Throughout the post $G = GL_{2}(\mathbb{F}_{q})$ where $q$ is a prime power with the prime not being 2.
Let $V_{1}$ and $V_{2}$ be cuspidal representations of $G$ over $\mathbb{C}$. I can understand that every cuspidal representation of $G$ is of dimension $q - 1$ and that every such representation comes with a non-decomposable (sometimes called regular) character $\tau$ of $\mathbb{F}_{q^{2}}^{\times}$. Recall that a character $\tau$, of $\mathbb{F}_{q^{2}}^{\times}$ is non-decomposable (or regular) if $\tau^{q} \neq \tau$.
I can also understand that 2 cuspidal representations $V_{1}$ and $V_{2}$ with corresponding non-decomposable characters $\tau_{1}$ and $\tau_{2}$ are isomorphic if and only if $\tau_{2} = \tau_{1}$ or $\tau_{2} = \tau_{1}^{q}$.
Here is where my question begins
Consider $V_{1} \otimes_{\hspace{1pt}\mathbb{C}} V_{2}$ as a representation of $G$ in the natural way. Is there anything we can say about how $V_{1} \otimes_{\hspace{1pt}\mathbb{C}} V_{2}$ can be broken down into irreducible representations of $G$?
I tried reading this paper https://www.tandfonline.com/doi/abs/10.1080/00927870008826973 which tries to answer the question in Theorem 3.1 point number 5. However it seems to assume (if I'm reading it correctly) that the product of 2 non-decomposable characters of $\mathbb{F}_{q^{2}}^{\times}$ is again non-decomposable which clearly shouldn't be the case. Any help on this would be greatly appreciated! Thank you! :)
