Suppose $f(x)=(x-a_1)...(x-a_n)$ be a polynomial with distinct roots in a splitting field $L$ of $\mathbb {Q}$. Consider the polynomial $$F(X,T_1,...,T_n)= \Pi_ { \sigma \in S_n} (X-(a_{\sigma(1)}T_1+...+a_{\sigma(n)}T_n))$$ Note that $F$ has rational coefficients,suppose $F= F_1...F_n$ be a factorization of $F$ into irreducibles over $\mathbb {Q}$. Then it is easy to see that
Each $F_i$ corresponds to cosets of $G$ in $S_n$ ?
I'm unable to understand what is the exact meaning of the statement that "each $F_i$ corresponds to cosets of $G$ in $S_n$". Could someone explain me the meaning and the idea of proof of the same.
Look at how $G$ acts on the $n!$ linear factors in $F$.
If you call $f_\sigma(X,T_1,\ldots,T_n) = (X-(a_{\sigma(1)}T_1 + \ldots a_{\sigma(n)} T_n))$ for $\sigma \in S_n$,
then if $\tau \in G$ is an automorphism sending $a_i$ to $a_{\tau(i)}$, you have :
$\tau.(f_\sigma) = (X - (\tau.a_{\sigma(1)}T_1+ \ldots + \tau.a_{\sigma(n)}T_n) \\ = (X - (a_{\tau(\sigma(1))}T_1+ \ldots + a_{\tau(\sigma(n))}T_n) = f_{\tau \circ \sigma}$
where $\circ$ is the composition law in $S_n$ and $\tau$ is identified to an element in $S_n$ (so $G$ is identified with a subgroup of $S_n$)
If $A \subset S_n$ and you look at the product $P_A = \prod_{\sigma \in A} f_\sigma$, then for $\tau \in G$ you have $\tau. P_A = \prod_{\sigma \in A} f_{\tau \circ \sigma} = P_{\tau A}$.
Moreover, because the roots are distinct, different subsets will always give you different products (or if $P_A = P_B$ then $A=B$)
But we know that $P_A$ is rational (has rational coefficients) if and only if $\tau.P_A = P_A$ forall $\tau \in G$, and so this is also equivalent to $\forall \tau \in G, A = \tau A \;\;( = \{\tau\circ\sigma \mid \sigma \in A\})$
Then it is quick to check that the smallest subsets of $S_n$ (so the irreducible rational factors) that have this property are the orbits of $S_n$ under the action of $G$ : they are of the form $A_\mu = \{\tau \circ \mu \mid \tau \in G \} = G\mu$, for some $\mu \in S_n$.
And those are exactly the right-cosets of $G$ in $S_n$, and so there is a correspondance between the irreducible factors and $G \setminus S_n$.