Proof Abelian Group: Consec Ints & No Follow w/2

[pedrommp]

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 basic group identities

The Attempt at a Solution

Well, the first part i made :

Let j be the smallest of the three consecutive integers. Then we have that

$$(a.b)^j = a^j.b^j$$

$$(a.b)^{j+1} = a^{j+1}.b^{j+1}$$ and

$$(a.b)^{j+2} = a^{j+2}.b^{j+2}$$ for all $$a, b \in G$$

Inverting the first equation and right multiplying it to the second equation implies

$$a.b = a^{j+1}.b^{j+1}.b^{-j}.a^{-j} = a^{j+1}.b.a^{-j}$$ hence

$$b.a^j = a^j.b$$

Inverting the second equation and right multiplying it to the third equation gives

$$a.b = a^{j+2}.b^{j+2}.b^{-j-1}.a^{-j-1} = a^{j+2}.b.a^{-j-1}$$ hence

$$b.a^{j+1} = a^{j+1}.b$$

Therefore $$a^{j+1}.b = b.a^{j+1} = b.a^{j}.a = a^{j}.b.a$$ and left multiplying by

$$a^{-j}$$ yields $$a.b = b.a$$ for arbitrary $$a, b \in G$$

Hence G is abelian.

But the second i tried to find an example of non-abelian group that satisfies the relation for two consecutive integers with no success.

 

[Robert1986]

I might be missing something, but have you tried the contrapositive for the second part?

That is, I am interpreting this question as: The relation (the one you described for 3 consecutive integers) holds if and only if G is abelian.

Did I interpret correctly? If so, I'd try to prove the contrapositive.

 

[micromass]

Would it be cheating to assume i=0 and i=1? This would provide a counterexample...

 

[pedrommp]

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.