Let $G$ be a finite group and $K \subset G$ a subgroup. Then $(G,K)$ is a Gelfand pair if the double coset Hecke algebra $\mathbb{C}(K \backslash G / K)$ is commutative.

Let $H$ be a subgroup of $Aut(G)$. We will consider $H$ and $G$ as subgroups of $G\rtimes H$.

*Lemma*: $(G\rtimes H,H)$ is a Gelfand pair iff $\forall g,g' \in G$, $HgHg'H = Hg'HgH$.

*Proposition*: If $G$ is abelian, then $(G\rtimes H,H)$ is a Gelfand pair.

*proof*: $\forall h_1, h_2, h_3 \in H$ and $\forall g,g' \in G$, we have:

$$h_1gh_2g'h_3 = h_1 g \sigma_{h_2}(g') h_2 h_3 = h_1 \sigma_{h_2}(g') g h_2 h_3 = h_1 h_2 g' h_2^{-1} g h_2h_3$$ which means that $HgHg'H \subseteq Hg'HgH$. Idem, $Hg'HgH \subseteq HgHg'H$. $\square$

*Question*: Is the converse true [i.e. $(G\rtimes H,H)$ Gelfand pair $\Rightarrow$ $G$ abelian]?