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Thread: Galois groups

  1. #1
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    Galois groups

    Let $\displaystyle G \leq Gal(K/F)$ be a subgroup of the Galois group of the extension $\displaystyle K/F$ and suppose $\displaystyle \sigma_1, \sigma_2, ..., \sigma_n$ are generators for $\displaystyle G$. Show that the subfield $\displaystyle E/F$ is fixed by $\displaystyle G$ if and only if it is fixed by the generators $\displaystyle \sigma_1, \sigma_2, ..., \sigma_n$.

    I think the forward direction is clear. Since $\displaystyle G$ fixes $\displaystyle E$ and since the generators $\displaystyle \sigma_1, \sigma_2, ..., \sigma_n$ are in $\displaystyle G$, then these generators must fix $\displaystyle E$.
    I am not sure how to prove the backward direction. Some help please.
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  2. #2
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    Quote Originally Posted by dori1123 View Post
    Let $\displaystyle G \leq Gal(K/F)$ be a subgroup of the Galois group of the extension $\displaystyle K/F$ and suppose $\displaystyle \sigma_1, \sigma_2, ..., \sigma_n$ are generators for $\displaystyle G$. Show that the subfield $\displaystyle E/F$ is fixed by $\displaystyle G$ if and only if it is fixed by the generators $\displaystyle \sigma_1, \sigma_2, ..., \sigma_n$.

    I think the forward direction is clear. Since $\displaystyle G$ fixes $\displaystyle E$ and since the generators $\displaystyle \sigma_1, \sigma_2, ..., \sigma_n$ are in $\displaystyle G$, then these generators must fix $\displaystyle E$.
    I am not sure how to prove the backward direction. Some help please.
    i think the reason that you can't solve an easy problem like this is that you don't know what is meant by "G is generated by $\displaystyle \{\sigma_1, \cdots , \sigma_n \}.$ well, here is the meaning:

    every element of G is in the form $\displaystyle x_1^{r_1}x_2^{r_2} \cdots x_m^{r_m},$ where $\displaystyle m$ is any natural number, $\displaystyle r_i \in \mathbb{Z},$ and $\displaystyle x_i \in \{\sigma_1, \cdots , \sigma_n \}.$
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