1. ## 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.

2. Originally Posted by dori1123
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 $\displaystyle x_i \in \{\sigma_1, \cdots , \sigma_n \}.$