Show that set of cube root of unity is a finite abelian group under multiplication

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Show that set of cube root of unity is a finite abelian group under multiplication

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Rylee 7 months 2021-10-13T17:19:11+00:00 2 Answers 0 views 0

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    0
    2021-10-13T17:20:32+00:00

    Answer:

    Step-by-step explanation:

    Cube roots of unity are (1,ω,ω²)

    Let G be the set  (1,ω,ω²).

    ×     1      ω      ω²

    ===============

    1      1      ω      ω²

    ω    ω     ω²     1

    ω²   ω²     1      ω

    To verify whether G is a group or not,

    1) Closure: Clearly, for all a,b belonging to G, a*b also belongs to G.

    (illustrated in the table)

    2)Associative: Clearly for all a,b and c belonging to G,

    a*(b*c) = (a*b)*c

    3)Existence of Identity: ‘e’ is called the identity of the group if for all a belong to G, a*e = e = e*a,

    Now from the table, it is clearly evident that e = 1 is the identity.

    4)Existence of Inverse:

    If for all a belonging to G, there exists s a b such that a*b=e, then b is called the inverse of a.

    Clearly, Inverse of 1 is 1, Inverse of ω  is ω² and Inverse of ω² is ω.

    Since all 4 properties of the group are satisfied G is a group.

    Also, it is evident from the table that for all a,b belonging to G, we have a*b = b*a

    Thus, G is commutative.

    A commutative group is called as an abelian group.

    Thus, cube roots of unity form a finite abelian group under multiplication.

    0
    2021-10-13T17:21:07+00:00

    Answer:

    ᖶᕼᓰᘉᖽᐸ ᕼᓰ ᖶᕼᗩᓰ ᗷᗩᕲᘿ ᗷᕼᗩᓰᖻᗩ Sᕼᗩᘺᕼᑘᗷ

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