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Abstract algebra question

  1. Aug 29, 2007 #1
    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. jcsd
  3. Aug 29, 2007 #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?
  4. Aug 29, 2007 #3
    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.
  5. Aug 29, 2007 #4

    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
  6. Aug 29, 2007 #5


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    Hi Benzoate,

    Do you agree that this sentence is incomplete?
  7. Aug 30, 2007 #6
    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.
  8. Aug 30, 2007 #7


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    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.
  9. Aug 30, 2007 #8
    yes. sorry about that
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