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Proof for a group

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Let (R,+,*) be a commutative ring with identity. Show that (R,+) is a group but (R,*) is never a group.

    3. The attempt at a solution
    This question confuses me because I thought a group was defined for a set with a binary operation, i.e. a set that uses multiplication.

    Also, does binary operation imply multiplication and addition are defined?
     
  2. jcsd
  3. Sep 15, 2008 #2

    tiny-tim

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    Hi fk378! :smile:

    "binary operation" means any operation on only two things …

    nearly every operation we use is binary …
    "binary operation" means one operation … and multiplication and addition are two operations … so nooo. :frown:

    "binary operation" only means that a-operation-b is defined for all a and b (in that order).

    It needn't be a group, or commutative, or anything … it could be totally random!

    We can call it anything we like … we normally call an operation "multiplication", just because that's the way we write it … but that doesn't mean it has any of the properties of "usual" multiplication.

    If a set has two binary operations, obviously we can't call both of them "multiplication".

    Anyway … how would you show that (R,*) isn't a group? :smile:
     
  4. Sep 15, 2008 #3
    A set is not a group if
    1) it is not closed under the operation
    2) not associative
    3) no identity element e
    4) no inverse.

    But can't this also apply to (R,+)? How can (R,*) --never-- be a group?
     
  5. Sep 15, 2008 #4

    tiny-tim

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    Hi fk378! :smile:

    Yes … and that's what the question asks you to prove!
    i] what is the identity under * ?

    ii] does every element of R have an inverse under * ? :smile:
     
  6. Sep 15, 2008 #5
    True. I see that 0 does not have an inverse. But can't you define the set to not include 0? [1, infinity)?
     
  7. Sep 15, 2008 #6

    Dick

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    i) No, the set is (R,*). You can't just redefine it. ii) Even if you do, there still may be elements without inverses. Look (Z,+,*), the ring of integers. 2 doesn't have an inverse.
     
  8. Sep 15, 2008 #7
    Oh I see. Thank you. Dick, for your example of (Z,+,*), does any number have an inverse (besides 1)?
     
  9. Sep 15, 2008 #8

    Dick

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    -1 does. That's about it.
     
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