I have the following question:
I am reading Serre's book "A Course in Arithmetic" (see http://www.math.purdue.edu/~lipman/MA598/Serre-Course%20in%20Arithmetic.pdf).
On page 75, it is stated that the set $\{p\in \mathbb{P}\ \big|\ (\frac{a}{p})=1\}$ has Dirichlet density $\frac{1}{2}$, if $a$ is an integer which is not a square. Here, $(\frac{a}{p})$ denotes the Legendre symbol.
In the proof, I don't unserstand, why and how it follows from theorem 2 on page 73 that the set of prime numbers verifying the condition $\overline{p}\in H:=$ker$(\chi_a)\leq G(m)$ has for Dirichlet density $\mathbf{the\ inverse\ of\ the\ index\ of}$ $H$ $\mathbf{in}$ $G(m)$.
I would be grateful for any hints.
Thanks for the help!
$H$ is a subgroup of $G(m)=(\mathbb{Z}/m\mathbb{Z})^*$, which can be thought of as a collection of residue classes modulo $m$. If $H$ has size $k$, then $\bar p$ will be in $H$ iff $p$ is in one of these $k$ residue classes. Each residue class has Dirichlet density $1/\phi(m)$ by Theorem 2. Therefore, the union of these classes has Dirichlet density $k/\phi(m)$. Since $G(m)$ has size $\phi(m)$, this is the reciprocal of the index of $H$ in $G(m)$, which is $\phi(m)/k$.