TITLE: How many primes stay inert in a finite (non-cyclic) extension of number fields?
QUESTION [9 upvotes]: In the following suppose L/K is a finite Galois extension of number fields, (maybe it works for other cases also, I don't know) By the Chebotorev density theorem when Gal(L/K) is cyclic,  there are infinitely many primes in K that stay inert during this extension (cf Janus p136, Algerbaic Number Fields.) When L/K is non cyclic, an exercise from Neukirch (somewhere in Chap I) says there are at most finitely many primes that stay inert. I want to say that there are none. The reason is by a cycle description from Janus, p101, Prop 2.8, 
In short, that proposition says when $\delta:=Frob(\frac{L/K}{\beta})$, $\beta|p$ is a prime in L, consider $\delta$ act on the cosets of H in G, H=Gal(L/E), $K\subset E\subset L$, then every cycle of length i corresponds to a prime factor in E with residue degree i. In particular, for inert guys we want there is only one cycle in the action. When we take H to be trivial, E=L is Galois over K, and the cosets are just the elements of G themselves. 
So we want that there exist an element (the Frobenius element above p) act transitively on G, thus G is cyclic. 
I wonder if this is true, then more people should have been aware of it. If it is not, is there a counter example?

REPLY [5 votes]: Does "inert" mean that there is only one prime over $p$, or does it mean that there is only one prime over $p$ and that prime is unramified? As the other answers have explained, unramified primes can not remain inert. However, it is possible that there is only prime over $p$. For example, let $p$ be an odd prime, take $K = \mathbb{Q}$ and $L= \mathbb{Q}(\sqrt{p}, \sqrt{a})$ where $a$ is not a quadratic residue modulo $p$.