the proof is concluded.

From the statement of the previous theorem, we see that $\rho^+\geq \rho_\infty$. However, $\rho\geq \rho_\infty$ is not necessarily true. In the following result, a condition is given that implies $\rho\geq \rho_\infty$ (see also [@hughes], Section 3).

\[thm-oriat\] Let $L/{\mathbb Q}$ be a finite abelian extension with Galois group of odd exponent $n$, and suppose that $-1$ is congruent to a power of $2$ modulo $n$. Then, in the notation of Theorem \[thm-rhos\], we have $\rho=\rho^+$. In particular, $\rho\geq \rho_\infty$.

We obtain the following corollary.

\[cor-order2iseven\] Let $L/{\mathbb Q}$ be a cyclic extension of odd prime degree $p$, and suppose that the order of $2$ in $({\mathbb Z}/p{\mathbb Z})^\times$ is even. Then, $\rho=\rho^+$. In particular, $\dim_{{\mathbb F}_2} {{\rm Cl}}(L)[2] = \dim_{{\mathbb F}_2} {{\rm Cl}}^+(L)[2]$.

Suppose that ${\operatorname{Gal}}(L/{\mathbb Q})\cong {\mathbb Z}/p{\mathbb Z}$ for some prime $p>2$, such that the order of $2$ in $({\mathbb Z}/p{\mathbb Z})^\times$ is even (since ${\operatorname{Gal}}(L/{\mathbb Q})$ is cyclic of order $p$, this is equivalent to $-1$ being congruent to a power of $2$ modulo $p$). Hence, Theorem \[thm-oriat\] applies, and $\rho=\rho^+$.

The odd primes below $100$ such that the order of $2$ is odd modulo $p$ are $7$, $23$, $31$, $47$, $71$, $73$, 