Does the unit index divide the degree of an extension of number fields? Let $L/K$ be a finite extension of algebraic number fields of degree prime $p$. Is it true that the index $(U_K:\text{Norm}(U_L))$ divides $[L:K]$,
where $U_K$ denotes the unit group and Norm denotes the ideal norm map of the relative extension of $L/K$?
 A: No, not necessarily. For example, you can take $K\neq \mathbb{Q}$ to be totally real, and take $L$ to be a quadratic CM extension of $K$ whose unit group is equal to that of $K$ (e.g. make sure that $L/K$ ramifies at some place that is not above $2\infty$). Then ${\rm Norm}(U_L) = U_K^2$, and the unit index is $2^{{\rm rk}U_K+1}>2$.
More generally, if $L/K$ is Galois with Galois group $G$, then the unit index is an invariant of the structure of $U_L$ as a Galois module. Suppose for example that $G$ has prime order $p$. Then the unit index depends only on the Galois module structure of $\mathbb{Z}_p\otimes_{\mathbb{Z}} U_L$ as a $\mathbb{Z}_p[G]$-module. If $K$ has no $p$-th roots of unity, then the latter is, in general, isomorphic to some number $r_1$ of copies of the trivial module $\mathbb{Z}_p$, some number of modules isomorphic to the augmentation ideal of $\mathbb{Z}_p[G]$, and a free module of some rank. Each copy of the trivial module contributes $p$ to the unit index, while the other two modules do not contribute, so that $(U_L:{\rm Norm}U_K)=p^{r_1}$. To determine $r_1$, i.e. to tell apart a free module of rank one from the direct sum of a trivial and an augmentation ideal, is difficult, in general (if I were to have to do this, I would in fact compute the unit index).
