Finding the Galois group of a quintic I'm trying to find the Galois group of $P(X)=X^5+3X^2+1 \in \mathbb{Q}[X]$, but I have no idea how to proceed.
First, the polynomial is irreducible in $\mathbb{Q}[X]$ (we can reduce modulo $2$ for instance). By checking the extrema, we verify there's only one real root. Let $R$ denote the set of roots. Since the polynomial has real coefficients, taking complex conjugates show the $5$ distinct (since $char\ \mathbb{Q}=0$ and $P$ is irreducible) roots can be written as $R=\{x,\alpha,\bar{\alpha},\beta,\bar{\beta}\}$, with $x\in\mathbb{R}$.
Let $K$ be the splitting field of the polynomial, and $G$ the associated Galois group, with $|G| = n$. Since the polynomial is a quintic, we know that $5| n$ and $n|5!=120$
By Cauchy's Theorem, $G$ contains an element of order $5$ i.e. a $5$-cycle $(12345)$. Complex conjugation defines a double transposition $(23)(45)$ that leaves fixed the real root; this implies, in particular, that $2|n \Rightarrow 10|n$.
This is where I start having doubts. 
These two elements seem generate the $3$-cycle $(345)$ as follows: $$12345 \rightarrow 23451 \rightarrow 24315 \rightarrow 15243\rightarrow 12534$$
Since a $3$-cycle has order $3$, we now have $3|n \Rightarrow 30|n$.
However, this seems to be wrong according to this list. Where do I go from here?
 A: We know that the transitive automorphism(the one of of order 5) and the complex conjugation (the double transposition) are part of the Galois group of $x^5 + 3x^2 + 1$ over $\mathbb{Q}$. Unfortunatelly this isn't enough for us to distinguish whether the Galois group is $A_5$ or $S_5$, as the two forementioned automorphisms generate an automorphism group isomorphic to $A_5$, but that doesn't mean that there isn't another automorphism on the roots, which combined with these two automorphisms spans over $S_5$. So as I mentioned combining all possible combinations of $(12345)$ and $(23)(45)$ won't help us much, except leaving us with two candidates $A_5$ or $S_5$.
But thankfully here the Dedekind's Theorem comes in very handy. Factoring the polynomial in $\mathbb{Z}_{19}$ we have that:
$$x^5 + 3x^2 + 1 \equiv (x+2)(x+9)(x+13)(x^2+14x+16) \pmod{19}$$
Therefore by the Dedekind's Theorem the Galois group of $x^5 + 3x^2 + 1$ over $\mathbb{Q}$ contains an automorphism isomorphic to the cycle of the type $(1,1,1,2)$, which is a transposition. Now using the fact that if a subgroup of $S_n$, $n$ a prime, contains a $n-$cycle and transposition it's actually $S_n$ itself we can conclude that the Galois group of $x^5 + 3x^2 + 1$ over $\mathbb{Q}$ is $S_5$
