# Dirichlet Characters modulo $260$

I want to count the number of Dirichlet characters with given properties:

• Number of Dirichlet characters modulo $260$
• Number of quadratic Dirichlet characters modulo $260$
• Number of primitive Dirichlet characters modulo $260$
• Number of primitive quadratic Dirichlet characters modulo $260$

The answer of the first question is $96$ because $\mathfrak{D}_N\cong\Bbb{Z}_N^*$ (here we notate the group of Dirichlet characters modulo N with $\mathfrak{D}_N$). Because of this argument we can also solve the second question, because we want to find the number of elements in $\Bbb{Z}_N^*$ with order $2$. But how can I count such elements? For the third question I don't know how to solve it. I think that we have to look up to subgroups of $\Bbb{Z}_n^*$, but how?

Can someone help me with this topic? Thanks

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For $n \in \mathbb{Z}_+$ we have

$$(\mathbb{Z}/n\mathbb{Z})^{*} \cong \bigoplus_{p|n} (\mathbb{Z}/p^{e_p}\mathbb{Z})^{*},$$

where $e_p = \max \{e: p^e|n\}.$

Furthermore, for $p > 2,$

$$(\mathbb{Z}/p^e\mathbb{Z})^{*} \cong \mathbb{Z}/(p-1)\mathbb{Z} \times \mathbb{Z}/p^{e-1}\mathbb{Z}.$$

If $p = 2,$

$$(\mathbb{Z}/p^e\mathbb{Z})^{*} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2^{(e-2)_+}\mathbb{Z}.$$

Everything you want to know should be able to read off from these decompositions. For example, $Hom((\mathbb{Z}/n\mathbb{Z})^{*},\mu_2)$ is an elementary abelian 2 group of rank equal to the number of prime divisors of $n$ if $n \not\cong 0 \mod 8$ and the number of prime divisors of $n$ plus one, otherwise. Denoting this rank $r,$ it follows that the number of quadratic characters is equal to $2^r - 1.$

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Thus we get $2^3-1=7$ quadratic characters? Can we this also conclude without such a decomposition? –  user55367 Jan 6 '13 at 17:01
Yes 7. You could count quadratic subfields of $\mathbb{Q(\mu_{260})}.$ –  jspecter Jan 6 '13 at 17:12
What do you notated with $\mu_{260}$? –  user55367 Jan 6 '13 at 17:17
$260^{th}$ roots of unity. –  jspecter Jan 6 '13 at 17:19
Oh okay :D yes in our course we only work with quadratic field extensions ... –  user55367 Jan 6 '13 at 17:20