Counterexample so that (ab)^i=a^ib^i for two consecutive integers

[SiddharthM]

counterexample so that (ab)^i=a^ib^i for two consecutive integers for any a and b in a group G does not imply that G is abelian.

this is a problem in herstein and I'm struggling to find an example. The previous problem to show that if (ab)^i=a^ib^i for 3 consecutive integers then G is abelian is a starred problem but seems to be easier.

 

[morphism]

Hint: It's trivial if you pick the right two consecutive integers.

 

[ice109]

im sorry but is i the square root of 1 here?

 

[LukeD]

no, i is an integer. It's a group theory problem

 

[morphism]

I wonder if we can find an easy nontrivial counterexample (i.e. where we choose a pair of values for i other than 0 and 1). I scribbled down some stuff and managed to convince myself that the symmetric and dihedral groups aren't going to be good candidates, but I haven't had the energy to pursue this any further.

Edit:
I did some searching and found http://www.groupsrv.com/science/about12002.html . Apparently the quaternion group Q works with i=4 and i=5.