On p832 of Coombes, Harbater - Hurwitz familes and arithmetic Galois groups, the following is claimed:
Let $K$ be a number field, take $1 \neq \omega \in \mathrm{Gal}(\bar{\mathbb{Q}}/K)$, and let $L_{\omega}$ be the fixed field of $\omega$. By Artin-Schreier the case when $\omega$ has finite order is boring, so suppose it has infinite order. Then $\Gamma = \mathrm{Gal}(\bar{\mathbb{Q}}/L_{\omega})$ is a free profinite group.
Question: Why is $\Gamma$ free profinite?
$\Gamma$ must be the smallest closed subgroup containing $\omega$, or the closure of the cyclic subgroup $<\omega>$. So it will be a procyclic group, a quotient of $\hat{\mathbb{Z}}$. So in order to be free, we need $\Gamma \cong \hat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$. How do we know this will be the case?
This doesn't mesh with another result I thought to be true: Suppose $K$ is a field, $\alpha \in K^{sep} \backslash K$ an algebraic element, and $F$ a field maximal with respect to not containing $\alpha$ such that $K^{sep}/F/K$. Then $\mathrm{Gal}(K^{sep}/F)$ is either isomorphism to $\mathbb{Z}/2\mathbb{Z}$ or to some $\mathbb{Z}_p$.
So what's the deal?