- #1
matticus
- 107
- 1
if G is a group such that (xy)[tex]^{3}[/tex] = x[tex]^{3}[/tex]y[tex]^{3}[/tex] 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)[tex]^{n}[/tex] = x[tex]^{n}[/tex]y[tex]^{n}[/tex]
(xy)[tex]^{n+1}[/tex] = x[tex]^{n+1}[/tex]y[tex]^{n+1}[/tex]
(xy)[tex]^{n+2}[/tex] = x[tex]^{n+2}[/tex]y[tex]^{n+2}[/tex] 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)[tex]^{5}[/tex]= x[tex]^{5}[/tex]y[tex]^{5}[/tex]. I went through proving it for different orders of G, but no patterns arose. any suggestions?
I proved an earlier result that said if there exists an n such that
(xy)[tex]^{n}[/tex] = x[tex]^{n}[/tex]y[tex]^{n}[/tex]
(xy)[tex]^{n+1}[/tex] = x[tex]^{n+1}[/tex]y[tex]^{n+1}[/tex]
(xy)[tex]^{n+2}[/tex] = x[tex]^{n+2}[/tex]y[tex]^{n+2}[/tex] 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)[tex]^{5}[/tex]= x[tex]^{5}[/tex]y[tex]^{5}[/tex]. I went through proving it for different orders of G, but no patterns arose. any suggestions?