[proofplan]
We first verify that the formula $\chi_{\bar a}(\bar x)=e(ax/q)$ is independent of the integer representatives $a$ and $x$. Then we prove that each $\chi_{\bar a}$ is an additive character. For the converse, an arbitrary character is determined by its value at the generator $\bar 1$, and the relation $q\bar 1=\bar 0$ forces that value to be a $q$-th root of unity. Finally, the classification of $q$-th roots of unity identifies that value uniquely as $e(a/q)$, and pointwise multiplication of characters corresponds to addition of residues.
[/proofplan]
[step:Verify that the displayed formula is well-defined on residue classes]
Fix $\bar a,\bar x\in G$, and choose representatives $a,x\in\mathbb Z$. Define
\begin{align*}
\chi_{\bar a}(\bar x):=e\left(\frac{ax}{q}\right).
\end{align*}
We check that this does not depend on the representatives. Let $a',x'\in\mathbb Z$ be another pair of representatives, so there exist $r,s\in\mathbb Z$ such that
\begin{align*}
a'=a+rq,\qquad x'=x+sq.
\end{align*}
Then
\begin{align*}
\frac{a'x'-ax}{q}=as+rx+rsq\in\mathbb Z.
\end{align*}
Since $e(t+n)=e(t)$ for every $t\in\mathbb R$ and every $n\in\mathbb Z$, it follows that
\begin{align*}
e\left(\frac{a'x'}{q}\right)=e\left(\frac{ax}{q}\right).
\end{align*}
Thus $\chi_{\bar a}:G\to\mathbb C^\times$ is well-defined.
[/step]
[step:Show that each $\chi_{\bar a}$ is an additive character]
Fix $\bar a\in G$. For $\bar x,\bar y\in G$, choose representatives $x,y\in\mathbb Z$. Since $\bar x+\bar y=\overline{x+y}$ in $G$, the definition gives
\begin{align*}
\chi_{\bar a}(\bar x+\bar y)=e\left(\frac{a(x+y)}{q}\right).
\end{align*}
Using $e(u+v)=e(u)e(v)$ for [real numbers](/page/Real%20Numbers) $u,v$, we obtain
\begin{align*}
\chi_{\bar a}(\bar x+\bar y)=e\left(\frac{ax}{q}\right)e\left(\frac{ay}{q}\right)=\chi_{\bar a}(\bar x)\chi_{\bar a}(\bar y).
\end{align*}
Also $\chi_{\bar a}(\bar 0)=e(0)=1$. Hence $\chi_{\bar a}$ is a [group homomorphism](/page/Group%20Homomorphism) from the additive group $G$ to the multiplicative group $\mathbb C^\times$.
[/step]
[step:Recover an arbitrary character from its value at $\bar 1$]
Let $\chi:G\to\mathbb C^\times$ be an additive character. Set
\begin{align*}
z:=\chi(\bar 1)\in\mathbb C^\times.
\end{align*}
Because $q\bar 1=\bar 0$ in $G$, the homomorphism property gives
\begin{align*}
z^q=\chi(\bar 1)^q=\chi(q\bar 1)=\chi(\bar 0)=1.
\end{align*}
Thus $z$ is a $q$-th root of unity.
By the elementary classification of $q$-th roots of unity, there is a unique residue class $\bar a\in\mathbb Z/q\mathbb Z$ such that
\begin{align*}
z=e\left(\frac{a}{q}\right).
\end{align*}
For every $\bar x\in G$, choose $x\in\mathbb Z$ with $\bar x=x\bar 1$. If $x\ge 0$, repeated use of the homomorphism property gives
\begin{align*}
\chi(\bar x)=\chi(x\bar 1)=\chi(\bar 1)^x=z^x=e\left(\frac{ax}{q}\right)=\chi_{\bar a}(\bar x).
\end{align*}
If $x<0$, then $\chi(x\bar 1)=\chi((-x)\bar 1)^{-1}=z^x$, so the same formula holds. Therefore $\chi=\chi_{\bar a}$.
