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!


1 Answer 1


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?

  • $\begingroup$ Thanks! I found the subgroups -- am now trying to write the subfield lattice. I've got that if $(a,b) \in \mathbb{Z}_4 \times \mathbb{Z}_2$ then $Fix(\{(0,0), (0,1)\}) = Q(\zeta_5)$ and $Fix(\{(0,0), (2,0)\}) = Q(\sqrt{7})$ though am quite unsure on the rest. For example, for $Fix(\{(2,1), (0,0)\})$ I see that the negative sign on the switching of the $\sqrt{7} \rightarrow - \sqrt{7})$ is cancelled out by the permuting of the roots of unity, though am stuck on concluding what the fixed field is. Is there a systematic way to go about this? $\endgroup$
    – Struggles
    Mar 20, 2019 at 2:52
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    $\begingroup$ Pick a field, see how the elements in the field look, use the groups on them to see which one fixes them, it is the easiest approach in my opinion. For example the elements in $\Bbb{Q}(\sqrt7 \zeta_5)$ have the form $a+b\sqrt7 \zeta_5$ with $a,b \in \Bbb{Q}$. You have $8$ automorphisms and the degree of the extension is equal to the amount of automorphisms, because the extension is finite, normal, separable and algebraic. $\endgroup$
    – B.Swan
    Mar 20, 2019 at 3:19
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    $\begingroup$ As a general hint: For an extension $\Bbb{Q}(\sqrt p, \zeta_k)/ \Bbb{Q}, \sqrt p$ irrational, the intermediate fields are the ones with $\zeta_k^n$ with $1\leq n \leq k-1 $ and $gcd(k,n) \neq 1$ and products $\sqrt p \zeta_k^n$ for all $0 \leq n \leq k-1$ $\endgroup$
    – B.Swan
    Mar 20, 2019 at 3:26
  • 1
    $\begingroup$ You might benefit from this thread: math.stackexchange.com/questions/625816/… $\endgroup$
    – B.Swan
    Mar 20, 2019 at 3:33

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