I am trying to prove some statements about specific Gelfand pairs, when considering representations of some finite groups over field $k$ which can be of positive characteristic. For this I study some classical properties of Hecke algebras of finite groups. Here is a definition I'm working with. Let $G$ be a finite group with a subgroup $H$.
We define a $k$-module of bi-invariant $k$-valued functions: \begin{equation} \mathcal{H}_{k}(G, H) = \left\{\Delta: G \longrightarrow k \mid \forall h_1 ,h_2\in H, \forall g\in G: \Delta(h_1 g h_2) = \Delta(g)\right\} \end{equation}
To provide it with an algebra structure one defines a convolution product on it. And in a zero characteristic case one does it as \begin{equation} (\Delta_{1}*\Delta_{2})(g ) = \frac{1}{|H|}\sum_{x\in G} \Delta_{1}(x)\Delta_{2}(x^{-1}g) \end{equation} This normalisation factor $1/|H|$is important, then showing that $\mathcal{H}_{k}(G, H) \simeq \mathrm{End}_{G}(\mathrm{Ind}_{H}^{G}(\mathrm{triv}_{H},k_{\mathrm{triv}_{H}}))$ (endomorphisms of induction from trivial representation of $H$). The problem is, that this factor brings several restrictions when going to modular setting. A possible solution of this issue, that I saw in some text by D. Bump (here for example http://sporadic.stanford.edu/bump/finite.pdf) is to try to consider a sum over the representatives of cosets: \begin{equation} (\Delta_{1}*\Delta_{2})(g) = \sum_{Hx \in H\backslash G} \Delta_{1}(x)\Delta_{2}(x^{-1}g) \end{equation} however, in this case I faced a lot of troubles showing that the convolution is associative, that the indicator-function of $H$ is a unit and showing that $\mathcal{H}_{k}(G, H)\simeq k[H\backslash G/ H] \simeq \mathrm{End}_{G}(\mathrm{Ind}_{H}^{G}(\mathrm{triv}_{H},k_{\mathrm{triv}_{H}}))$. I tried playing with the order of multiples in this sum and choice of right/left cosets or choice of definition of induced representation, but every time I'm facing the problem, which is that re-indexing the sum is not always preserving the cosets.
I want to understand if it is actually impossible to define convolution this way, or maybe I am just missing something with the definitions.