If $G$ is a group in which $(ab)^i = a^ib^i$ for two consecutive integers $i$, for all $a,b \in G$, can we conclude that $G$ is abelian? Problem: 

If $G$ is a group in which $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a,b \in G$, then $G$ is abelian. If we assume the relation $(ab)^i = a^ib^i$ for just two consecutive integers can we conclude that $G$ is abelian? 

Thank all!
 A: Two consecutive integers is not enough to conclude that $G$ is abelian.
Let $G$ be any non-abelian group and $e$ the exponent of $G$ (so the least positive integer such that $g^e=1$ for all $g \in G$). Then $(ab)^e=1=a^eb^e$ and $(ab)^{e+1} = ab=a^{e+1}b^{e+1}$ for all $a,b \in G$.
A: Three is enough.
First we have that 
$$
a^{i+2} b^{i+2} = (ab)^{i+2} = (ab)^{i+1} (ab) = a^{i+1} b^{i+1} ab,
$$
that is,
$$ a^{i+2} b^{i+2} = a^{i+1} b^{i+1} ab, $$
implying that 
$$ a b^{i+1} = b^{i+1} a,$$
meaning that $i+1$ powers commute with all elements. 
Now,
$$
a^{i+2} b^{i+2} = (ab)^{i+2} = (ab)^{i} (ab)^2 = a^i b^i (ab)^2 = a^i b^i (ab)(ab) = a^i b^i ab ab ,
$$
that is,
$$ a^{i+2} b^{i+2} = a^i b^i a b a b, $$
from which it follows that 
$$ a^2 b^{i+1} = b^i a b a.$$
Reorder the terms on the LHS and conclude 
$$ b^{i+1} a^2 = b^i a b a, $$
so that 
$$ b a = ab, $$ 
as required.
A: We have:
$$(ab)^iab=(ab)^{i+1}=a^{i+1}b^{i+1}=aa^ib^ib=a(ab)^ib$$
Right-cancelling with $b^{-1}$, we get:
$$(ab)^ia=a(ab)^i$$
By a similar argument (starting with $ab(ab)^i$), we have that $b$ commutes with $(ab)^i$. This will be true in any group where $g^i=e$ for all $g$. However, this is true in any finite group, Abelian or not, if we set $i=k|G|$ for any integer $k$, for example.
