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Math Help - Lagranges theorem

  1. #16
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam_leeds View Post
    the inverse of -1 is -1 and is in the set

    1 times -1 = -1 which is in the set which shows closure

    but i thought you said 1 couldnt be in the subset?
    No - the subset cannot be precisely \{1\} if it is trivial, but 1 must be in it.
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  2. #17
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    Quote Originally Posted by Swlabr View Post
    No - the subset cannot be precisely \{1\} if it is trivial, but 1 must be in it.
    so the non trivial subgroups are

    {1,-1}
    {1,i}
    {1, -i}

    {1,-1,i}
    {1,-1,-i}
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  3. #18
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam_leeds View Post
    so the non trivial subgroups are

    {1,-1}
    {1,i}
    {1, -i}

    {1,-1,i}
    {1,-1,-i}
    No. Remember what the theorem you are trying to show says. Which of your sets don't adhere to what it says?

    The problem with some of your given sets is closure: a, b \in H \Rightarrow ab \in H.
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  4. #19
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    Quote Originally Posted by Swlabr View Post
    No. Remember what the theorem you are trying to show says. Which of your sets don't adhere to what it says?

    The problem with some of your given sets is closure: a, b \in H \Rightarrow ab \in H.
    bottom two because -1 * 1 * i = -i which is not in the subgroup

    -1 * 1 * -i = i which is not in the subgroup
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  5. #20
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam_leeds View Post
    bottom two because -1 * 1 * i = -i which is not in the subgroup

    -1 * 1 * -i = i which is not in the subgroup
    Yes, but you just multiply two elements together and not all of them. The product of any two elements in the set must be in the set.
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  6. #21
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    Quote Originally Posted by Swlabr View Post
    Yes, but you just multiply two elements together and not all of them. The product of any two elements in the set must be in the set.
    i love you, cheers for your help
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