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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')

enter image description here

According to the textbook, I know that the relationship between the extension of field and subgroups is like the graph below:

enter image description here

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)):

enter image description here

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: $\begin{cases}r_1=\frac{\sqrt[5]{2500 \sqrt{2}+2500}}{2^{2/5} 5^{4/5}}+\frac{-500 \sqrt{2}-500}{2^{4/5} \left(5 \left(2500 \sqrt{2}+2500\right)\right)^{3/5}}-\frac{2^{2/5} 5^{4/5}}{\sqrt[5]{2500 \sqrt{2}+2500}}-\frac{5\ 2^{4/5} 5^{3/5}}{\left(2500 \sqrt{2}+2500\right)^{2/5}}\\ r_2=-\frac{\sqrt[5]{-2500 \sqrt{2}-2500}}{2^{2/5} 5^{4/5}}+\frac{(-1)^{2/5} \left(-500 \sqrt{2}-500\right)}{2^{4/5} \left(5 \left(2500 \sqrt{2}+2500\right)\right)^{3/5}}-\frac{(-5)^{4/5} 2^{2/5}}{\sqrt[5]{2500 \sqrt{2}+2500}}+\frac{5 (-5)^{3/5} 2^{4/5}}{\left(2500 \sqrt{2}+2500\right)^{2/5}}\\ r_3=2^{2/5} 5^{4/5} \sqrt[5]{-\frac{1}{2500 \sqrt{2}+2500}}-5\ 2^{4/5} 5^{3/5} \left(-\frac{1}{2500 \sqrt{2}+2500}\right)^{2/5}+\frac{(-1)^{4/5} \sqrt[5]{2500 \sqrt{2}+2500}}{2^{2/5} 5^{4/5}}-\frac{\left(-500 \sqrt{2}-500\right) \left(-\frac{1}{5 \left(2500 \sqrt{2}+2500\right)}\right)^{3/5}}{2^{4/5}}\\ r_4=-\frac{(-1)^{3/5} \sqrt[5]{2500 \sqrt{2}+2500}}{2^{2/5} 5^{4/5}}-\frac{(-2)^{2/5} 5^{4/5}}{\sqrt[5]{2500 \sqrt{2}+2500}}-\frac{5 (-2)^{4/5} 5^{3/5}}{\left(2500 \sqrt{2}+2500\right)^{2/5}}-\frac{\sqrt[5]{-1} \left(-500 \sqrt{2}-500\right)}{2^{4/5} \left(5 \left(2500 \sqrt{2}+2500\right)\right)^{3/5}}\\ r_5=\frac{(-1)^{2/5} \sqrt[5]{2500 \sqrt{2}+2500}}{2^{2/5} 5^{4/5}}+\frac{(-1)^{4/5} \left(-500 \sqrt{2}-500\right)}{2^{4/5} \left(5 \left(2500 \sqrt{2}+2500\right)\right)^{3/5}}+\frac{(-1)^{3/5} 2^{2/5} 5^{4/5}}{\sqrt[5]{2500 \sqrt{2}+2500}}+\frac{5 \sqrt[5]{-1} 2^{4/5} 5^{3/5}}{\left(2500 \sqrt{2}+2500\right)^{2/5}} \end{cases}$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Jan 4, 2023 at 21:38

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