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Group definition (checks)

  1. Feb 8, 2015 #1


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    I was wondering, if we take a "group" [itex]G[/itex] (so multiplication is defined among the elements) it forms a group if it has the following properties:
    Contains the identity element
    Contains the inverse elements
    follows associativity.

    I was wondering if associativity is not a must though... like it can be contained in the previous properties.

    Take for example [itex] g_1, g_2, g_3 \in G[/itex], then:
    [itex]g_1 g_2 g_3 = g \in G [/itex] (closure).

    Let's take the :

    [itex]g_1 (g_2 g_3) = g_1 g_k = g_h [/itex]
    [itex](g_1 g_2) g_3 = g_m g_3 = g_i [/itex]
    associativity holds if [itex]g_h = g_i [/itex] or doesn't if [itex]g_h \ne g_i [/itex]. Let's take the last assumption, that is [itex]g_h \ne g_i [/itex]
    Since the elements [itex]g_{1,2,3}[/itex] are taken arbitrarily, I can take [itex]g_{2}=g_{3}^{-1} [/itex] as well as [itex]g_1=e[/itex]. If I do that, I'm getting that [itex]g_i =g_h[/itex] which is a contradiction.

    Is that a correct thinking?
  2. jcsd
  3. Feb 8, 2015 #2


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    No. You have shown that associativity will hold as a consequence of the other properties for some elements of G, but in order to be a group, associativity must hold for all elements of G. If I can find any g1, g2, g3 for which [itex]g_h \ne g_i[/itex], then G is not a group.
  4. Feb 8, 2015 #3


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    A set with a binary operation which has an identity element and inverse elements needs not be associative. As an example, consider the set {1,a,b} with the following composition table for the binary operation:

    * | 1 a b
    1 | 1 a b
    a | a 1 b
    b | b b 1

    1 is the identity element, 1-1=1, a-1=a, b-1=b, but associativity doesn't hold, since e.g. a(bb)=a1=a but (ab)b=bb=1.
  5. Feb 8, 2015 #4

    Stephen Tashi

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    The notation for the elements is arbitrary, but if you say that a product involving [itex] g_1,g_2,g_3 [/itex] can represent an arbitrary product of 3 elements in the group then the assumption [itex] (g_1)(g_2 g_3) \ne (g_1 g_2) g_3 [/itex] becomes a claim about each combination of 3 elements in the group. For an indirect proof, you would only be allowed the assumption that there exist 3 elements whose multiplication is not associative - rather than the assumption that each set of 3 elements has a multiplication that is non-associative.
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