I Proving a property when elements of a group commute

[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?

 

[blue_leaf77]

Isn't it the case that anana^n and bmbmb^m are in fact elements of the group?

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.

 

[Mr Davis 97]

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.

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

 

[mathman]

The proof you want seems trivial.

 

[WWGD]

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?

Why do you believe you need to prove it separately?

 

[lavinia]

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?

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.

 

[mathman]

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