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Math Help - Algebra, element orders

  1. #1
    Kai
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    Algebra, element orders

    Hi, need quick help with these questions, have exams on Tuesday

    1) Let G be a group. Prove that
    (xax^-1)^n=xa^nx^-1

    I have been able to prove it, first with n=1, then assuming it holds for n=k and proving for n=k+1 and n=-m for some +ve int m

    But the lecturer used n=0 for the first one,

    For n=0, he got (xax^-1)^0=e and for RHS xa^0x^-1=xex^-1

    I dont understand how a^0=e or (xax^-1)^0=e

    2) Let G be a group and a is an element of G. Prove o(a)=o(a^-1)

    For second question i have no idea at all, what is o(a) ??

    Third question i have been able to do, needs to verify if its corect

    3) A relation R is defined on Z by aRb iff a-b is divisible either by 5 or by 7(a,b in Z). Is R an equivalence relation on Z? Justify your answer.

    For reflexive, 5|a-a or 7|a-a = 5|0 or 7|0
    Hence aRa.

    For symmetric, aRb implies 5|a-b or 7|a-b = 5|b-a or 7|b-a which implies bRa.

    For transitivity, aRb and bRc implies
    5|a-b or 7|a-b and 5|b-c or 7|b-c
    = 5|a-b+b-c or 7|a-b+b-c
    = 5|a-c or 7|a-c
    =aRc

    Hence R is an equivalence relation on Z


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  2. #2
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    Quote Originally Posted by Kai View Post
    Hi, need quick help with these questions, have exams on Tuesday

    1) Let G be a group. Prove that
    (xax^-1)^n=xa^nx^-1

    I have been able to prove it, first with n=1, then assuming it holds for n=k and proving for n=k+1 and n=-m for some +ve int m

    But the lecturer used n=0 for the first one,

    For n=0, he got (xax^-1)^0=e and for RHS xa^0x^-1=xex^-1

    I dont understand how a^0=e or (xax^-1)^0=e
    I haven't done groups in ages, but as far as I recall we used to use e as to symbolise the identity, which for multiplication would be 1, which would make sense for your a^0 = 1.

    For the second question o(a) is the order of a...
    And o(a) = smallest m such that a^m = e.

    Don't know the answer to your question, but hope that helps...
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  3. #3
    Kai
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    Quote Originally Posted by Unenlightened View Post
    I haven't done groups in ages, but as far as I recall we used to use e as to symbolise the identity, which for multiplication would be 1, which would make sense for your a^0 = 1.
    Yes under multiplication it would make sense, but the binary operation on G has not been defined. it could be addition or any other type
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Kai View Post
    Hi, need quick help with these questions, have exams on Tuesday

    1) Let G be a group. Prove that
    (xax^-1)^n=xa^nx^-1

    I have been able to prove it, first with n=1, then assuming it holds for n=k and proving for n=k+1 and n=-m for some +ve int m

    But the lecturer used n=0 for the first one,

    For n=0, he got (xax^-1)^0=e and for RHS xa^0x^-1=xex^-1

    I dont understand how a^0=e or (xax^-1)^0=e
    Thats a debatable topic. Some may call it a convenient notation.
    Convenient because:

    a\cdot a^{-1} = e \text{ and } a\cdot a^{-1} = a^{(1-1)} = a^0


    Quote Originally Posted by Kai View Post
    2) Let G be a group and a is an element of G. Prove o(a)=o(a^-1)

    For second question i have no idea at all, what is o(a) ??
    o(a) is the order of a.

    The order of an element a in a group is the least positive integer k such that
    a^k is the identity.


    Quote Originally Posted by Kai View Post
    Third question i have been able to do, needs to verify if its corect

    3) A relation R is defined on Z by aRb iff a-b is divisible either by 5 or by 7(a,b in Z). Is R an equivalence relation on Z? Justify your answer.

    For reflexive, 5|a-a or 7|a-a = 5|0 or 7|0
    Hence aRa.

    For symmetric, aRb implies 5|a-b or 7|a-b => 5|b-a or 7|b-a which implies bRa.

    For transitivity, aRb and bRc implies
    5|a-b or 7|a-b and 5|b-c or 7|b-c
    Isomorphism: What if 5|a-b and 7|b-c?
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  5. #5
    Kai
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    But if u see it this way, then for reflexive, 5|0 AND 7|0 not 5|0 or 7|0
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