This follows from everything we have said so far, plus Artin reciprocity, in case
\(L/K\) is abelian. In the general case, let
\(f\) be the order of
\(g\text{,}\) and let
\(K'\) be the fixed field of
\(g\text{;}\) then we know that the set of primes of
\(K'\) with Frobenius
\(g \in \Gal(L/K') \subset G\) has Dirichlet density
\(1/f\text{.}\) The same is true if we restrict to primes of absolute degree 1 (see
Exercise 2).
Let
\(Z\) be the centralizer of
\(g\) in
\(G\text{;}\) that is,
\(Z = \{z \in G\colon zg = gz\}\text{.}\) The order of the conjugacy class of
\(g\) is
\(\#G/\#Z\text{,}\) so we are trying to prove that the set of primes of
\(K\) (of absolute degree 1) with Frobenius in the conjugacy class of
\(g\) has Dirichlet density
\(\frac{\#G/\#Z}{\#G} = \frac{1}{\#Z}\text{.}\)
To begin with, we apply the abelian case to the extension
\(L/K'\text{,}\) to see that the set of primes of
\(K'\) (of absolute degree 1) with Frobenius
\(g\) equals
\(1/f\text{.}\) Next, note that a prime
\(\gothp\) of
\(K\) (of absolute degree 1) has Frobenius in the conjugacy class of
\(g\) if and only if there is a prime
\(\gothq\) of
\(L\) with Frobenius
\(g\text{;}\) note that
\(\gothq\) has inertia degree
\(f\) over both
\(K'\) and
\(K\text{,}\) so
\(K' \cap \gothq\) is a prime of
\(K'\) of absolute degree 1. Finally, note that for each
\(h \in G\text{,}\) the Frobenius of
\(\gothq^h\) is
\(hgh^{-1}\text{,}\) so only the elements of
\(Z\) correspond to cases where this Frobenius is again equal to
\(g\text{.}\) However, the
\(f\) elements of the subgroup generated by
\(g\) do not even move
\(\gothq\text{,}\) so there are in all
\(\#Z/f\) primes of
\(K'\) above
\(\gothp\) with Frobenius
\(g\text{.}\) Consequently, the density of primes of
\(K\) with Frobenius in the conjugacy class of
\(g\) is
\(\frac{1}{f} \frac{1}{\#Z/f} = \frac{1}{\#Z}\) as desired.
