I have a question here that might be very difficult, but perhaps someone who knows about groups and/or exponential sums might be able to help with it.
Let $\Gamma$ be a finite group and $G_p\subset \Gamma$ an abelian subgroup of order $N_p$. As an abelian subgroup
$$G_p\cong \mathbb{Z}_{n_1}\times\cdots \times\mathbb{Z}_{n_a}.$$
Therefore, using the identification above, $x,\,t\in G_p$ can be written as $(x_1,\dots,x_a),\,(t_1,\dots,t_a)\in G_p$.
Let $x\in G_p$. Define an element of the group algebra $\mathbb{C}\Gamma$: $$x_{(p)}=\frac{1}{N_p}\sum_{t\in G_p}\left(\prod_{j=1}^a\exp\left(\frac{2\pi i x_jt_j}{n_j}\right)\right)t.$$
Similarly, for another abelian subgroup $G_q\subset \Gamma$, of order $N_q$ identified with $$G_q\cong \mathbb{Z}_{m_1}\times \cdots\times\mathbb{Z}_{m_b}$$ define: $$y_{(q)}=\frac{1}{N_q}\sum_{s\in G_q}\left(\prod_{l=1}^b\exp\left(\frac{2\pi i y_ls_l}{m_l}\right)\right)s.$$
Say we also know all about $G_p\cap G_q$.
Question: Are there necessary and sufficient conditions, perhaps related to where $xy$ sits in $\Gamma$, on $x\in G_p$ and $y\in G_q$ for $$x_{(p)}y_{(q)}\neq 0?$$
For an example of $G_p,G_q\subset \Gamma$, these can just be computed directly. The only general result I (seem to) have is:
Proposition Let $G_pG_q:=\{ts\in \Gamma\,\colon\,t\in G_p,\,s\in G_q\}$. If the factorisation of $ts\in \Gamma$ is unique: $$x\in G_p,\,y\in G_q\text{ and }xy=ts\implies x=t\text{ and }y=s,$$ then for all $x\in G_p$, $y\in G_q$, we have $x_{(p)}y_{(q)}\neq 0$.
Corollary If $G_p\cap G_q=\{e\}$, then for all $x\in G_p$ and $y\in G_q$, we have $x_{(p)}y_{(q)}\neq 0$.
I have another partial result (14 July 2023):
Proposition Let $G_p=\mathbb{Z}_p$ and $G_q=\mathbb{Z}_q$ and $G_p\subseteq G_q$ so that $G_q=\langle g\rangle$ and $G_p=\langle g^k\rangle$ where $q=kp$. Let $x=g^a$, in $G_q$, and $y=(g^k)^b$, in $G_p$. Then: $$x_{(p)}y_{(q)}\neq 0\text{ if and only if } a\,\mathrm{mod}\, p=b.$$