I think i got it, if i take S_3 then for all a , b \in S_3 we have
(a.b)^0=e=e.e=a^0.b^0 and
(a.b)^1=a^1.b^1
Hence (a.b)^i=a^i.b^i for two consecutive integers but S_3 is not abelian.
Homework Statement
If G is a group in which (a.b)^i=a^i.b^i for three consecutive integers i for all a,b \in G , show that G is abelian.
Show that the conclusion does not follow if we assume the relation (a.b)^i=a^i.b^i for just two consecutive integers.
Homework Equations
The...