I can get the Hasse diagram(the subgroup lattice) easily by maple, such as for polynomial $x^3 - 2$:
DrawSubgroupLattice(GaloisGroup(x^3 - 2, x), 'indices')
According to the textbook, I know that the relationship between the extension of field and subgroups is like the graph below:
Here $r_1$,$r_2$ and $r_3$ are $\sqrt[3]{2}$,$\frac{\sqrt[3]{2}(-1+\sqrt{3} i)}{2}$ and $\frac{\sqrt[3]{2}(-1-\sqrt{3} i)}{2}$ respectively,which are the roots of $x^3 - 2$. But how about this Hasse diagram about polynomial $x^5+15x+44$(its Galois Group is smallgroup(20,3) or we call it F(20)):
What is the field extension represented by each of these subgroups? This information is not available from soft maple(I'm actually asking the same question here), so I'd like to ask for some advice here.
I can get The roots of the $x^5+15x+44$ by this Wolfram Mathematica code, they are: $\begin{cases}r_1=\frac{\sqrt[5]{2500 \sqrt{2}+2500}}{2^{2/5} 5^{4/5}}+\frac{-500 \sqrt{2}-500}{2^{4/5} \left(5 \left(2500 \sqrt{2}+2500\right)\right)^{3/5}}-\frac{2^{2/5} 5^{4/5}}{\sqrt[5]{2500 \sqrt{2}+2500}}-\frac{5\ 2^{4/5} 5^{3/5}}{\left(2500 \sqrt{2}+2500\right)^{2/5}}\\ r_2=-\frac{\sqrt[5]{-2500 \sqrt{2}-2500}}{2^{2/5} 5^{4/5}}+\frac{(-1)^{2/5} \left(-500 \sqrt{2}-500\right)}{2^{4/5} \left(5 \left(2500 \sqrt{2}+2500\right)\right)^{3/5}}-\frac{(-5)^{4/5} 2^{2/5}}{\sqrt[5]{2500 \sqrt{2}+2500}}+\frac{5 (-5)^{3/5} 2^{4/5}}{\left(2500 \sqrt{2}+2500\right)^{2/5}}\\ r_3=2^{2/5} 5^{4/5} \sqrt[5]{-\frac{1}{2500 \sqrt{2}+2500}}-5\ 2^{4/5} 5^{3/5} \left(-\frac{1}{2500 \sqrt{2}+2500}\right)^{2/5}+\frac{(-1)^{4/5} \sqrt[5]{2500 \sqrt{2}+2500}}{2^{2/5} 5^{4/5}}-\frac{\left(-500 \sqrt{2}-500\right) \left(-\frac{1}{5 \left(2500 \sqrt{2}+2500\right)}\right)^{3/5}}{2^{4/5}}\\ r_4=-\frac{(-1)^{3/5} \sqrt[5]{2500 \sqrt{2}+2500}}{2^{2/5} 5^{4/5}}-\frac{(-2)^{2/5} 5^{4/5}}{\sqrt[5]{2500 \sqrt{2}+2500}}-\frac{5 (-2)^{4/5} 5^{3/5}}{\left(2500 \sqrt{2}+2500\right)^{2/5}}-\frac{\sqrt[5]{-1} \left(-500 \sqrt{2}-500\right)}{2^{4/5} \left(5 \left(2500 \sqrt{2}+2500\right)\right)^{3/5}}\\ r_5=\frac{(-1)^{2/5} \sqrt[5]{2500 \sqrt{2}+2500}}{2^{2/5} 5^{4/5}}+\frac{(-1)^{4/5} \left(-500 \sqrt{2}-500\right)}{2^{4/5} \left(5 \left(2500 \sqrt{2}+2500\right)\right)^{3/5}}+\frac{(-1)^{3/5} 2^{2/5} 5^{4/5}}{\sqrt[5]{2500 \sqrt{2}+2500}}+\frac{5 \sqrt[5]{-1} 2^{4/5} 5^{3/5}}{\left(2500 \sqrt{2}+2500\right)^{2/5}} \end{cases}$