in one of my recent exams in abstract! How would you rank such a problem in an exam in a first undergrad course in abstract algebra?
    Last edited: Nov 2, 2008
  6. Nov 2, 2008 #5
    Suppose G is a group with the property that, for all a and b, [itex](ab)^2 = a^2 b^2[/itex]. Show that G is Abelian.

    Let [tex]x,y \in G[/tex] we want to show that [tex] xy=yx ?[/tex]

    [tex]x(yx)y=(xy)(xy)=(xy)^2=x^2y^2=(xx)(yy)=x(xy)y[/tex] So:

    [tex] x(yx)y=x(xy)y /*y^{-1}=> x(yx)yy^{-1}=x(xy)yy^{-1}=>x(yx)=x(xy)[/tex]

    Now multiplying by the inverse of x from the left we get:


    So, since x,y were arbitrary, we conclude that G is abelian.

    Nice exercise!
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