# Subgroup

• Jan 9th 2010, 06:46 AM
Sampras
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 $H$?
• Jan 9th 2010, 09:25 AM
NonCommAlg
Quote:

Originally Posted by Sampras
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 $F$?
• Jan 9th 2010, 09:30 AM
Sampras
Quote:

Originally Posted by NonCommAlg
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 $H$.
• Jan 9th 2010, 07:18 PM
NonCommAlg
Quote:

Originally Posted by Sampras
$G(E/F)$ 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 $H=.$ now since $x \in E_S,$ we have $\sigma_{i_k}(x)=x,$

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