This is likely to be a stupid question, but I can't tell which part is going wrong.
Let $\chi:(\mathbb{Z}/N)^\times\to S^1$ be a Dirichlet character mod $N$. I found the definition is that $\chi$ is primitive if it is not induced by another character $(\mathbb{Z}/M)^\times\to S^1$ with $M|N$ and $M<N$. I also see another definition says $\chi$ is primitive if the lifted arithmetic function (sending $x$ to $0$ if $(x,N)\neq 1$) has conductor $N$.
Now let's consider $\chi: (\mathbb{Z}/5)^\times\to S^1$, $2\mapsto i,3\mapsto -i,4\mapsto -1$. It induces a Dirichlet character $\chi':(\mathbb{Z}/10)^\times\to S^1$, $\chi'(3)=-i,\chi'(7)=i,\chi'(9)=-1$. But isn't this implies the minimal natural number such that $\chi'(1+q)=\chi'(1)$ is $10$, that is $\chi'$ has conductor $10$, hence being primitive? What is wrong with my steps?
Any help is much appreciated.
Best Answer
$\chi'$ is $N$-periodic (here $N=10$).
Its conductor is the minimal $q$, which divides $N$, such that $\chi'(n)= f(n)$ for all $\gcd(n,N)=1$, with $f$ a $q$-periodic function.
ie. $\gcd(a,N)=\gcd(b,N)=1$ and $q|a-b$ implies $\chi'(a)=\chi'(b)$.
It is really a rephrasing of your first definition!
We can set $f(n)=0$ for $\gcd(n,q)=1$ in which case $f$ is a Dirichlet character: the unique primitive character underlying $\chi'$.
Here it is $f=\chi$, $q=5$.
A character is non-primitive iff it is of the form $1_{\gcd(n,k)=1} \psi(n)$ with $\psi$ a character $\bmod m$ coprime with $k$. A character $\bmod p^2$ can be primitive with conductor $p$.