I am doing some independent study on Galois Theory using Lisl Gaal's Classical Galois Theory textbook. I didn't completely understand the proof given in the book for the theorem that 'The general equation of degree $n$ is not solvable for $n\geq 5$ so I did some research online. I have come to understand that for a field $F$ of characteristic zero a polynomial, $f(x)\in F[x]$ is solvable by radicals if and only if its Galois group $G(E/F)$ is solvable where $E$ is the splitting field of $f(x)$ over $F$. Then we see that for the general polynomial of degree $5$ the Galois Group is $S_5$ which is not solvable since the only composition series of $S_5$ is $S_5>A_5>{e}$ and $A_5/{e}$ is not abelian and thus the general quintic is not solvable by radicals.
I am fine with all of this but I struggle with showing that the Galois Group of the general quintic is in fact $S_5$. I have the following, though I have a feeling it is fundamentally flawed but can't figure out precisely why;
Let $F$ be a field with zero characteristic and let $f(x)\in F[x]$ be a general quintic. So we have $$f(x)=x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$$ where the $a_i$ do not satisfy any algebraic relations over $F$. Let $f(x)$ have distinct roots $x_1,...,x_5$. Then the roots do not satisfy any algebraic relations over $F$. Let $E=F(x_1,..,x_5)$ be the splitting field of $f(x)$ over $F$. Then $E/F$ is separable and normal and thus is a Galois Extension and so $G(E/F)$ is the Galois group associated with $f(x)$. ($G(E/F)$ is the group of automorphisms of $E$ which fix $F$). We have $$ f(x)=(x-x_1)...(x-x_5)=x^5-e_1x^4+e_2x^3-e_3x^2+e_4x-e_5 $$ where the $e_is$ are the elementary symmetric polynomials in the roots $x_is$.
If we define $\sigma(x_i)=x_{\sigma(i)}$ for all $\sigma\in S_5$ together with $$ \sigma (\phi(x_1,...,x_5)=\phi(x_{\sigma(1)},...,x_{\sigma(5)}) $$ for all $\phi\in E[x]$. It is easy to show that $\sigma$ is an automorphism of $E$. Moreover it is clear from the first expression of $f(x)$ above that any permutation of the 5 distinct roots, $x_i$ leaves $f(x)$ unchanged. So $\sigma$ fixes $F$.
Hence $\forall\sigma\in S_5, \sigma \in G(E/F)$. Thus $S_5\subset G(E/F)$. But we know that for any polynomial of degree $n$ the order of its Galois Group must divide $n!$ and since $\mid S_5 \mid =5!$ it follows that $S_5= G(E/F)$.
Any help in understanding where I've gone wrong would be greatly appreciated.
This argument looks fundamentally correct to me. It also generalizes to arbitrary $n$: the Galois group of a general polynomial of degree $n$ is $S_n$.
I'm missing one (important) detail, though. Where are you using the assumption that $f$ is a general quintic?
The critical step for this is where you extend $\sigma$ from a permutation of $\{x_1, \dots, x_5\}$ to a map from $E$ to itself. That should use the fact that the $x$'s have no algebraic relations to conclude that the extension of $\sigma$ is well-defined.