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Low Level Abstract Algebra Question

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ab=a and ba=b, show that a^2 = a and that b^2 = b

    2. Relevant equations
    none

    3. The attempt at a solution
    Not sure if I did this correct.. but here is what I did.

    Given:
    ab = a. Multiply both by left hand multiplication by a^-1
    a^-1*a*b = 1. where a^-1*a is obviously the identity.
    so b = 1.

    Given:
    ba = b. Multiply both by left hand multiplication by b^-1.
    b^-1*b*a = 1
    1a=1
    a=1
    so a =1.

    If b = 1 and a=1, then b^2 = 1 and a^2 = 1, so a^2 = a and b^2 = 2. Did I do this correctly?
     
  2. jcsd
  3. Oct 27, 2014 #2

    RUber

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    Homework Helper

    What if you combined them. a=ab and b=ba.
    ab=abba, ba=baab.
    Can you do anything with those?
    (Edit) Or maybe other combinations might work to show the point.
     
  4. Oct 27, 2014 #3

    Dick

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    Science Advisor
    Homework Helper

    What kind of algebraic structure are you dealing with? If it's a group under multiplication then it's as easy as you say. If it's not a group (and I suspect it's not) then you'd better tell us.
     
  5. Oct 27, 2014 #4
    If ##a## and ##b## are elements of a ring, then there is no guarantee they have multiplicative inverses. So you can't necessarily do what you did. For example, take ##R## to be the ring of ##2\times2## matrices with $$a=\begin{pmatrix}1&1\\0&0\end{pmatrix}\ \text{ and }\ b=\begin{pmatrix}0&0\\1&1\end{pmatrix}$$ You get ##ab=a## and ##ba=b##, but neither matrix is invertible, and neither is the identity.

    I recommend you start with ##a^2=abab## and ##b^2=baba## and see if you can't use ##ab=a## and ##ba=b## to whittle down the right-hand sides of those equalities until you're left with what you need.
     
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