Proving a property when elements of a group commute

In summary: This is easily seen to be true by induction on m. m=1: a^1b^1=a a^0b^0=a. m=2: a^2b^2=a^1b^1+a^0b^0. m=3: a^3b^3=a^2b^2+a^1b^1+a^0b^0. . . . m=n: a^n b^m=a^{n-1}bab^{m-1}.
  • #1
Mr Davis 97
1,462
44
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?
 
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  • #2
Mr Davis 97 said:
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
blue_leaf77 said:
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
The proof you want seems trivial.
 
  • #5
Mr Davis 97 said:
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
Mr Davis 97 said:
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
Proof: [tex]a^nb^m=a^{n-1}bab^{m-1}[/tex], etc.
 

1. What is a group in mathematics?

A group in mathematics is a set of elements that follow a specific set of rules or operations. These rules include closure, associativity, identity, and inverses. Groups are commonly used to study algebraic structures and symmetries.

2. How do elements of a group commute?

In a group, elements commute if their order does not matter when they are multiplied together. This means that a * b = b * a for any elements a and b in the group. Commutativity is an important property in groups as it allows for easier manipulation and calculation of elements.

3. Why is proving commutativity important in a group?

Proving commutativity in a group is important because it allows for the simplification of calculations and the identification of patterns within the group. It also helps in understanding the structure and properties of the group, which can lead to further insights and applications in mathematics and other fields.

4. What are some common techniques for proving commutativity in a group?

There are several techniques that can be used to prove commutativity in a group. These include using the properties of the group, such as associativity and identity, to manipulate the elements and show that they commute. Another technique is to use the definition of a group to show that the elements commute. Induction and direct proof are also commonly used methods for proving commutativity in a group.

5. Can a group have elements that do not commute?

Yes, it is possible for a group to have elements that do not commute. In fact, many groups do not have the property of commutativity. However, even in such cases, it is still possible to prove commutativity for subsets of the group or for specific operations within the group. The lack of commutativity does not necessarily affect the other properties and structure of the group.

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