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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.

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    $\begingroup$ $K$ is called a $\textit{splitting field}$ of $f(x)$. $\endgroup$ – Ken Duna Jun 16 '16 at 1:23
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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 $\bigcap_{g\in G} gHg^{-1}$ 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 $b\in E$ 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 $f(x)$ $a\in F$ 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.

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

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  • $\begingroup$ In fact, this is exactly what I' m trying to prove, so that' s makes it an example instead of a counter one. $\endgroup$ – richarddedekind Jun 16 '16 at 20:05
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To answer to the question of the title. Consider the group $G=\mathbb{Z}/6\mathbb{Z}$, we have $\mid G \mid = 3\times 2$ and $\gcd(2,3)=1$. Since $G$ is abelian then $H=\langle 2 \rangle \simeq \mathbb{Z}/3\mathbb{Z}$ is a non trivial normal subgroup of $G$.

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    $\begingroup$ This does not answer the question. In fact the statement is not true in general. Consider $S_3$. Its order is $6=3\times 2$. But each subgroup of order 2 is not normal. $\endgroup$ – Ken Duna Jun 16 '16 at 13:15
  • $\begingroup$ If $G$ is a Galois group of order 6 isomorphic to $\mathbb{Z}_6$ you are right and it is a valid counterexample. But, if there are no splitting fields with Galois group isomorphic to $\mathbb{Z}_6$? In fact, I tried to find one but I always end up into the symmetric group $S_3$. $\endgroup$ – richarddedekind Jun 17 '16 at 3:40

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