Let $\mathbb F_q$ be a finite field of $q$ elements. Let $G$ be a group with two generators $\{a,b\}$. Let $A$ be a non-trivial one-dimensional representation of $G$ over $\Bbb F_q$.
Consider the set of maps \begin{equation} \begin{aligned} S:=\{f:G\longrightarrow A\mid f(g) - h\cdot f(g) = f(h) - g \cdot f(h),\ \ \ \forall g,h\in G \}. (\star) \end{aligned} \end{equation}
Here the "$\cdot$" denotes the action of $G$ on $A$.
Claim: $\operatorname{codim}_{A^2}S = 1$.
proof attempt: Consider the embedding
\begin{equation} \begin{aligned} S &\hookrightarrow A^{2}\\\\ f &\mapsto (f(a), f(b)). \end{aligned} \end{equation} Intuitively, I can say that $S$ is an affine subspace of $A^2$ of codimension one over $\Bbb F_q$ because $S$ is defined by one non-trivial relation.
But it seems difficult to formally get $f(a)$ from $f(b)$ using the relation ($\star$), so I find this proof unsatisfactory. Is the claim correct? Is there a better way to prove it?
Thank you.