[guided]
The key point is that $G=\mathbb Z/q\mathbb Z$ is generated by the single element $\bar 1$. Thus a homomorphism out of $G$ is forced once we know where $\bar 1$ goes.
Let $\chi:G\to\mathbb C^\times$ be an additive character, and define
\begin{align*}
z:=\chi(\bar 1).
\end{align*}
The relation that makes $G$ finite is $q\bar 1=\bar 0$. Applying the homomorphism $\chi$ to this relation converts addition in $G$ into multiplication in $\mathbb C^\times$:
\begin{align*}
\chi(q\bar 1)=\chi(\bar 1)^q.
\end{align*}
Since $q\bar 1=\bar 0$ and every group homomorphism sends the identity to the identity, we get
\begin{align*}
z^q=\chi(\bar 1)^q=\chi(q\bar 1)=\chi(\bar 0)=1.
\end{align*}
So $z$ must be a $q$-th root of unity.
We now use the elementary classification of $q$-th roots of unity: the solutions of $w^q=1$ in $\mathbb C^\times$ are precisely
\begin{align*}
e\left(\frac{a}{q}\right)
\end{align*}
with $\bar a\in\mathbb Z/q\mathbb Z$, and the residue class $\bar a$ is unique. Hence there is a unique $\bar a\in G$ such that
\begin{align*}
z=e\left(\frac{a}{q}\right).
\end{align*}
It remains to show that this value at $\bar 1$ determines the whole character. Take any $\bar x\in G$, and choose an integer representative $x\in\mathbb Z$, so $\bar x=x\bar 1$. If $x\ge 0$, repeated application of the homomorphism property gives
\begin{align*}
\chi(\bar x)=\chi(x\bar 1)=\chi(\bar 1)^x=z^x.
\end{align*}
Substituting the value of $z$ gives
\begin{align*}
\chi(\bar x)=e\left(\frac{a}{q}\right)^x=e\left(\frac{ax}{q}\right)=\chi_{\bar a}(\bar x).
\end{align*}
If $x<0$, the same conclusion follows because homomorphisms preserve inverses:
\begin{align*}
\chi(x\bar 1)=\chi((-x)\bar 1)^{-1}=z^x=e\left(\frac{ax}{q}\right).
\end{align*}
Thus every additive character is exactly one of the characters $\chi_{\bar a}$.
[/guided]
[/step]
[step:Identify the character group with $\mathbb Z/q\mathbb Z$]
Define
\begin{align*}
\Phi:G\to\operatorname{Hom}(G,\mathbb C^\times),\qquad \Phi(\bar a):=\chi_{\bar a}.
\end{align*}
The previous steps prove that $\Phi$ is well-defined and surjective. If $\Phi(\bar a)=\Phi(\bar b)$, then evaluating at $\bar 1$ gives
\begin{align*}
e\left(\frac{a}{q}\right)=\chi_{\bar a}(\bar 1)=\chi_{\bar b}(\bar 1)=e\left(\frac{b}{q}\right).
\end{align*}
By uniqueness in the classification of $q$-th roots of unity, $\bar a=\bar b$ in $G$. Hence $\Phi$ is injective.
Finally, for $\bar a,\bar b,\bar x\in G$, choose representatives $a,b,x\in\mathbb Z$. Pointwise multiplication gives
\begin{align*}
(\chi_{\bar a}\chi_{\bar b})(\bar x)=\chi_{\bar a}(\bar x)\chi_{\bar b}(\bar x)=e\left(\frac{ax}{q}\right)e\left(\frac{bx}{q}\right)=e\left(\frac{(a+b)x}{q}\right)=\chi_{\bar a+\bar b}(\bar x).
\end{align*}
Therefore
\begin{align*}
\Phi(\bar a+\bar b)=\Phi(\bar a)\Phi(\bar b).
\end{align*}
Thus $\Phi$ is a group isomorphism from $\mathbb Z/q\mathbb Z$ to the character group of $\mathbb Z/q\mathbb Z$.
[/step]
