Let $M$ normal in $H\subseteq G$ with $[G:H]<\infty$ and $H/M$ abelian, and let $T$ be a right transversal for $H$ in $G$, there exists a subset $T_0\subseteq T$ and positive integers $n_t$ for $t\in T_0$ such that $$\displaystyle \pi(g)≡\prod_{t\in T_0}tg^{n_t} t^{-1}\mod M$$
It’s from page 119 of M.Isaacs Algebra about transfer theory
I would be so grateful if anybody could help obtain a proof for this.
Choose some $t \in T$. Then $tg=h_1t_1$ for some $t_1 \in T$, $h_1 \in H$. If $t_1 \ne t$, then $t_1g = h_2t_2$ for some $t_2 \in T$, $h_2 \in H$. Keep doing this until some $t_i$ is repeated, which must happen, because $T$ is finite. In fact the $t_i$ must eventaully cycle, so the first one to repeat must be $t$ itself.
Let $n_t \ge 1$ be minimal such that $t = t_{n_t}$. Then $h_1h_2\cdots h_{n_t} = gt^{n_t}t^{-1}$, so you get one term from the product formula you are trying to prove for $\pi(g)$. We put $t$ into the set $T_0$, but we do not put the other $t_i$ in this cycle, because we have included those already in the product.
If $n_t<n=|G:H|$, then we choose another $t \in T$ that did not arise in the previous cycle and do the same thing, getting another term in the product. Keep going until $\sum n_t = n$.