Is it true that $Gal(K/F)\cong S_{n_1}\times \cdots S_{n_k}$?

Question

I was reading galois theory and galois group from Dummit Foote and while reading Galois groups of polynomial a sudden question came into my mind that if $f(x)$ is an irreducible separable polynomial of degree $n$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F) \cong S_n$?

Moreover, can we generalize this question that if irreducible factorization of $f(x)=f_1(x)....f_k(x)$ where $f_i(x)$ is an irreducible separable polynomial of degree $n_i$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F)\cong S_{n_1}\times \cdots S_{n_k}$? If so then what is the arguement?

For $\Bbb Q(\sqrt 2, \sqrt 3)$ we see that $f(x)=(x^2-2)(x^2-3)$ and the $Gal(K/F)\cong S_2 \times S_2 $ again for Galois group of $x^3-2$ we see that $Gal(K/F)\cong S_3$.

Answer 1

This is not always true as shows the answer of this question for $X^n-2$, the order of the Galois group is $n\phi(n)$ or $n\phi(n)/2$.

https://mathoverflow.net/questions/143739/galois-group-of-xn-2

Answer 2

After reading one general answer given by @Tsemo Aristide one trivial answer came into my mind which is for the splitting field of $x^p-1$ over $\Bbb Q$ we have $\Bbb Q(\psi_p)$ which is Galois and of the degree of extension $p-1$. Basically, the element of Galois group will be completely determined by the image of $\psi_p$ so arbitrary permutation is not possible at all.