# Finite Groups as Galois Groups

So my old proof isn’t really working on wordpress for some reason, so I’ve taken it as a sign to do it in a different way. This method is far more complicated than the old way (in which I just call upon some theorems and look at orders and then am done), but I think it better gets at what is going on.

Anyways, since we were on the topic of Galois theory, here is a fact I found astonishing the first time I heard it (maybe it is quite obvious to you). Every finite group arises as the Galois group of a field extension, moreover we can choose the two fields to be number fields. Recall that a number field is just a subset of the complex numbers that is algebraic over $\mathbb{Q}$.

Proof: Let G be a finite group. Then by Cayley’s theorem $G\cong H< S_n$ for some n. But then there is a prime p, such that n<p, meaning $S_n. Let’s find a Galois extension $K/\mathbb{Q}$ such that $S_p\cong Gal(K/\mathbb{Q})$.

Our natural choice is the splitting field of $f(x)=x(x^2+1)(x^2-1)(x^2-4)\cdots (x^2-m^2)+1/p$, (note the sarcasm) where we chose our prime to be of the form $p=2m+3$. Thus we have that $deg(f(x))=p$, it is irreducible by Eisenstein’s criterion, and it has exactly 2 roots in $\mathbb{C}\setminus\mathbb{R}$. Thus, $Gal(K/\mathbb{Q})\cong S_p$. The details of this just amounts to playing around with cycles.

Now we can just invoke the Fundamental Theorem of Galois Theory. We have the subgroup $H < S_p$ which corresponds to the fixed field, say L, where $K\supset L\supset \mathbb{Q}$ and $Gal(K/L)\cong H\cong G$, which is what we wanted.

Now that this is typed out, I think the other way is better since I didn’t have to skip over the cycle argument. This method is just as mysterious overall, but in fact the other is probably more believable, since I only use things that are standard.

Lastly, let me just clear up some mystery about $f(x)$ at least. The way we know it has $2m+1$ real roots is that $f(x)$ alternates signs between pairs $-m-1/2, -m+1/2$, then $-m+3/2, -m+5/2$, all the way to $m-1/2, m+1/2$. Thus by the IVT, there is a real root between each of those. Thus there are 2 complex roots left over. We can even locate them to be near $\pm i$ if p is sufficiently large by Rouche’s Theorem. Also, $f(x)$ is irreducible iff $pf(x)$ is irreducible, and $pf(x)=px(x^2+1)(x^2-1)\cdots (x^2-m^2)+1$ and here it is clearly seen that p divides all the coefficients except the constant term and $p^2$ does not divide the highest term (Eisenstein).

Oh, and while I’m at it, I might as well lead you in the right direction if you want to check that $Gal(K/\mathbb{Q})\cong S_p$. By order, we know that $Gal(K/\mathbb{Q})\leq S_p$ (at least there is an isomorphic copy in there). So let the copy of $Gal(K/\mathbb{Q})$ act on $S_p$. The action is transitive, so there is a 2-cycle, and since p divides the order there is a p-cycle. These three things should get you there (that it is inside $S_p$, that there is a 2-cycle, and that there is a p-cycle).

So although they exist, they aren’t always the easiest to find. In fact, if we want our base field to be $\mathbb{Q}$, then this is known as the “Inverse Galois Problem” and is still open. Some cases have been resolved. For instance, it is known that every finite solvable group arises as a Galois group of an algebraic extension of $\mathbb{Q}$.

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### Author: hilbertthm90

I write about math, philosophy, literature, music, science, computer science, gaming or whatever strikes my fancy that day.