Galois Group and Subfield lattice of $(x^2 – 7)(x^4 + x^3 + x^2 + x + 1)$ over $\mathbb{Q}$

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If $K$ is the splitting field of $f(x)$, then $K$ is Galois Closure of $Q(\sqrt{7}, \zeta_5)$ where $\zeta_5 = \alpha$ is a primitive root of unity.

There's 6 roots of $f(x)$: $\pm \sqrt{7}, \alpha, \alpha^2, \alpha^3, \alpha^4$. I've explicitly found the Galois group automorphisms as the following permutations in $S_6$ where $1 = \sqrt{7}, 2 = -\sqrt{7}, 3 = \alpha, \dots, 6 = \alpha^4$.

  1. identity permutation
  2. (3465)
  3. (3564)
  4. (36)(45)
  5. (12)
  6. (12)(3465)
  7. (12)(3564)
  8. (12)(36)(45)

These are all 8 of the automorphisms since it's a Galois extension and the degree of the extension is 8.

I'm tasked with finding out what group this is isomorphic to, and then drawing the subfield lattice of $K$. I'm not sure what group this is. I did some work and found that (3465) generates the first four permutations in the list, so I think this group is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$? Even given this though, I'm not very familiar with groups of this type to know where to begin with thinking of what the possible subgroups are. Any help is appreciated, thanks!

Best Answer

You're right about the group isomorphy to $\Bbb{Z}_2 \times \Bbb{Z}_4$.

You need to find the subgroups of $\Bbb{Z}_2 \times \Bbb{Z}_4$ (not hard, just regard the elements as tuples), see which elements they fix, adjunct those elements to $\Bbb{Q}$ and you get your intermediate fields. Then you need to arrange them in a lattice.

The fields are $\Bbb{Q}(X)$, $X \in \{\emptyset,\sqrt{7},\zeta_5,\sqrt{7}\zeta_5, \sqrt{7}\zeta_5^2, \sqrt{7}\zeta_5^3, \sqrt{7}\zeta_5^4, \{\sqrt7, \zeta_5\}\}$, do you see the pattern?

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