Let $K$ be an finite abelian extension of $\mathbf{Q}$ conductor $p$, where $p$ is an odd prime. That is, $K \subset \mathbf{Q}(\mu\_ p)$, the $p$-th cyclotomic field. Let $h\_K$ be the class number of $K$. 


If $[K:\mathbf{Q}]=2$, we know that $K=\mathbf{Q}(\sqrt{p})$ or $\mathbf{Q}(\sqrt{-p})$. Gauss's genus theory tells us that $h\_K\equiv 1 \pmod 2$.


If $[K:\mathbf{Q}]=3$, a paper of G.Gras <https://eudml.org/doc/151629> (Page 94 line 1-2) says that $h\_K \equiv 1 \pmod 3$. Unfortunately, I did not find the proof of this result. I am very appreciated if someone gives references or ideas.


So my question is that 


if $[K:\mathbf{Q}]=n$ is a prime number, then do we have that $h\_K \equiv 1 \pmod n$?



class number of prime degree field with prime conductor
Maybe I am making a mistake here, but let me try:


Let $H$ be the Hilbert class field of $K$. Then $H\cap \mathbb{Q}(\zeta\_p)=K$ as otherwise one prime in there should be totally ramified and unramified at the same time. This shows the triviality of the maximal quotient of the class group $C$ on which the Galois group $G=\operatorname{Gal}(K/\mathbb{Q})\cong \mathbb{Z}/n\mathbb{Z}$ acts trivially. Then the action of $G$ has no non-trivial fixed points on $C$, which means each orbit of $G$ on $C$, apart from the neutral element, is of size $n$, because $n$ is prime. Hence $\vert C\vert \equiv 1 \pmod{n}$.


