I am stuck at the following exercise:
Given a character $\chi$ modulo $q$, define the Gauss sum
$$\tau(\chi) := \sum_{a=1}^q\chi(a)e\bigg(\frac{a}{q}\bigg) $$
where $e(x) := e^{2\pi ix}$. Show that
$$\frac{1}{\varphi} \cdot \sum_{\chi} \overline{\chi}(a)\tau(\chi) = \begin{cases}&e\bigg(\frac{a}{q}\bigg) &\text{ if } gcd(a,q)=1\\&0 &else \end{cases}$$
The case $gcd(a,q) \ne 1$ is clear, so let us assume $gcd(a,q) = 1$. If I am not mistaken we then have:
$$\sum_{\chi} \overline{\chi}(a)\tau(\chi) = \sum_{\chi} \sum_{a=1}^q \overline{\chi}(a) \cdot \chi(a) \cdot e\bigg(\frac{a}{q}\bigg)= \sum_{\chi} \sum_{a=1}^q e\bigg(\frac{a}{q}\bigg)$$
So I do not see how I should continue from here. Could you help me?
The problem here is that you are using the variable $a$ in two different ways (both as a free variable and as a bound one). I would write $$\sum_\chi\overline\chi(a)\tau(\chi)=\sum_\chi\overline\chi(a)\sum_{b=1}^q \chi(b)e(b/q)$$ and then reverse the order of summation.