What is the Galois group of $x^6+x^5-8x^4-5x^3+19x^2+4x-11=0$? A while ago I happened on this equation: $$f(x)=x^6+x^5-8x^4-5x^3+19x^2+4x-11=0.$$
$f$ does not have linear factors, as can be seen by drawing its graph. I am not certain about whether it has quadratic or cubic factors over $\mathbb{Q}$, but over $\mathbb{Q}(\sqrt{5}$) it factors into $$(x^3+\frac{1}{2}(1-\sqrt{5})x^2+\frac{1}{2}(-\sqrt{5}-7)x+\frac{1}{2}(1+3\sqrt{5}))*$$ $$(x^3+\frac{1}{2}(1+\sqrt{5})x^2+\frac{1}{2}(\sqrt{5}-7)x+\frac{1}{2}(1-3\sqrt{5})). $$
This seems to say that one can first extend $\mathbb{Q}$ to $\mathbb{Q}(\sqrt{5})$, then extend somehow to include the roots of the two cubics. I do note that the polynomial $f(x^2-3)$ is divisible by $f(x)$ with no remainder. This means that if $r$ is a root of $f(x)$; i.e. a root of either of these two cubics, then $r^2-3$ is also a root. In fact I constructed the polynomial by evaluating $g(g(g(x)))-x$, where $g(x)=x^2-3$, and throwing out the quadratic factor. 
I evaluated numerically the roots of the equation and came up with
(1) -2.22918
(2) 1.96925
(3) 0.877962
(4) -2.10553
(5) 1.43326
(6) -0.94576
$g$ permutes these roots; in fact, $g$ as a permutation is (123)(456), using the labels above. So $g$ is an element of the Galois group. There is also an order-2 mapping of the roots, namely the mapping $h$ that takes 1 to 1 and $\sqrt{5}$ to $-\sqrt{5}$; the non-zero element of the Galois group of $\mathbb{Q}(\sqrt{5})$ over $\mathbb{Q}$. I have tried generating subgroups of $S_6$ by using $g$ and some triple transposition (for $h$) for several possibilities of the triple transposition, and in each case I came up with $\mathbb{Z}_6$ as the Galois group. But if that's the case, then a cubic extension of $\mathbb{Q}(\sqrt{5})$ to an extension field $E$ would split the equation; $E$ would be the splitting field. But that then says the roots of one of the cubics above can be written as combinations of the roots of the others using only numbers from $\mathbb{Q}(\sqrt{5})$ as coefficients, but I don't see how this is done. It still seems to me the Galois group could be a group of order 18. How does one show that this is not the case?
 A: Paper & pencil progress:
The sextic is irreducible because it remains irreducible modulo two. If you don't want to test it with quadratic or cubic irreducibles of $\Bbb{F}_2[x]$ (three of those), then you can do the following. By freshman's dream $$g(g(g(x)))=x^8+1\in\Bbb{F}_2[x].$$ 
Therefore $g(g(g(x)))-x=x^8-x+1$ falls under this umbrella, allowing us to conclude that
the modulo two factorization into irreducibles is
$$
g(g(g(x)))-x=(x^2+x+1)(x^6+x^5+x^3+x^2+1).
$$ 
By a theorem of Dedekind this implies that the Galois group $G\le S_6$ of $f(x)$ has a 6-cycle $\sigma$ (when viewed as a subgroup of permutation of roots).
The polynomial $f(x)$ has the obvious zero $x=0$ over $\Bbb{F}_{11}$. Hence $g(0)=-3$ and $g(g(0))=6$ are also zeros of $f(x)$ in that field. This gives us the factorization
$$
f(x)=x(x+3)(x+5)(x^3+4x^2+1)\in\Bbb{F}_{11}[x].
$$
A bit tedious but straightforward check shows that the cubic has no zeros in $\Bbb{F}_{11}$. Therefore that cubic is irreducible, and Dedekind tells us that $G$ also contains a 3-cycle $\tau$.
If $L$ is the splitting field of $f(x)$ over $\Bbb{Q}$ we clearly have
$L=\Bbb{Q}(\alpha,\beta)$, where $\alpha$ is some zero of $f$ and $\beta$ is one of the three zeros that cannot be gotten from $\alpha$ by iterating $g$. Clearly $\beta$ generates at most a cubic extension of $\Bbb{Q}(\alpha)$, so
$[L:\Bbb{Q}]\le 18$. But, $\tau$ cannot be an element of the cyclic group generated by $\sigma$, so it easily follows that $|G|\ge6\cdot3=18$. Therefore we can conclude that
$$
|G|=[L:\Bbb{Q}]=18.
$$
These pieces of information already allow us to identify $G$ uniquely. Clearly $G$ normalizes a Sylow $3$-subgroup $P\simeq C_3\times C_3$ of $S_6$, so $G$ is a semidirect product $G\simeq(C_3\times C_3)\rtimes C_2$. Inside $S_6$ there is no room for $\sigma$ and $\tau$ to commute, so the product is not direct. Equivalently, the conjugation action of the element $\delta=\sigma^3$ of order two on $P$ is non-trivial. But, $\sigma^2$ clearly
commutes with $\delta$. So when we think of $P$ as a two-dimensional vector space over $\Bbb{F}_3$, the conjugation action has eigenvalues $\pm1$ both with multiplicity one. The resulting group is, as already identified by both Will and Robert, isomorphic to the wreath product $G\simeq C_3\wr C_2$.
By Sylow theory and Galois correspondence there is only a single quadratic subfield of $L$, namely the fixed field of $P$. Your factorization implies that this must be $\Bbb{Q}(\sqrt5)$.

