Does (ab)^2 = a^2b^2 Imply Commutativity in Group G?

[Benzoate]

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)

 

[ZioX]

If $(ab)^2=a^2b^2$ 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 $(ab)^2=a^2b^2$ .

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?

 

[Benzoate]

If $(ab)^2=a^2b^2$ 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 $(ab)^2=a^2b^2$ .

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.

 

[Benzoate]

If $(ab)^2=a^2b^2$ 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 $(ab)^2=a^2b^2$ .

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

 

[quasar987]

Homework Statement

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

Hi Benzoate,

Do you agree that this sentence is incomplete?

 

[d_leet]

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?

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.

 

[HallsofIvy]

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.

 

[Benzoate]

Hi Benzoate,

Do you agree that this sentence is incomplete?

yes. sorry about that