How can we use induction to prove that (xy)^{3} = y^{3}x^{3} for all x,y in G?

[matticus]

if G is a group such that (xy)^{3} = x^{3}y^{3} 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)^{n} = x^{n}y^{n}
(xy)^{n+1} = x^{n+1}y^{n+1}
(xy)^{n+2} = x^{n+2}y^{n+2} 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)^{5}= x^{5}y^{5}. I went through proving it for different orders of G, but no patterns arose. any suggestions?

 

[rasmhop]

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:
(xy)^{3k} = (yx)^{3k}= x^{3k}y^{3k}
(xy)^{3k+1} = x^{3k+1}y^{3k+1}
(xy)^{3k+2} = y^{3k+2}x^{3k+2}
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:
y^{3k+1} = x^{3k+1}y^{3k+1} = (xy)^{3k+1} = xy(xy)^{3k} = xyx^{3k}y^{3k}
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.

 

[matticus]

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:
(xy)^{3k} = (yx)^{3k}= x^{3k}y^{3k}

can you be more clear on how you got this?
the inductive step as far as a i can see would be:
suppose (xy)^{3k} = x^{3k}y^{3k}.
then (xy)^{3(k+1)}= x^{3k}y^{3k}x^{3}y^{3}. but then what...

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

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