# Property of Dirichlet character (Apostol 8.17)

We may use the following theorem,

Let $$\chi$$ be a Dirichlet character modulo $$k$$ and assume $$d|k , d. 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.

• 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 $1_{gcd(n,k) = 1}$. – reuns Jan 23 '17 at 20:19

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.
• 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. – Dietrich Burde Jan 23 '17 at 19:22