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.