Sums of products of average character values on cosets Consider a finite group $G$, a subgroup $H\leq G$, and a transversal $G/H = \{t_1H, t_2H,\ldots,t_rH\}$. Given three characters $\chi_1,\chi_2$ and $\chi_3$ of $G$, I'd like to compute:
$$
\sum_{i}^r \left[\left(\sum_{\sigma_1\in t_iH}\chi_{1}(\sigma_1)\right)\left(\sum_{\sigma_2\in t_iH}\chi_{2}(\sigma_2)\right)\left(\sum_{\sigma_3\in t_iH}\chi_{3}(\sigma_3)\right)\right]
$$
Actually, I just need to know if it is zero or not.
In the case where $H=\{e\}$ (the trivial group) we get
$$
\sum_{\sigma\in G}\chi_1(\sigma)\chi_2(\sigma)\chi_3(\sigma),
$$
and we can reinterpret this as the multiplicity of the trivial rep of $G$ in the (tensor) product rep $\chi_1\otimes \chi_2\otimes \chi_3$ of $G\times G\times G$ restricted to $G$ via $g \mapsto (g,g,g
)$. 
In the case where $H=G$ we get
$$
\left(\sum_{\sigma_1\in G}\chi_{1}(\sigma_1)\right)\left(\sum_{\sigma_1\in G}\chi_{2}(\sigma_2)\right)\left(\sum_{\sigma_3\in G}\chi_{3}(\sigma_3)\right),
$$
and we can reinterpret this as the product of the multiplicities of the trivial rep of $G$ in each of $\chi_1$, $\chi_2$ and $\chi_3$. Hence if $\chi_1$, $\chi_2$ and $\chi_3$ are irreducible we get a non-zero answer iff each character is trivial.
In the intermediate cases with $H$ a proper subgroup, I was hoping it would also just be a matter of reinterpreting using inner products of induced characters $H\uparrow G$, but this doesn't seem to work out (unless I am missing something).
As per my question title you see it comes down to average values of the characters on each coset, but I can't see a nice way of understanding what this is telling me.
Failing a theoretical answer, I've started coding up in GAP to compute specific examples. Any suggestions on how this could be achieved efficiently would also be very much appreciated.
Thanks!
 A: The product will be zero (for any choice of coset representative $t_{i}$ and for each coset) if any of $\chi_{1},\chi_{2},\chi_{3}$ contain the trivial character with multiplicity zero on restriction to $H.$ However, I do not think the converse ( for a particular choice of $t_{i}$) need be true.
 It suffices to note that if the complex (matrix) representation $\sigma$ of the finite group $H$ does not contain the trivial representation, then $\sum_{h \in H} (h \sigma) = 0.$ It suffices to check this for each irreducible constituent of $\sigma,$ by Maschke's Theorem. So it suffices to check the case that $\sigma$ is irreducible, but non-trivial. Then $\sum_{h \in H} h\sigma$ commutes with $x\sigma$ for each $x \in H.$ Hence it is a scalar matrix by Schur's Lemma. But it has trace zero by the orthogonality relations, so it is the zero matrix.
Note then, that for general $\sigma$ not contaig the trivial representation on restriction to $H,$ and any $t \in G,$ we have $\sum_{h \in H}(th\sigma)
= (t\sigma)\sum_{h \in H}h\sigma = [0].$
