Let $f$ be irreducible over $\mathbb{Q}$ with splitting field $F$. Suppose $\text{Gal}(F/\mathbb{Q})$ is either $D_8$ or $Q_8$. What are the possibilities for $\deg f$?

I'm using Dummit & Foote, and they characterize possibilities for the Galois groups of irreducible polynomials of degree $\le 4$. There are quartics with Galois group $D_8$, such as $x^4\pm 2$, but no polynomial of degree $<4$ has Galois group $Q_8$. I have no experience with finding Galois groups of polynomials of degree $>4$ by hand (this problem was not given in my class, so please refer me to the proper readings if there is any literature that deals with such questions).

For $Q_8$ I have an example of an 8th degree polynomial whose Galois group is $Q_8$, namely, $x^8−24x^6+144x^4−288x^2+144$ (splitting field is $\mathbb{Q}((2+\sqrt 2)(3+\sqrt 3))$, but this is my go-to example and I do not know of any others nor do I know how to prove or disprove that there are others of higher degree with the same Galois group.

For $D_8$ I would bet that there is some degree 8 out there with this as its Galois group, but I can't construct one, nor can I classify which degrees can give a Galois group $D_8$.

How does one proceed with this type of question (please suggest how this problem can be approached for some other well-known groups as well)?

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    $\begingroup$ How are F and E related? $\endgroup$ – Adam Hughes Jun 20 '16 at 22:54
  • $\begingroup$ They're equal, that was an error. $\endgroup$ – cap Jun 21 '16 at 5:04
  • $\begingroup$ The irreducible polynomial could be $p(x)=x^8-2$.The Galois group is $D_8$ $\endgroup$ – Upstart Jun 21 '16 at 5:39

The two key facts you need are the following:

1) The Galois group of the splitting field of an irreducible polynomial $f$ acts transitively on the roots of $f$.

In particular, when $f$ has $n=\deg f$ roots, $\mathrm{Gal}(E/\mathbb Q)$ must be a transitive subgroup of $S_n$.

2) The order of the Galois group of the splitting field of an irreducible polynomial $f$ is divisible by $\deg f$.

This follows because $E$ must have a subfield generated by a single root of $f$.

The second fact tells us the that only possible degrees of $f$ are $1,2,4$ and $8$.

The first fact tells us that for $D_8$, the only possible degrees are $4$ and $8$, and for $Q_8$, the only possible degree is $8$.

An example of a degree $8$ polynomial with Galois group $D_8$ is $X^8 + 6X^4+1$.

  • $\begingroup$ Can you expand on how the first fact is being applied? Why isn't $Q_8$ a transitive subgroup of $S_4$? $\endgroup$ – cap Jun 21 '16 at 5:19
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    $\begingroup$ @cap: Because $|S_4|=24=2^3\cdot3$, the Sylow 2-subgroups of $S_4$ have order $8$. At least one of them, hence all (they are conjugates), is isomorpfic to $D_4$ (some people call it $D_8$ - all depending on whether $|D_n|=2n$ or $n$ for them). Because $Q_8$ and $D_4$ are not isomorphic, the former is not isomorphic to any subgroup of $S_4$. $\endgroup$ – Jyrki Lahtonen Jun 21 '16 at 7:20

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