First of all, I know that there are other questions similar to this, but these use other definitions or don't have a complete answer.
I'm trying to prove that if $\chi_{1}$ and $\chi_{2}$ are two Dirichlet characters mod $a$ and $b$ respectively, with $(a,b)=1$, then $$c(\chi_{1}\cdot\chi_{2})=c(\chi_{1})\cdot c(\chi_{2})$$ where $c(\chi)$ is the conductor of the character $\chi$.
The definitions that I'm using are the following:
A Dirichlet character $\chi$ mod $q$ is a function $\chi:\mathbb{N}\to\mathbb{C}$ such that $\chi(mn)=\chi(m)\chi(n)$ for all $m,n\in\mathbb{N}$, $\chi$ is periodic of period $q$, and $\chi(n)=0$ if $(n,q)>1$.
The conductor $c(\chi)$ of a Dirichlet character $\chi$ is the least period of $\chi$ restricted to values of $n$ such that $(n,q)=1$. In particular $c(\chi)\mid q$. I will use the notation $\chi^{*}$ for the function ''$\chi$ restricted for the corresponding values'', so the conductor of $\chi$ is simply the least period of $\chi^{*}$.
I know that if $(n,ab)=1$ then $(n,a)=1$ and $(n,b)=1$, so $(\chi_{1}\cdot\chi_{2})^{*}=\chi_{1}^{*}\cdot\chi_{2}^{*}$. Now $\chi_{1}^{*}$ has period $c(\chi_{1})$ and $\chi_{2}^{*}$ has period $c(\chi_{2})$, so $(\chi_{1}\cdot\chi_{2})^{*}$ has period $\text{lcm}(c(\chi_{1}),c(\chi_{2}))=c(\chi_{1})\cdot c(\chi_{2})$, therefore $c(\chi_{1}\cdot\chi_{2})\mid c(\chi_{1})\cdot c(\chi_{2})$.
My problem is that I'm unable to prove that $c(\chi_{1})\cdot c(\chi_{2})\mid c(\chi_{1}\cdot\chi_{2})$, and therefore these two numbers are equal.
I tried this:
For given $n$ such that $(n,ab)=1$ we can find $m$ such that $m\equiv 1\mod{a}$ and $m\equiv n\mod{b}$, so $$\chi_{1}(n)\cdot\chi_{2}(n)=\chi_{1}(1)\cdot\chi_{2}(m)=\chi_{2}(m)$$ therefore $$\{\chi_{1}\cdot\chi_{2}(n)\mid (n,ab)=1\}\subseteq\{\chi_{2}(m)\mid (m,b)=1\text{ and }m\equiv 1\mod{a}\}\subseteq\{\chi_{2}(m)\mid (m,b)=1\}$$ from where (I think) we conclude that $(\chi_{1}\cdot\chi_{2})^{*}$ has period $c(\chi_{2})$, and similarly one proves that it has period $c(\chi_{1})$, but this would imply that it has period $(c(\chi_{1}),c(\chi_{2}))=1$ which doesn't have any sense at all.
Any help would be appreciated.
Use that $$\chi_1(n) = \chi_1\chi_2(un+(1-u))\qquad where\quad u=1\bmod a,u= 0\bmod b$$ If $\chi_1\chi_2$ is $r$-periodic then so is $\chi_1$.
Thus the least period of $\chi_1$ divides that of $\chi_1\chi_2$.
The idea is the same when restricting to the $n$ coprime with $a,b$.