Fix a number field $K$.  For an integer $m$, let $S_1(m,K)$ be the congruence classes $a$ mod $m$ that contain infinitely many primes $p$ such that $p \mid \mathfrak p$ for some prime $\mathfrak p$ of $K$ satisfying $f(\mathfrak p|p) = 1$. (That was a mouthful: $p$ is lying below some prime of $K$ with residue field degree 1.) 
If $K/\mathbf Q$ is Galois, then such $p$ are the primes splitting completely in $K$, up to finitely many exceptions (among the ramified primes).  That is, when $K/\mathbf Q$ is Galois, $S_1(m,K)$ is the set of congruence classes mod $m$ containing infinitely many primes which split completely in $K$. (The prime numbers that split completely in a number field are identical to the prime numbers that split completely in its Galois closure over $\mathbf Q$, so attempting to describe such "split sets" by congruence conditions could just as well assume the number field is Galois over $\mathbf Q$.  I am working over base field $\mathbf Q$ throughout.)


As Kevin has suggested, it is not obvious at first that these sets $S_1(m,K)$ have much structure, particularly that they contain $1$ mod $m$.  By the pigeonhole principle, any $S_1(m,K)$ is certainly a nonempty set, and it is a subset of the unit group $(\mathbf Z/m)^\times$ rather than just $\mathbf Z/m$, but this is kind of superficial.  

A good reason (the right reason?) that $1$ mod $m$ is in $S_1(m,K)$ is that $S_1(m,K)$ is actually a subgroup of $(\mathbf Z/m)^\times$.  In fact, under the usual identification of $\mathrm{Gal}(\mathbf Q(\zeta_m)/ \mathbf Q)$ with $(\mathbf Z/m)^\times$, $S_1(m,K)$ is the image of the restriction homomorphism $\mathrm{Gal}(K(\zeta_m)/K) \longrightarrow \mathrm{Gal}(\mathbf Q(\zeta_m)/\mathbf Q)$.  For a proof, see Theorem 3 at 

http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf

and Theorem 4 there is a generalization where $(\mathbf Z/m)^\times$ is replaced with any Galois group of number fields.