Abstract algebra question

1. Aug 29, 2007

Benzoate

1. The problem statement, all variables and given/known data

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

2. Relevant equations
No equations necessary for this proof

3. 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)

Last edited: Aug 30, 2007
2. Aug 29, 2007

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?

3. Aug 29, 2007

Benzoate

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. Aug 29, 2007

Benzoate

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

Last edited: Aug 29, 2007
5. Aug 29, 2007

quasar987

Hi Benzoate,

Do you agree that this sentence is incomplete?

6. Aug 30, 2007

d_leet

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?

(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. Aug 30, 2007

HallsofIvy

Staff Emeritus
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. Aug 30, 2007