On $\ker\chi $ , $\: \chi :{\rm Gal}(E,F)\rightarrow S_n$

Assume $E$ is a splitting field of a separable polynomial of degree $n$,so $E$ must have $n$ district roots $(a_1,...,a_n)$ of the polynomial, if $\phi\in{\rm Gal}(E,F)$ then $\phi (a_i)$ must be one of the $a_j$, now

Consider the homomorphism $\: \chi :{\rm Gal}(E,F)\rightarrow S_n$

$\chi(\phi_i)=j$

Question1 : what's the $\ker\chi $?

The kernel is consisting of all those $\phi_i$ that send $a_i$ to $a_1$ and so $i\rightarrow 1$?

Now if I apply the $1st$ isomorphism theorem, I get that ${\rm Gal}(E,F)/\ker\chi $ is isomorphic to a subgroup of $S_n$

Question2: can I get that ${\rm Gal}(E,F) $ is isomorphic to a subgroup of $S_n$, with the $\chi$ homorphism?

(This maybe simple but I am a bit confused right now)

Solution 1:

The kernel of $\chi$ consists of all those automorphisms of $E$ over $F$ that leave all $a_i$ fixed. But then these also leave the field $F(a_1,\ldots,a_n)$ fixed. By definition of $E$ as a splitting field (as opposed to just some - perhaps bigger - field where the polynomial splits), this is all of $E$. In other words, the only element of the kernel is the identity, $\chi$ is injective. This also gives a "yes" on the second question.