Finding the value of the square of the Gauss sum

Question

While finding the value of the square of the Gauss sum $G^2$, at some point we make a substitution as follows: $$G^2 = \sum_{m_1=1}^{q-1}\sum_{m_2=1}^{q-1} \left(\frac{m_1 m_2}{q}\right) e_q(m_1+m_2) = \sum_{m_1=1}^{q-1}\sum_{n=1}^{q-1} \left(\frac{n}{q}\right) e_q(m_1+m_1 n),$$ where $$m_2 \equiv m_1 n\pmod q.$$ I understand that by the equivalence condition, the terms of the inner sum are one by one equal after the substitution. But I do not understand how we are allowed to interchange the variable of the inner sum from $m_2 = 1$ to $q-1$ to $n = 1$ to $q-1$ by just having $m_2 \equiv m_1 n \pmod q$. Note: Here, $q$ is a prime.

Answer 1

As discussed in the comments, the sum over $m_2$ can be replaced by a sum over $n$ because by the group properties of the multiplicative group $\mathbb Z_q^\times$ the equation

$$ m_2=m_1n\pmod q $$

establishes a bijection between the values of $m_2$ and $n$. For a given value of $m_2$, the corresponding value of $n$ is $m_1^{-1}m_2$.