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

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 $$Gal(K/F)\cong S_3$$.

• $X^4-2$ is the minimal polynomial of $2^{1/4}$ over $F=\mathbb{Q}(i)$ and $F(2^{1/4})/F$ is Galois with Galois group $C_4$ (sending $2^{1/4}$ to $i^m2^{1/4}$). In general for $K/F$ the splitting field of $f \in F[X]$ then $Gal(K/F)$ is a subgroup of $S_n, n = \deg(f)$ and iff $f$ is irreducible separable then $Gal(K/F)$ is a transitive subgroup of $S_n$ – reuns Jan 30 at 0:11
• What is true is that the Galois group of a splitting field of a degree $n$ polynomial is a subgroup of $S_n$. Better, if the polynomial is irreducible, then the Galois group is a transitive subgroup of $S_n$. – Mathmo123 Feb 25 at 15:29

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 $$n\phi(n)/2$$.
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.