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

1. ## 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...

2. ## 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:

3. ## 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.