# Proving a group is abelian

1. Aug 26, 2009

### 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?

2. Aug 26, 2009

### 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.

3. Aug 26, 2009

### matticus

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?

Last edited: Aug 26, 2009