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Math Help - Group Theory

  1. #1
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    Group Theory

    Can someone explain why the following are groups:

    (R, +)
    ({0}, +)

    and the following are not:
    (N, +)
    (R, *)
    (Z\{0}, *)

    * is denoting multiplication
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  2. #2
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    Quote Originally Posted by TexasGirl
    Can someone explain why the following are groups:

    (R, +)
    ({0}, +)

    and the following are not:
    (N, +)
    (R, *)
    (Z\{0}, *)

    * is denoting multiplication
    For a set and binary operation "ordered pair" to be a group we must meet several properties:

    For an ordered pair (G,*) to be a group we must have:
    (a,b,c,e \epsilon G)
    1) The set G must be closed under *: For any a, b we must have that a*b is also a member of G.
    2) a*(b*c) = (a*b)*c (i.e. * is associative)
    3) \exists an element e such that for any a, a*e=e*a=a (i.e. an identity element exists)
    4) For every element a, \exists an element a^{-1} such that a*a^{-1}=a^{-1}*a=e (i.e. each element has an inverse.)

    So.
    (R,+) is a group because it meets all these requirements.
    ({0}, +) is a group because it meets these requirements...I would say "vacuously" because + doesn't actually do anything. Either way, e=0 and the inverse of 0 is 0. 0+(0+0)=(0+0)+0, etc.

    (N, +): Why is this not a group? What is the inverse of, say, 6? Is it an element of the set N?

    (R, *): Again, look at inverses. We may easily find the inverse of any element but 0. What is the inverse of 0?

    (Z\{0}, *): I'm not clear on the notation here. I take it that Z\{0} is the set {...,-3, -2, -1, 1, 2, 3, ...}? If so then what is the inverse of 2?

    -Dan
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