Normal extension with simple, not cyclic Galois group Suppose that the extension $K/\mathbb{Q}$ is normal and has a Galois group which is simple, but not cyclic. Show that there is no rational prime $p$ such that $(p)$ remains prime in $K$.
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I am confused by the term simple but not cyclic. It would help me a lot to understand the concept more deeply if somenone is willing do this case. Thanks in advance!
 A: I have a proof like this:
We claim that there is no rational prime $p$ such that $(p)$ remains prime in $K$, i.e. $p$ is not inert in $K$. Otherwise, we have 
$$\left\{
\begin{array}{c}
 \mathfrak {P}_D=p, \\
 \mathfrak{P}_I=\mathfrak{P},
\end{array}\right.\quad(*)$$
where $D\triangleq  D_{\mathfrak{P}}$ and $I\triangleq I_{\mathfrak{P}}$ denotes the decomposition group and inertia group of $\mathfrak{P}$ respectively. $(*)$ is equivalent to $$\left\{
\begin{array}{c}
 G=D, \\
 I=1_G.
\end{array}\right.\quad(**)$$
From Galois theory, we know that $D/I$ is a cyclic group of order $f$. Therefore, we deduce from $(**)$ that $G$ and $D$ are cyclic, which is a contradiction.
See if it is true.
A: If a prime $p$ remained prime in $K$, then $K_p/\mathbb{Q_p}$ would be an extension of local fields with Galois group $G$. But Galois groups of finite extensions of local fields are soluble, because they can be filtered by the ramification subgroups with subsequent quotients being abelian.
So more generally, the decomposition group at any prime in a Galois extension of number fields is always soluble. Further, the precise structure of the higher ramification groups gives more restrictions on the structure of these decomposition groups.
I don't know where this question was quoted from, so I don't know whether this is the intended solution. But it is the conceptual explanation.
