Orders of products of group elements.

  • #1

Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
Hurkyl
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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!
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.
 
  • #3
CompuChip
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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!)
 
  • #4
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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]

Now:

[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]

[tex]=a^{-1}*(aba^{-1})*(aba^{-1})*(aba^{-1})*(aba^{-1})*.....*(aba^{-1})*a=a^{-1}*(aba^{-1})^m*a=a^{-1}*e*a=e=>n|m[/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:
  • #5
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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.
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:

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

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

Nice exercise!
 

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