Let $G$ be finite abelian group of type $(m_1, \dots, m_d)$: $G = \langle a_i \rangle \times \cdots \times \langle a_d \rangle$, where $|\langle a_i \rangle| = m_i$
And let $S \subseteq G$: $k = |S|$, $S = \{s_1, \dots, s_k\}$. $$ \begin{eqnarray*} s_1 & = & a_1^{k_{11}} \cdot a_2^{k_{12}} \dots a_d^{k_{1d}} \\ s_2 & = & a_1^{k_{21}} \cdot a_2^{k_{22}} \dots a_d^{k_{2d}} \\ & \dots \\ s_k & = & a_1^{k_{k1}} \cdot a_2^{k_{k2}} \dots a_d^{k_{kd}} \\ \end{eqnarray*} $$ It is convenient for me that the set $S$ is given by the $d \times k$ matrix $(K_{li}) = \frac{k_{li}}{m_i}$
Let $\Gamma$ be a Caley graph of group $G$ and generating set $S$. I want to count triangles in $\Gamma$.
Well know, that spectrum of this Caley graph is ${\rm Spec}(\Gamma) = \{\chi_i(S) | i=1,\dots,n\}$, where $\chi(S) = \sum_{s\in S}\chi(s).$ Thus, number of triangles $\Delta(\Gamma)$ is $$ \Delta(\Gamma) = \frac{1}{6} \sum\limits_{i=1}^n \chi_i(S)^3 $$
There is a canonical bijection between the set of characters $\{\chi_i \}$ and $G$.
If $a \in G$ and $a = a_1^{x_1} \cdots a_d^{x_d}$, $\zeta_i$ be a primitive $m_i$-th root ofunity, then we can define $$ \chi_{a}(a_1^{y_1} \cdots a_d^{y_d})= \chi_{x_1, x_2, \dots, x_d}(a_1^{x_1} \cdots a_d^{x_d}) =\zeta_1^{x_1 y_1} \cdots \zeta_d^{x_d y_d} $$ And $$ \Delta(\Gamma) =\frac{1}{6} \sum\limits_{x_1 = 1}^{m_1} \sum\limits_{x_2=1}^{m_2} \cdots \sum\limits_{x_d=1}^{m_d} \left(\sum_{l = 1}^k \zeta_1^{x_1 k_{l1}} \cdots \zeta_d^{x_d k_{ld}}\right)^3 $$ By $$ \zeta_1^{x_1 k_{l1}} \cdots \zeta_d^{x_d k_{ld}} = e^{\frac{2i\pi}{m_1}x_1k_{1l}} \cdots e^{\frac{2i\pi}{m_d}x_dk_{dl}} $$ and $$ {\frac{2i\pi}{m_1}x_1k_{1l}} + \cdots +{\frac{2i\pi}{m_d}x_dk_{dl}} = 2i\pi\sum\limits_{i=1}^d \frac{x_i k_{li}}{m_i} $$ Finally we get $$ \Delta(\Gamma) =\frac{1}{6} \sum\limits_{x_1 = 1}^{m_1} \sum\limits_{x_2=1}^{m_2} \cdots \sum\limits_{x_d=1}^{m_d} \left( e^{2\pi i F(x_1,x_2,\dots,x_d)}, 1_k \right)^3 $$ where $F(x_1,x_2,\dots,x_d)$ is vector-valued function $F(x_1,x_2,\dots,x_d) = K\cdot {\rm x}$, ${\rm x} = (x_1,x_2,\dots,x_d)^{T}$, $1_k$ - all-$1$ vector, and $(\cdot, \cdot)$ - dot product.
This formula looks awful, can it be simplified? For example, using additionally that $S^{-1} = S$.
Is it possible to get a nontrivial upper bound of triangles number in terms of $m_i$ and $|S|$?
Is it possible to get simple criteria that $\Gamma$ is triangle free?