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Math Help - this is used to prove cauchy's theorem(james mcKay)

  1. #1
    Senior Member abhishekkgp's Avatar
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    this is used to prove cauchy's theorem(james mcKay)

    Let G be finite group and let p be a prime dividing |G|. Let S denote the set of p-tuples of elements of G the product of whose coordinate is 1:

    S={(x1,x2,x3,...,xp)|and x1*x2*x3*...*xp=1}
    where x1,x2,x3... belong to G. and '*' represents the binary operation under which G is a group. Note that x1,x2,x3... need not be distinct.

    PROVE THAT: S has |G|^(p-1) elements.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    Let G be finite group and let p be a prime dividing |G|. Let S denote the set of p-tuples of elements of G the product of whose coordinate is 1:

    S={(x1,x2,x3,...,xp)|and x1*x2*x3*...*xp=1}
    where x1,x2,x3... belong to G. and '*' represents the binary operation under which G is a group. Note that x1,x2,x3... need not be distinct.

    PROVE THAT: S has |G|^(p-1) elements.
    you have |G| choices for the first p-1 elements. How many choice does this leave for the last element?
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Swlabr View Post
    you have |G| choices for the first p-1 elements. How many choice does this leave for the last element?

    thanks.. that helped.. although now it seem really simple.

    3987^{12}+4365^{12}=4472^{12} ?? isnt this wrong? doesnt satisfy divisibility by 3(as well as invalidates FLT)
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Yes, it is wrong. However, they match for the first 10 digits, so at a glance...
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    Senior Member abhishekkgp's Avatar
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    whoa!!
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