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Why a set is not a group?

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Why does the set {[tex]\overline{1}[/tex],[tex]\overline{2}[/tex],[tex]\overline{3}[/tex]} of nonzero classes modulo 4 fail to be a group under addition? Under multiplication?


    2. Relevant equations



    3. The attempt at a solution

    Under addition it fails because there is no identity element such that e+s=s for s [tex]\epsilon[/tex] S. (ie 0)
    And because theres no additive inverse (ie -1, -2, -3) such that -a + a = e (0)

    Right?

    And under multiplication if fails because theres no multiplicative inverse such that a-1[tex]\cdot[/tex]a=e

    Right?
     
  2. jcsd
  3. Nov 6, 2009 #2

    lanedance

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    Homework Helper

    be a little careful, you need to show/check what you're stating, which in the examples you've given means testing alot of cases.. a counter example may be easier...

    for the identity element, this works, but you need to pick an element & test all otehrs to show they're not identities
    for the additive inverse, note 3+2 = 1, so some additive inverses exist
    for a counter example, is it closed under addition...?

    for the multiplticative group
    1*1 = 1
    3*3 = 1
    so some inverses exist... so that argument isn't solid, but can be used for 2
    once again, might be worth checking closure...
     
  4. Nov 6, 2009 #3

    lanedance

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    edited & updated above for clarity
     
  5. Nov 6, 2009 #4
    Ahhhh, I see.

    I still have to do a mod 4 after performing the addition and multiplication operations. I knew I was wrong but wasn't sure why. Thanks.
     
  6. Nov 6, 2009 #5
    So under addition: 1+3 = 0 and zero is not in the set, not closed
    Under multiplication 2x2 = 0, not closed
     
  7. Nov 6, 2009 #6
    Correct
     
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