Resorting to Mathematica:-(
The permutation $\eta:x\mapsto x^2-3$ of the roots of $f$ is an element of order three in the center of $G$. Because $|Z(G)|=3$ it follow that $Z(G)$ is generated by $\eta$. We can identify the fixed field of $Z(G)$ as follows. Let $\alpha,\beta$ be roots of $f$ as above. Consider the elements
$$
\begin{aligned}
\gamma_1&=\alpha \beta+g(\alpha)g(\beta)+g(g(\alpha))g(g(\beta)),\\
\gamma_2&=\alpha g(\beta)+g(\alpha)g(g(\beta))+g(g(\alpha))\beta,\\
\gamma_3&=\alpha g(g(\beta))+g(\alpha)\beta+g(g(\alpha))g(\beta).
\end{aligned}
$$
They are all obviously in the fixed field of $\eta$. By plugging in the numerical values of the zeros (this is cheating, I know) we see that
$$
r(x)=\prod_{j=1}^3(x-\gamma_j)=x^3+x^2-46x-36.
$$
Because $r(x)$ has no integral zeros it is irreducible over $\Bbb{Q}$. Its discriminant is $d(r)=2^2\cdot5\cdot139^2$ is not a square, so the Galois group of $r$ is $S_3$. Hence the splitting field $K$ of $r$ is the fixed field of $Z(G)$.
Unless I made a mistake $Z(G)$ and $P$ are the only non-trivial normal subgroups of $G$, so $K$ and its subfield $\Bbb{Q}(\sqrt5)$ are the only intermediate fields that are Galois over $\Bbb{Q}$.
A: Maple gives the Galois group of this polynomial as 
"6T5", {"3 wr 2", "F_18(6)", "[3^2]2"}, "-", 18, {"(2 4 6)", "(3 6)(1 4)(2 5)"}

That is, it is a group of order 18 generated by the permutations (in disjoint cycle notation) $[2,4,6]$ and $(3 6) (1 4) (2 5)$.
A: Irreducible over the rationals. Galois group as Robert said; I added two degree six polynomials where the roots are also real of modest absolute value, but come out very pretty, method originally due to Gauss.
================================================================
? f = x^6 + x^5 - 8 * x^4 - 5 * x^3 + 19 * x^2 + 4 * x - 11
%13 = x^6 + x^5 - 8*x^4 - 5*x^3 + 19*x^2 + 4*x - 11
? factor(f)
%14 = 
[x^6 + x^5 - 8*x^4 - 5*x^3 + 19*x^2 + 4*x - 11 1]

? polgalois(f)
%2 = [18, -1, 1, "F_18(6) = [3^2]2 = 3 wr 2"]
? poldisc(f)
%3 = 2415125
? factor(poldisc(f))
%4 = 
[5 3]

[139 2]
===================================================================
? f = x^6 + x^5 - 5 * x^4 - 4 * x^3 + 6 * x^2 + 3 * x - 1
%5 = x^6 + x^5 - 5*x^4 - 4*x^3 + 6*x^2 + 3*x - 1
? 
? polgalois(f)
%6 = [6, -1, 1, "C(6) = 6 = 3[x]2"]
? poldisc(f)
%7 = 371293
? factor(poldisc(f))
%8 = 
[13 5]

===========================================================
? 
? 
? f = x^6 + x^5 - 15 * x^4 - 28 * x^3 + 15 * x^2 + 38 * x - 1
%9 = x^6 + x^5 - 15*x^4 - 28*x^3 + 15*x^2 + 38*x - 1
? polgalois(f)
%10 = [6, -1, 1, "C(6) = 6 = 3[x]2"]
? poldisc(f)
%11 = 8390618797
? factor(poldisc(f))
%12 = 
[11 2]

[37 5]
============================================================

One of them has roots $2  \cos \left( \frac{2\pi}{13} \right),2  \cos \left( \frac{4\pi}{13} \right), $ and so on. The last one has a root $2  \cos \left( \frac{2\pi}{37} \right) +2  \cos \left( \frac{20\pi}{37} \right) +2  \cos \left( \frac{22\pi}{37} \right), $ the other five are also sums of three cosine terms.
