# Question concerning the Dirichlet density of a subset of the set of primes

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!

• Thank you for your comment. Unfortunately, I am still unable to see, why the Dirichlet density is the inverse of the index of $H$ in $G(m)$. Isn't the inequality $\mathscr{P}$ $(\{p\in\mathbb{P}\ |\ p\equiv 1\ \text{mod}\ m\})$ $=\frac{1}{\phi (m)}\neq \frac{1}{2}=\frac{1}{[G(m):H]}$ correct in general? Mar 29 '15 at 19:38
• Note that $m=4a$ in the above proof, and the $m$ in Theorem $2$ is not the same $m$. The density is $1/[G(m):H]=1/\phi(4)=1/2$. Mar 29 '15 at 20:09
• Thanks again for the help. Sorry, but I'm really at a loss...how does $\phi (4)$ come into play? Why don't we have to use $\phi (m)=\phi (4|a|)$? But we don't know much about $a$... Mar 29 '15 at 20:53
• Because, as I said, the $m=4|a|$ is not the $m$ you have to use in Theorem $2$. Mar 29 '15 at 21:03
• I'm sorry, but I still don't see, why and how it follows from thm. 2 that the Dirichlet density of $\{p \in \mathbb{P}\ |\ \overline{p}\in H\}$ is given by the inverse of the index of $H$ in $G(m)$. Mar 29 '15 at 23:08

$H$ 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 $\phi(m)/k$.