# Proof that if a Dirichlet character has two induced moduli then their gcd must also be an induced modulus

Let $\chi$ be a Dirichlet character mod $k$ and let $d$ be any positive divisor of $k$. The number $d$ is called an induced modulus of $\chi$ if we have

$\chi(a) = 1$ whenever $gcd(a, k) = 1$ and $a \equiv 1$ (mod $d$)

I wanted to construct an intuitively straightforward proof for myself that if $\chi$ is a character mod $k$, and if $k_1$ and $k_2$ are induced moduli for $\chi$, then $d = gcd(k_1, k_2)$ must also be an induced modulus for $\chi$, and $\chi(a) = 1$. I managed to come up with a relatively simple proof as follows.

Suppose $a \equiv 1$ (mod $d$), with $gcd(a, k) = 1$. Since $k_1$ and $k_2$ are induced moduli for $\chi$, $\chi$ must take equal values at numbers which are relatively prime to $k$ and congruent modulo either $k_1$ or $k_2$. In other words we must have

$\chi(a) = \chi (a + xk_1) = \chi(a + yk_2)$

for any integers $x, y$ such that

$gcd(a + xk_1, k) = gcd(a + yk_2, k) = 1$

This is only possible in general if $\chi$ takes equal values at numbers which are relatively prime to $k$ and congruent modulo some common divisor of $k_1$ and $k_2$ (otherwise in some cases it would be possible to find some $x, y$ such that $\chi (a + xk_1) \neq \chi(a + yk_2)$). Since any common divisor of $k_1$ and $k_2$ must also divide their greatest common divisor $d$, it follows that $\chi$ must take equal values at numbers which are relatively prime to $k$ and congruent modulo $d$. Thus, since $a \equiv 1$ (mod $d$), we have

$\chi(a) = \chi(1) = 1$

and $d = gcd(k_1, k_2)$ is an induced modulus for $\chi$ if $k_1$ and $k_2$ are. QED