Math Help - Show G is not necessarily abelian

1. Show G is not necessarily abelian

Let a,b be elements of a group G. Given that (ab)^i = (a^i)(b^i) for two consecutive integers i. Show that G is not necessarily abelian. I'm am unable to find an counter example to show this.

2. Re: Show G is not necessarily abelian

$D_4$?

3. Re: Show G is not necessarily abelian

Originally Posted by wellfed
Let a,b be elements of a group G. Given that (ab)^i = (a^i)(b^i) for two consecutive integers i. Show that G is not necessarily abelian. I'm am unable to find an counter example to show this.
Hi wellfed!

That is S3, the permutations of 3 elements.

Now consider which values "i" are interesting.
For instance, i=1 is okay, since you'll get a trivial identity.
But for instance, i=2 is a problem, since abelian follows immediately (can you show that?).

What are other interesting values for "i" in the sense that you can say something immediately (as related tot S3)?

4. Re: Show G is not necessarily abelian

Originally Posted by ILikeSerena
Hi wellfed!

That is S3, the permutations of 3 elements.

Now consider which values "i" are interesting.
For instance, i=1 is okay, since you'll get a trivial identity.
But for instance, i=2 is a problem, since abelian follows immediately (can you show that?).

What are other interesting values for "i" in the sense that you can say something immediately (as related tot S3)?
Hi ILikeSerena,

Thanks for the hint. I guess I simply didn't have the patience to examine the S3 carefully before.
I found that (2 3)^6(1 2)^6=[(2 3)(1 2)]^6 = e , (2 3)^7(1 2)^7=[(2 3)(1 2)]^7 = (1 2 3) , and (2 3)(1 2) ne (1 2)(2 3).

5. Re: Show G is not necessarily abelian

Yep. Patience helps. ;-)

Did you notice that you've selected the order of the group, which is equivalent to i=0?
And also i=7, which is equivalent to i=1, since 6 is the order of the group?