Find Galois Group I want to know how to find a polynomial $f(x)$ of degree $6$ in $\mathbb{Q}[x]$ with Galois groups $G_f=\mathbb{S}_6$. I have a criterion to find a polynomial wich Galois group is $\mathbb{S}_p$ with $p$ prime, but i don´t now if it work in this case
 A: Consider the polynomial
$$
f(x)=x^6-8x^4-5x^3+4x^2+5x-12.
$$
We view its Galois group $G$ as a subgroup of $S_6$ as permutations of its roots. I shall be using the fact that if a polynomial in $\mathbb{Z}[x]$ only has simple factors modulo a prime $p$, then the degrees of the irreducible factors give lengths of cycles in an element of a Galois group.
Modulo $p=2$ it factors as $f(x)\equiv x(x^5+x^2+1)$, so $G$ contains a 5-cycle. Furthermore we know that either $f(x)$ is irreducible or it has a rational integer as a root. The latter case can be excluded with the high school level rational root test,
so we know that $f(x)$ is irreducible and $G$ is a transitive subgroup of $S_6$. (The presence of that 5-cycle proves that a point stabilizer of $G$ is also transitive, so we could also already conclude that $G$ is doubly transitive.)
Modulo $p=3$ it factors as $f(x)\equiv (x^3-x)(x^3-x+1)=x(x-1)(x+1)(x^3-x+1)$ so the Galois group has a 3-cycle.
Modulo $p=5$ it factors as $f(x)\equiv (x^4-1)(x^2+2)\equiv (x+1)(x+2)(x-1)(x-2)(x^2+2)$
so we can conclude that there is also a 2-cycle in $G$. 
By transitivity of $G$ we can conclude that the point stabilizer $G_1$ of a fixed root $x_1$ of $f(x)$ in $G$ also contains a 5-cycle, a 3-cycle and a 2-cycle. (If the permutations produced above don't give one right away, then conjugate them by a suitable element of $G$). Therefore we can conclude that $2\cdot3\cdot5\mid |G_1|$. We can naturally view $G_1$ as a subgroup of $S_5$. The presence of a 2-cycle tells that $G_1$
is not a subgroup of $A_5$. The index of $G_1\cap A_5$ in $A_5$ is thus at most 4. But $A_5$ cannot have a proper subgroup of index $\le 4$, because that would give rise to a non-trivial homomorphism from $A_5$ to $S_k, k\le 4$, violating the known fact that $A_5$ is simple. Thus $G_1$ must contain all of $A_5$. Again, the presence of a 2-cycle in $G_1$ tells us that we must have $G_1=S_5$. From this it follows that $G=S_6$, because the only transitive subgroup of $S_6$ that has full $S_5$ as a point stabilizer is obviously $S_6$.
This was a bit ad hoc. At this point it is probably best that I admit having constructed $f(x)$ by prescribing those three factorizations modulo $p=2,3,5,$ and then using the Chinese Remainder Theorem. The rest was just wishfully making deductions hoping that they would take me to the finish line. The thing guiding me was that I knew enough about subgroups of $S_5$ to hope that the presence of those short cycles would force the group to be quite big.
