Abstract algebra question

In summary, this proof states that if (ab)^2=a^2*b^2, then (ab)^2=a^2*b^2. This is straightforward, and you almost had it. However, e was invoked to show that this is also true for any a and b in the group, which is what the proof is trying to show.
  • #1
Benzoate
422
0

Homework Statement



Prove that if (ab)^2=a^2*b^2, in a group G, then ab =ba

Homework Equations


No equations necessary for this proof

The Attempt at a Solution



Suppose (ab)^2=a^2*b^2. Then (ab)^2=(ab)(ab)=(abba)=(ab^2*a)=a^2 *b^2=> (ab)(ba)=(ba)(ab) = e

By cancellation, (ab)=(ba) <=> (ba)=(ab)
 
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  • #2
If [itex](ab)^2=a^2b^2[/itex] what? Is the group commutative? Then this holds, for all a and b.

I think the question is supposed to be posed as thus: Suppose G is an abelian group. Then for any a and b in G we have [itex](ab)^2=a^2b^2[/itex] .

This is fairly straightforward, and you almost had it. If you hadn't written anything after a^2 *b^2 you would've been done. Everything after that is nonsense. Why was e invoked?

Are you self-learning algebra, by any chance?
 
  • #3
ZioX said:
If [itex](ab)^2=a^2b^2[/itex] what? Is the group commutative? Then this holds, for all a and b.

I think the question is supposed to be posed as thus: Suppose G is an abelian group. Then for any a and b in G we have [itex](ab)^2=a^2b^2[/itex] .

This is fairly straightforward, and you almost had it. If you hadn't written anything after a^2 *b^2 you would've been done. Everything after that is nonsense. Why was e invoked?

Are you self-learning algebra, by any chance?

No, but with my class schedule , its almost impossible to meet my professor for extra help during his office hours. In addition, the textbook (contemporary abstract algebra by joseph gallian ) doesn't do a good job of showing each part of the proof , step by step. Someone once said that in order to be good proof writer , you have to read and observe a lot of proof , just as a good writer has to read a lot of books. the textbooks does have examples , but you are expected to connect the dots for the proof.
 
  • #4
ZioX said:
If [itex](ab)^2=a^2b^2[/itex] what? Is the group commutative? Then this holds, for all a and b.

I think the question is supposed to be posed as thus: Suppose G is an abelian group. Then for any a and b in G we have [itex](ab)^2=a^2b^2[/itex] .

This is fairly straightforward, and you almost had it. If you hadn't written anything after a^2 *b^2 you would've been done. Everything after that is nonsense. Why was e invoked?

Are you self-learning algebra, by any chance?
What do you mean I almost had it if I did not write anything after a^2*b^2 ? I barely wrote anything before I wrote a^2 *b^2. a^2*b^2 was only an assumption that had to be made for the proof
 
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  • #5
Benzoate said:

Homework Statement



Prove that if (ab)^2=a^2*b^2

Hi Benzoate,

Do you agree that this sentence is incomplete?
 
  • #6
Benzoate said:

Homework Statement



Prove that if (ab)^2=a^2*b^2

If that then what? Is the rest of that sentence supposed to be that if this is true in a group then the group is abelian?

Benzoate said:

The Attempt at a Solution



Suppose (ab)^2=a^2*b^2. Then (ab)^2=(ab)(ab)=(abba)=(ab^2*a)=a^2 *b^2=> (ab)(ba)=(ba)(ab) = e

By cancellation, (ab)=(ba) <=> (ba)=(ab)

(ab)(ab) is not equal to (abba) unless the group is abelian, which I suspect is what you are trying to prove, so you cannot assume this.
 
  • #7
I thought I had posted this before.

In any group, it is easy to prove that (ab)-1= b-1a-1.
You are given that (ab)-1= a-1b-1. In other words, since the inverse is unique, a-1b-1= b-1a-1. You should be able to manipulate that to get ab= ba.
 
  • #8
quasar987 said:
Hi Benzoate,

Do you agree that this sentence is incomplete?

yes. sorry about that
 

1. What is abstract algebra?

Abstract algebra is a branch of mathematics that deals with algebraic structures such as groups, rings, and fields. It studies the properties and relationships between these structures, rather than focusing on specific numerical values.

2. What are the applications of abstract algebra?

Abstract algebra has many practical applications in fields such as cryptography, coding theory, and physics. It also provides a foundation for higher-level mathematics such as topology and algebraic geometry.

3. What are some basic concepts in abstract algebra?

Some basic concepts in abstract algebra include groups, which are sets of elements with a defined operation and properties, and rings, which are sets of elements with two operations (usually addition and multiplication) and specific properties. Other important concepts include homomorphisms, substructures, and isomorphisms.

4. What are the differences between abstract algebra and traditional algebra?

Traditional algebra focuses on manipulating specific numerical values and solving equations, while abstract algebra deals with algebraic structures and their properties. Abstract algebra also uses more abstract concepts and notation, such as groups and rings, rather than specific numbers and variables.

5. Is abstract algebra difficult to learn?

Like any branch of mathematics, abstract algebra can be challenging to learn, but it ultimately depends on the individual. Some people may find it easier to grasp abstract concepts, while others may struggle with the more abstract nature of the subject. With dedication and practice, anyone can learn abstract algebra.

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