I Proving a property when elements of a group commute

1. Mar 29, 2017

Mr Davis 97

By commutative, we know that $ab = ba$ for all a,b in G. Thus, why do we need to prove separately that $a^n b^m = b^ma^n$? Isn't it the case that $a^n$ and $b^m$ are in fact elements of the group? So shouldn't the fact that they commute automatically be implied?

2. Mar 29, 2017

blue_leaf77

Yes but each of $a^n$ and $b^m$ might equal another element of the group and they need not commute if the group is non-Abelian.

3. Mar 29, 2017

Mr Davis 97

Oh, I see. I neglected to see that the group is not actually abelian

4. Mar 30, 2017

mathman

The proof you want seems trivial.

5. Apr 1, 2017

WWGD

Why do you believe you need to prove it separately?

6. Apr 2, 2017

lavinia

If you know that the group is commutative then there is nothing to prove. $a^nb^m = b^ma^n$ by definition. If you do not know that the group is commutative then you need a proof.

Last edited: Apr 2, 2017
7. Apr 3, 2017

mathman

Proof: $$a^nb^m=a^{n-1}bab^{m-1}$$, etc.