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Math Help - Proving a group G is abelian by assuming N_G(A) = C_G(A), where A<G

  1. #1
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    Proving a group G is abelian by assuming N_G(A) = C_G(A), where A<G

    "If G is a finite group such that, for each abelian subgroup A, N_G(A) = C_G(A), then G is abelian"

    I know that in an abelian group, the conjugacy classes are singleton sets, but I'm not sure how to use that to show N_G(A) = C_G(A). Actually I know I'm not supposed to assume G is abelian, so I'm guessing I'm supposed to show that N_G(A) = C_G(A) implies the conjugacy classes are singletons which implies G is abelian? This question looks like it should be easy but my textbook and prof doesn't explain the theorems very clearly...
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    Re: Proving a group G is abelian by assuming N_G(A) = C_G(A), where A<G

    I tried to think of an "elementary" proof of this result, but was unsuccessful. I had to rely on an old (1911) theorem of Burnside. Here's a solution using Burnside's Theorem:

    Proving a group G is abelian by assuming N_G(A) = C_G(A), where A&lt;G-mhfgroups8.png
    Thanks from HowDoIMath
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    Re: Proving a group G is abelian by assuming N_G(A) = C_G(A), where A<G

    Thanks for the help! It's gonna take me some time to understand it thoroughly, but this was very useful.
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