Htykuut

In fact, the exercise I'm in is this:

Suppose you have an irreducible polynomial $f(x)in mathbb{Q}[x]$ of degree $p$, where $p$ is a prime. Also suppose that $K$ is a splitting field of $f(x)$ over $mathbb{Q}$.

Prove that:

1. $[K:mathbb{Q}]=pm$, where $gcd(p,m)=1$


2. If $H$ is a normal subgroup of $mathrm{Gal}(K,mathbb{Q})$ of order $|H|=m$ then $m=1$.

The first part was easy but the second part has something to do with Sylow theorems in which I'm not very keen on.

Answer to your second question: Since $H$ is normal in $G=Gal(K/mathbb{Q})$, $H$ corresponds to a field $E$ such that $mathbb{Q} subseteq E subseteq K$ (more preciously $E=K^{H}$). Now $H$ being normal in $G$ we must have that $E$ is a normal extension of $mathbb{Q}$. (Because $bigcap_{gin G} gHg^{-1}$ corresponds to $prod_{sigma in G}sigma(E)$ and $H$ being normal in $G$ we've $prod_{sigma in G}sigma(E)$=E, which forces $E/mathbb{Q}$ is anormal extension). And as $K/mathbb{Q}$ is itself separable (as it is a Galois extension), $E/mathbb{Q}$ is also separable and hence Galois of degree $p$.

So now by Primitive Element theorem there exists $bin E$ such that $E=mathbb{Q}(b)$. Now if $b$ is root of $f(x)$ then since $E/mathbb{Q}$ is normal $E$ would become a splitting field of $f(x)$ forcing $m=1$. So WLOG $b$ is not a root of $f(x)$. Now pick a root of $f(x)$ $ain F$ to and consider $mathbb{Q}(a,b)/mathbb{Q}$.

Note that $[mathbb{Q}(b):mathbb{Q}]=p$ so now if we can show $[mathbb{Q}(a,b):mathbb{Q(b)}]=p$ then we will get $p^2 mid [K:mathbb{Q}]$ which is a contradiction from first part. So, now we're just left to prove $[mathbb{Q}(a,b):mathbb{Q(b)}]=p$

Suppose not, which means $f(x)$ is reducible over $mathbb{Q}(b)$. Now note that $f(x)$ is irreducible over $mathbb{Q}$ and $mathbb{char}(mathbb{Q})=0$so $f(x)$ is seperable, so it has no multiple roots in any extension of $mathbb{Q}$ and so, $f$ must split into linear factors over $mathbb{Q}(b)[x]$ (if not then its has a factor of form $g(x)^r$ with $r>1$ and $g$ is irreducible over $mathbb{Q}(b)[x]$, but then any root of $g$ in some extension comes as a root of $f$ with multiplicity more than $1$, contradiction) and so $b$ is a root of $f$ which is a contradiction to our assumption that $b$ is not a root of $f$ and so we're done.

There is the simple counterexample of $S_3$. Its order is $6=3times2$ but its order 2 subgroups are not normal.

To answer to the question of the title. Consider the group $G=mathbb{Z}/6mathbb{Z}$, we have $mid G mid = 3times 2$ and $gcd(2,3)=1$. Since $G$ is abelian then $H=langle 2 rangle simeq mathbb{Z}/3mathbb{Z}$ is a non trivial normal subgroup of $G$.