# Orders of products of group elements.

1. Nov 1, 2008

### sairalouise

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. Nov 1, 2008

### Hurkyl

Staff Emeritus
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.

3. Nov 1, 2008

### CompuChip

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!)

4. Nov 2, 2008

### sutupidmath

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: Nov 2, 2008
5. Nov 2, 2008

### sutupidmath

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!