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Identity Element in Groups

  1. Nov 25, 2014 #1
    Hi
    I'm taking a math course at university that covers introductory group theory. The textbook's definition of the identity element of a group defines it as two sided; that is, they say that a group ##G## must have an element ##e## such that for all ##a \in G##, ##e \cdot a = a = a \cdot e## .

    Is it possible to define a group's identity element as one-sided, and then prove two-sidedness as a theorem? Or is it an intrinsic property of group identities?

    I started with a left-identity ##e \cdot a = a## and tried to prove that ##e \cdot a = a = a \cdot e## and kept hitting walls, so I though I'd better check if it's doable.

    Thanks.
     
    Last edited: Nov 25, 2014
  2. jcsd
  3. Nov 25, 2014 #2

    Stephen Tashi

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    If you change the assumption that there exists a two-sided identity, how are you going to modify the assumption that each element has an inverse? That assumption mentions the two-sided identity.
     
  4. Nov 26, 2014 #3
    Thanks for the reply Stephen Tashi. :)
    In my book, the assumption that each element has an inverse is stated as follows: for all ##a \in G## , there exists a ##b \in G## such that ##a \cdot b = e##. That doesn't seem to rely on the two-sidedness of the identity element (and neither it seems to me does the left-handed version of the statement, for all ##a \in G## , there exists a ##b \in G## such that ##b \cdot a = e##). So other than noting that in these definitions ##e## refers to a left identity, I'm not seeing how I would need to modify them.

    Could you please clarify this?
     
  5. Nov 26, 2014 #4

    rubi

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    You can weaken the group axioms to require only the existence of a left inverse and a left neutral element:
    (i) ## e g = g##
    (ii) ## g^{-1} g = e##

    Then every left inverse is a right inverse:
    ##g g^{-1} = e g g^{-1} = (g^{-1})^{-1} g^{-1} g g^{-1} = (g^{-1})^{-1} g^{-1} = e##
    And every left neutral element is a right neutral element:
    ## g e = g g^{-1} g = e g = e##.
    (The proof uses the fact that ##g^{-1}## is a right inverse, which I proved before.)

    You can of course replace "left" by "right" and get similar results. You can't mix them, though: Having left inverses and a right neutral element doesn't suffice.
     
  6. Nov 26, 2014 #5

    Stephen Tashi

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    The point is that the meaning of the statement [itex] a \cdot b = e [/itex] depends on what you mean by the notation "[itex] e [/itex]".
     
  7. Nov 26, 2014 #6
    Thank you both very much!
     
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