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.

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- December 22nd 2012, 05:03 AMwellfedShow 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.

- December 22nd 2012, 07:04 AMabenderRe: Show G is not necessarily abelian
?

- December 22nd 2012, 07:10 AMILikeSerenaRe: Show G is not necessarily abelian
Hi wellfed! :)

Well, let's start with the smallest group that is not abelian.

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)? - December 22nd 2012, 07:47 AMwellfedRe: Show G is not necessarily abelian
- December 22nd 2012, 07:59 AMILikeSerenaRe: 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?