# A finite group $G$ is isomorphic to $\operatorname{Gal}(f,K)$.

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 $degf < |G|$.

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.

• Please share your thoughts so far :) Jul 6 '14 at 10:25
• @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. Jul 6 '14 at 10:32
• 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? Jul 6 '14 at 11:59
• @MartinBrandenburg: I think that the not-Galois part means that $f$ is not irreducible. But what the maximal tells us? Jul 6 '14 at 13:12
• "I think that the not-Galois part means that f is not irreducible." Not at all. Jul 6 '14 at 13:26

• 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.
• 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. Jul 7 '14 at 7:52