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We may use the following theorem,

Let $\chi$ be a Dirichlet character modulo $k$ and assume $d|k , d<k$. Then the following two statement are equivalent:

  1. $d$ is an induced modulus for $\chi$

  2. There is a character $\psi$ modulo $d$ such that $\chi(n) = \psi(n)\chi_{1}(n)$ for all $n$, where $\chi_1$ is the principal character modulo $k$.

And I wanna show that, if $k$ and $j$ are induced moduli for $\chi$ then so is their gcd $(k, j)$.

Using the theorem presented above, I can roughly see that the statement is true. But I do not know how to start the proof.

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  • $\begingroup$ That's the definition : a non-primitive character is of the form $\chi(n) = \psi(n) 1_{gcd(n,k)=1}$ for some character $\psi \bmod d$ with $d | k$. Note that the product of two Dirichlet characters is a Dirichlet character (completely multiplicative, periodic)... Non-primtive means one of the factor is the trivial character $1_{gcd(n,k) = 1}$. $\endgroup$ – reuns Jan 23 '17 at 20:19
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Here is the proof of Theorem 8.17 given in Tom Apostol's book (page 170, fifth edition): Assume that $2.$ holds. Choose $n$ satisfying $(n,k)=1$ and $n\equiv 1 \bmod d$. Then $\chi_1(n)=\psi(n)=1$ so that $\chi(n)=1$ and hence $d$ is an induced modulus. Thus $2.$ implies $1.$ For the converse (which is much longer), see Apostol's text. One exhibits a character $\psi$ modulo $d$ for which $\chi(n)=\psi(n)\chi_1(n)$ holds for all $n$ using Dirichlet's theorem. More precisely, if $(n,d)>1$ we can just take $\psi(n)=0$. For $(n,d)=1$ we need to find an integer $m$ such that $m\equiv n \bmod d$ with $(m,k)=1$. This is where Dirichlet's theorem on infinitely many primes in arithmetic progressions is used.

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  • $\begingroup$ Look, my question is exercise6 in page 175. $\endgroup$ – Edgar.W Jan 23 '17 at 19:19
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    $\begingroup$ No, in my copy of Apostol's book exercise $6$ is: Let $\chi$ be a character mod $k$. If $k_1$ and $k_2$ are induced moduli then so is $(k_1,k_2)$. I suppose you have an old edition perhaps. Look into the $5$-th edition of 1998, where this is exactly Theorem 8.17. $\endgroup$ – Dietrich Burde Jan 23 '17 at 19:22
  • $\begingroup$ I can't believe Apostol is so bad for explaining Dirichlet characters. So many people don't understand what he writes ! $\endgroup$ – reuns Jan 23 '17 at 20:22

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