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