Proving a property when elements of a group commute

  • #1
1,456
44

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
blue_leaf77
Science Advisor
Homework Helper
2,629
784
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.
 
  • #3
1,456
44
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
 
  • #4
mathman
Science Advisor
7,800
430
The proof you want seems trivial.
 
  • #5
WWGD
Science Advisor
Gold Member
2019 Award
5,289
2,711
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?
 
  • #6
lavinia
Science Advisor
Gold Member
3,204
604
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.
 
Last edited:
  • #7
mathman
Science Advisor
7,800
430
Proof: [tex]a^nb^m=a^{n-1}bab^{m-1}[/tex], etc.
 

Related Threads on Proving a property when elements of a group commute

  • Last Post
Replies
4
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
5
Views
8K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
1
Views
493
Replies
4
Views
579
Top