Galois group of $X^5 - X^3 - 2X^2 - 2X - 1$ over $\mathbb{Q}$. So far I have found that this polynomial is irreducible, and its discriminant is a square. Therefore the Galois group it a transitive subgroup of $A_5$. 
I found out that the only transitive subgroups in $A_5$ are $A_5$ and $D_{10}$. Is there a way to tell the Galois group from these two possibilities?
 A: $C_5$ is also a transitive subgroup of $A_5$, but yes, the discriminant is a square, so you know it must be $A_5$, $C_5$, or $D_{10}$.
Factoring over small prime numbers can give an idea of the structure of the Galois group. For example, $f(x)=x^5-x^3-2x^2-2x-1=(x+4)(x^2+3)(x^2+x+2)$ over $\mathbb{F}_5$, which instantly rules out $C_5$.
To decide between $A_5$ and $D_{10}$, recall that the Galois group of $f$ permutes its roots. Let $\alpha_1,\dots,\alpha_5$ be the roots of $f$, and consider the connected pentagon with the roots at each vertex:
$\hskip1.7in$
Then we can look at the geometric effect of each subgroup on the roots. In particular, we look at the resolvent. Now calculating the resolvent does require knowing something about the roots of $f$ to some precision, but letting $\alpha_1,\dots,\alpha_5$ be the roots of $f$ we define the resolvent to be:
$$
p(x)=\prod_{1\le i,j\le 5}(x-(\alpha_i+\alpha_j))
$$
Then the roots of $p$ represent the ten edges of the connected pentagon, and in fact, $p$ has the same Galois group as $f$ (it's a separable polynomial over $\mathbb{Q}$ with the same splitting field), but exhibits much nicer qualities.
Alternatively, you can simplify the quintic into Bring-Jerrard form, say $g(x)=x^5+ax+b$, and then the resolvent can be written succinctly as:
$$
p(x)=x^{10}-3ax^6-11bx^5-4a^2x^2+4abx-b^2
$$
$A_5$ is the group of even permutations, and is generated by $\sigma_1=(1,2,3,4,5)$ and $\sigma_2=(1,2,3)$. If $\sigma_2$ acts on $\alpha_1,\dots,\alpha_5$, then only one of the edges of the pentagon can be fixed. By applying $\sigma_1$ before $\sigma_2$, we can change the edge that is fixed. Also note that all the edges of the connected pentagon can permute with one another, so $A_5$ is made up of a single orbit, and $p$ is irreducible over $\mathbb{Q}$.
$D_{10}$ is made up of the 5 rotations and 5 reflections of the pentagon, so the Galois group of $p$ should be made up of two orbits of length $5$: one for the rotations, and a second for the reflections. So $p$ should factor into two irreducible polynomials of degree $5$.

In your case, either approximating the roots of $f$ and calculating the resolvent, or reducing to Bring-Jerrard form and using the formula above, you'll see that it factors into two irreducible factors of degree $5$, and so the Galois group is $D_{10}$.

You could verify this with the following Maple code:
    galois(x^5-x^3-2*x^2-2*x-1);

This will give the following output:
    "5T2", {"5:2", "D(5)"}, "+", 10, {"(1 2 3 4 5)", "(1 4)(2 3)"}

So the Galois group is indeed $D_{10}$.
