solovei
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Registered User
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Apr 28 |
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Character free proof that Frobenius kernel is a normal subgroup? Showing it is a subgroup is a problem, but normality (invariant under conjugation) is easy. |
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Apr 28 |
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Why didn’t finite group theorists consider groups where all centralizers of non-identity elements are solvable? mathoverflow.net/howtoask#yourtitle |
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Apr 27 |
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Why didn’t finite group theorists consider groups where all centralizers of non-identity elements are solvable? @Geoff thanks. I just located the interesting page en.wikipedia.org/wiki/… |
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Apr 27 |
asked | Why didn’t finite group theorists consider groups where all centralizers of non-identity elements are solvable? |
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Apr 26 |
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Classification of groups in which the centralizer of every non-identity element is cyclic @Stefan Ok. But I have already asked a revised question for the class of amenable groups. |
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Apr 26 |
asked | Can one characterize amenable groups with $C_G(x)$ cyclic for all $x\neq 1$? |
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Apr 26 |
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Classification of groups in which the centralizer of every non-identity element is cyclic Well the question is closed, although I don't know why. Was it too open-ended? should I have said characterize rather than classify? Would it have been better if I had restricted the class of groups? say to subgroups of GL_n(Z)? |
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Apr 26 |
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Classification of groups in which the centralizer of every non-identity element is cyclic Andy, for finite groups I believe that groups where all centralizers are abelian can also be classified. |
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Apr 26 |
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Classification of groups in which the centralizer of every non-identity element is cyclic that's a partial answer |
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Apr 26 |
asked | Classification of groups in which the centralizer of every non-identity element is cyclic |
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Mar 4 |
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Cubic Fields Up to Isomorphism @Frank See page 77 of Elementary and Analytic Theory of Algebraic Numbers By Wladyslaw Narkiewicz. |
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Mar 4 |
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Cubic Fields Up to Isomorphism It is apparently known that the number of cubic fields with discriminant $D$ is unbounded, so by the bijection with the subgroups of the class group of $Q(\sqrt{D})$, that you mention, these subgroups are arbitrarily large. |
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Mar 3 |
asked | Cubic Fields Up to Isomorphism |
