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Math Help - is the set a group wrt to multiplication

  1. #1
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    Unhappy is the set a group wrt to multiplication

    Is the set a group wrt multiplication. If not, what all conditions does it fail? If so, what's the order.

    The set {[1], [3]} ⊆ Z(sub)8.

    The book says it's a group of order 2.

    I made a Cayley table but have no clue how to check associativity, and inverse. I know it's closed by theorem.
    But how do I check associativity? I'm confused.

    If {[1]} = {[1], [2], [3], [4], [5], [6], [7]}, and
    {[3]} = {[3], [6], [1], [4], [7], [2], [5]}, then is
    {[1], [3]} = [1]}? I don't know the relevance of this. It's just something that I am confused about.

    Identity is [1], that was clear in the table.

    Inverse: Rule [a] [b] = [1]. How do I check this?

    Thanks for all your help. This is blowing my mind!
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  2. #2
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    Quote Originally Posted by yvonnehr View Post
    Is the set a group wrt multiplication. If not, what all conditions does it fail? If so, what's the order.

    The set {[1], [3]} ⊆ Z(sub)8.

    The book says it's a group of order 2.

    I made a Cayley table but have no clue how to check associativity, and inverse. I know it's closed by theorem.
    But how do I check associativity? I'm confused.

    If {[1]} = {[1], [2], [3], [4], [5], [6], [7]}, and
    {[3]} = {[3], [6], [1], [4], [7], [2], [5]}, then is
    {[1], [3]} = [1]}? I don't know the relevance of this. It's just something that I am confused about.

    Identity is [1], that was clear in the table.

    Inverse: Rule [a] [b] = [1]. How do I check this?

    Thanks for all your help. This is blowing my mind!
    {[1], [3]} with multiplication is a subgroup of order 2. the identity element is [1] and note that [3].[3] = [9] = [1].
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  3. #3
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    huh?

    Thanks, but I don't know that means.
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