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I know following
G is group in which $a^ib^i=(ab)^i$ this hold for consecutive 3 integer then it is abelian .
I can prove this fact. I know that this is not true in case of 2 consecutive integer.I searched for example but I am not getting .
Any Help will be appreciated .
Thanks
Note: Similar question will be here An example of a non-abelian group $G$ where, for all $a,b\in G$, the equality $(ab)^{n}=a^{n}b^{n}$ holds for two consecutive integers $n$ with answer But I know trivially for 0 ,1 integer will work But I am searching nontrivial example that is power is more than 2.
Also there is answer that $D_4$ will work But I am not convincing with that as there are only 2 kind of elements in that Rotation and reflection
Consider $(rs)^3$=rs and $(rs)^4=e$ which is suggested by that answer but $r^3s^3=r^3s$ which is not both same .

Curious student
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    I know that But I am not convincing with that answer.Please Help me – Curious student Apr 30 '18 at 09:50
  • Why are you not convinced by that answer? I have edited it slightly to replace $i=0, 1$ with $i=6, 7$. You should work out why, in that question, these statements are equivalent. – user1729 Apr 30 '18 at 11:02

1 Answers1

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Try in $Q_8$, the Quaternion group!

Added: $Q_8=\{1,-1,i,-i,j,-j,k,-k\}$ where $i^2=j^2=k^2=-1$ and $ij=k$, $jk=i$, $ji=-k$, $ik=j$, $ki=-j$, $kj=-i$.

In this group $(a.b)^i=a^i.b^i$ holds for $i=4,5$. But $Q_8$ is non-Abelian, Since $ij \neq ji$