3
$\begingroup$

Let $G$ be a finite group with $1 \neq H < G $ a minimal subgroup which is not normal. Prove that there exists a field $K$ and a polynomial $f \in K[X]$ so that $G \cong \operatorname{Gal}(f,K)$ with $degf < |G|$.

From the comments:

I know that every finite G is isomorphic to some Galois extension. Let's say that every finite Galois extension is the splitting field of some polynomial (though I'm not sure about that). That's as far as I got.

$\endgroup$
7
  • 1
    $\begingroup$ Please share your thoughts so far :) $\endgroup$
    – Shaun
    Jul 6 '14 at 10:25
  • 1
    $\begingroup$ @Shaun: I know that every finite $G$ is isomorphic to some Galois extension. Let's say that every finite Galois extension is the splitting field of some polynomial (though I'm not sure about that). That's as far as I got. $\endgroup$
    – student
    Jul 6 '14 at 10:32
  • $\begingroup$ So we may write $G = \mathrm{Gal}(L/K) = \mathrm{Gal}(f,K)$ for some $f$ which may have too large degree. But the hypothesis tells us that there is a maximal intermediate field $L'$ which is not Galois over $K$. Can we use this somehow? $\endgroup$ Jul 6 '14 at 11:59
  • $\begingroup$ @MartinBrandenburg: I think that the not-Galois part means that $f$ is not irreducible. But what the maximal tells us? $\endgroup$
    – student
    Jul 6 '14 at 13:12
  • $\begingroup$ "I think that the not-Galois part means that f is not irreducible." Not at all. $\endgroup$ Jul 6 '14 at 13:26
2
$\begingroup$

Extended hints:

  • Realize $G$ as a group of permutations on a finite set of variables $S=\{x_1,x_2,\ldots,x_n\}$. For example you can use the Cayley construction, when $n=|G|$. Often you get away with smaller $n$, but let's not worry about that.
  • Let $L=\Bbb{Q}(S)$ be the purely transcendental extension of the rationals gotten be treating the elements of $S$ as algebraically independent variables. By the previous bullet identify $G$ as a (sub)group of automorphisms of $L$ simply by permuting the indeterminates $x_i$. Let $K$ be the fixed field of $G$, so by basic Galois theory the extension $L/K$ is Galois with Galois group $\cong G$.
  • Let $F$ be the fixed field of $H$. Thus $K\subset F\subset L$, and $[L:F]=|H|$. Show that $F/K$ is a finite separable algebraic extension and thus simple. Conclude that $F=K(\alpha)$ for some $\alpha\in F$.
  • Let $f(x)\in K[x]$ be the minimal polynomial of $\alpha$. Show that $L$ is the splitting field of $f$. Here is the beef. Clearly $L/K$ is Galois, but you need to show that no proper subfield $M$ such that $F\subseteq M\subset L$ is Galois. You need to make full use of the assumptions made about $H$ in this step.
  • Convince yourself, your mates and your teacher that this does it.
$\endgroup$
2
  • $\begingroup$ All: Sorry about getting in the way of your effort of engaging the OP. If you think that I'm giving too much, please comment/downvote as you wish. I felt that the need to first realize $G$ as a Galois group of some extension was a major obstacle holding the OP back, and without the means to do that it is difficult to start making any progress. If you see a different route, please prove me wrong. $\endgroup$ Jul 7 '14 at 7:52
  • $\begingroup$ See this question for more details about bullets 1&2. $\endgroup$ Jul 8 '14 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.