# 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.