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Math Help - Subgroup

  1. #1
    Senior Member Sampras's Avatar
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    Subgroup

    Let  E be an algebraic extension of a field  F . Let  S = \{\sigma_{i}: i \in I \} be a collection of automorphisms of  E such that every  \sigma_i leaves each element of  F fixed. Show that if  S generates a subgroup  H of  G(E/F) , then  E_S = E_H .

    The group of automorphisms of a field is cyclic and has the Frobenius automorphism as a natural generator. So  H is cyclic. Thus the elements left fixed by  S are the same as the elements left fixed by  H ?
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  2. #2
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    Quote Originally Posted by Sampras View Post
    Let  E be an algebraic extension of a field  F . Let  S = \{\sigma_{i}: i \in I \} be a collection of automorphisms of  E such that every  \sigma_i leaves each element of  F fixed. Show that if  S generates a subgroup  H of  G(E/F) , then  E_S = E_H .

    The group of automorphisms of a field is cyclic and has the Frobenius automorphism as a natural generator. So  H is cyclic. Thus the elements left fixed by  S are the same as the elements left fixed by  H ?
    what are E_S, E_H, G(E/F) ? i've seen your posts. most of them suffer from the lack of clarity! use standard notation or define clearly what you mean by them. for example, did you by G(E/F)

    mean \text{Gal}(E/F), the Galois group of E over F?
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  3. #3
    Senior Member Sampras's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    what are E_S, E_H, G(E/F) ? i've seen your posts. most of them suffer from the lack of clarity! use standard notation or define clearly what you mean by them. for example, did you by G(E/F)

    mean \text{Gal}(E/F), the Galois group of E over F?
     G(E/F) is the group of automorphisms of  E that leave  F fixed.  E_S and  E_H are the elements of  E left fixed by  S and  H .
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    Quote Originally Posted by Sampras View Post
     G(E/F) is the group of automorphisms of  E that leave  F fixed.  E_S and  E_H are the elements of  E left fixed by  S and  H .
    well, trivially E_H \subseteq E_S because S \subseteq H. conversely, let x \in E_S and \sigma \in H. then \sigma = \sigma_{i_1} \sigma_{i_2} \cdots \sigma_{i_n}, for some i_1, i_2, \cdots , i_n \in I, because H=<S>. now since x \in E_S, we have \sigma_{i_k}(x)=x,

    for all i_k. thus \sigma(x)=x, i.e. x \in E_H.
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