# Orders of products of group elements.

## 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!

Related Linear and Abstract Algebra News on Phys.org
Hurkyl
Staff Emeritus
Gold Member
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, $(ab)^2 = a^2 b^2$. Show that G is Abelian.

CompuChip
Homework Helper
Let n be the order of $a b (a^-1)$. Then
$$(a b a^{-1})^n = (a b a^{-1}) (a b a^{-1}) \cdots (a b a^{-1}) = e$$

(Hint: use that the group multiplication is associative -- i.e. you can put the brackets differently!)

Let $$o(aba^{-1})=m, o(b)=n$$ So it follows that :$$(aba^{-1})^m=e, b^n=e$$

now lets consider the following

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

$$ab^na^{-1}=a*e*a^{-1}=e=> m|n$$

Now:

$$b^m=e*b*e*b*e*b*e*......*b*e=(a^{-1}a)*b*(a^{-1}a)*b*(a^{-1}a)*b*......*(a^{-1}a)*b*(a^{-1}a)=$$

$$=a^{-1}*(aba^{-1})*(aba^{-1})*(aba^{-1})*(aba^{-1})*.....*(aba^{-1})*a=a^{-1}*(aba^{-1})^m*a=a^{-1}*e*a=e=>n|m$$

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:
Exercise: Suppose G is a group with the property that, for all a and b, $(ab)^2 = a^2 b^2$. Show that G is Abelian.
Suppose G is a group with the property that, for all a and b, $(ab)^2 = a^2 b^2$. Show that G is Abelian.

Let $$x,y \in G$$ we want to show that $$xy=yx ?$$

$$x(yx)y=(xy)(xy)=(xy)^2=x^2y^2=(xx)(yy)=x(xy)y$$ So:

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

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

$$x^{-1}*\x(yx)=x(xy)=>x^{-1}x(yx)=x^{-1}x(xy)=>yx=xy$$

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

Nice exercise!