1. ## Galois groups

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

I think the forward direction is clear. Since $G$ fixes $E$ and since the generators $\sigma_1, \sigma_2, ..., \sigma_n$ are in $G$, then these generators must fix $E$.
I am not sure how to prove the backward direction. Some help please.

2. Originally Posted by dori1123
Let $G \leq Gal(K/F)$ be a subgroup of the Galois group of the extension $K/F$ and suppose $\sigma_1, \sigma_2, ..., \sigma_n$ are generators for $G$. Show that the subfield $E/F$ is fixed by $G$ if and only if it is fixed by the generators $\sigma_1, \sigma_2, ..., \sigma_n$.

I think the forward direction is clear. Since $G$ fixes $E$ and since the generators $\sigma_1, \sigma_2, ..., \sigma_n$ are in $G$, then these generators must fix $E$.
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 $\{\sigma_1, \cdots , \sigma_n \}.$ well, here is the meaning:

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