Define a *Chebotarëv datum* over a number field $K$ to be a finite group $G$ together with a map $\mathfrak{p}\mapsto\gamma\_{\mathfrak{p}}$ from a cofinite set of primes of $K$ into the set of conjugacy classes of $G$ such that for every conjugacy class $c\subset G$, the proportion of $\mathfrak{p}$ with $\gamma\_{\mathfrak{p}}=c$ is $\operatorname{Card}c/\operatorname{Card}G$. 


Two Chebotarëv data $(G,\gamma)$, $(G',\gamma')$ over the same number field are said to be *equivalent* if there is an isomorphism $\varphi:G\to G'$ such that $\varphi(\gamma\_{\mathfrak{p}})=\gamma'\_{\mathfrak{p}}$ for almost all $\mathfrak{p}$. If so, we identify the two.


Every finite galoisian extension $L$ of $K$ gives rise to a Chebotarëv datum $(\operatorname{Gal}(L|K),\gamma\_{L|K})$ (*Chebotarëv's density theorem*).


Moreover, if $L\_1$ and $L\_2$ are two finite galoisian extensions of $K$ for which the associated Chebotarëv data $(\operatorname{Gal}(L\_i|K),\gamma\_{L\_i|K})$ are equivalent, then $L\_1=L\_2$ (see Lemma 1, p. 174, of [Mazur's recent article](http://www.ams.org/journals/bull/2011-48-02/S0273-0979-2011-01326-X/S0273-0979-2011-01326-X.pdf)).


**Question.** Does every Chebotarëv datum over a given number field $K$ arise from some finite galoisian extension $L$ of $K$ ?



Chebotarëv data over number fields
I think this is false for cardinality reasons. Take some exceedingly thin, but infinite, set of primes P. Then you can take any valid Chebotarev datum and change it arbitrarily at each prime in P, and it still remains a valid datum, since P is too thin to change the densities. If the group has at least two conjugacy classes, then there'll be uncountably many ways of changing it like this. Since the set of number fields is clearly countable, that's a contradiction.


