Are there relations among Frobenii? Let $G=\text{Gal}(\overline{\mathbf Q}/\mathbf Q)$, and for each prime $p$, choose an embedding $\overline{\mathbf Q} \hookrightarrow \overline{\mathbf Q_p}$. Let $\sigma_p$ be a choice of Frobenius in $\text{Gal}(\overline{\mathbf Q_p}/{\mathbf Q_p})$ and denote also by $\sigma_p$ its image in $G$ by the restriction map $\text{Gal}(\overline{\mathbf Q_p}/{\mathbf Q_p}) \hookrightarrow G$.
Let $H$ be the subgroup of $G$ generated by the $\sigma_p$'s. By the Chebotarev density theorem, $H$ is dense in $G$.

Question: Is $H$ free on the generators $\{\sigma_p\}$?

It seems likely to me that this is true. Any relation between the $\{\sigma_p\}$'s in $G$ would descend to a relation between the (chosen) Frobenii of any finite Galois extension of $\mathbf Q$. It seems unlikely that there might be such a universal relation.
(Interestingly, there is a relation if we include also the Frobenius at $\infty$, namely complex conjugation which satisfies $\sigma^2=1$.)
I am interested in the resulting morphism of profinite groups $\widehat{H} \to G$, where $\widehat{H}$ is the profinite completion of $H$ considered as a discrete group. If $H$ is indeed free on the $\{\sigma_p\}$, then $\widehat{H}$ is profinite-free on the $\{\sigma_p\}$. Is this homomorphism an isomorphism? It seems to me that it should be a bijective homomorphism of profinite groups, but that it is probably not a homeomorphism (since after all, in constructing $\widehat{H}$ we forgot the topology on $H$ induced from $G$)... 
 A: Let me alter things a little by taking $G$ to be the Galois group of the maximal
extension unram. outside $S$ (a fixed finite set of primes) and $H$ to be the
free group generated by symbols $\sigma_p$ for $p \not\in S$.  Then 
we have a morphism $H \to G$ with dense image (send $\sigma_p$ to Frobenius
at $p$).
A profinite group is compact, and the morphism $\widehat{H} \to G$ is continuous,
with closed and dense image, hence surjective.  If it is furthermore bijective,
then it is a homeomorphism (source and target are compact).
However, it is not a bijection by any means; it has enormous numbers of
relations.  For example, the group $G$ has (by Serre's conjecture) has
very constrained representations into $GL_2(\overline{\mathbb F_2})$.
(These have to correspond to mod $2$ modular forms of level prime to $S$,
by Serre's conj.)
These relations are subtle, which is what makes the reciprocity between
automorphic forms and Galois representations deep and difficult to prove,
but so powerful.
