Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Introduction to Group Theory - Abstract Algebra

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that if (ab)2 = a2b2 in a group G, then ab = ba.

    2. Relevant equations
    * For each element a in G, there is an element b in G (called the inverse of a) such that ab = ba = e (the identity).
    * For each element in G, there is a unique element b in G such that ab = ba = e.
    * The operation (under which the group is defined) is associative; that is, (ab)c = a(bc).

    3. The attempt at a solution
    [tex](ab)^{2} = (a)^{2}(b)^{2}[/tex]
    [tex](a)^{-1}(ab)^{2}(b)^{-1} = (a)^{-1}(a)^{2}(b)^{2}(b)^{-1}[/tex]
    [tex](a)^{-1}abab(b)^{-1} = (a)^{-1}aabb(b)^{-1}[/tex]
    ebae = eabe
    ba = ab.

    Is this allowed? (Keep in mind, the operation which the group is defined over is not necessarily multiplication. it may be composition.)
    Last edited: Jan 16, 2009
  2. jcsd
  3. Jan 16, 2009 #2
    Sure is but your pre last line should be ba = ab I believe.

    Also your 2nd "given" is the same as your first, you probably meant something else?
  4. Jan 16, 2009 #3
    The second given is the uniqueness statement. The first given is the rule that an element in a group has at least an inverse (not necessarily unique). I said "the" when I meant "an".

    edit: Thanks for correcting my mistake. It's a typo. I had it right on paper, I just didn't type out what I wrote down for some reason. . .
    Last edited: Jan 16, 2009
  5. Jan 16, 2009 #4
    Then that seems redundant, line 2 is stronger than line 1.
  6. Jan 16, 2009 #5
    Your work is fine, where I read it as (line 1) => (line 2) => ... .

    However, depending on the level of pendency of your instructor you may be required to justify your use of associativity. E.g., the left-hand side is [tex](ab)^2 = (ab)(ab) = a(b(ab)) = a((ba)b)[/tex], and the right-hand side is [tex]a^2b^2 = (aa)(bb) = a(a(bb)) = a((ab)b)[/tex]; then multiply both sides on the left by [tex]a^{-1}[/tex], and so on.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook