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]?


No, let $G$ be an arbitrary group and take $H=G$ acting on itself by conjugation. Then $H$ becomes the diagonal in $G\rtimes H\cong G\times G$. This is well known to be a Gelfand pair.

PS: The Hecke algebra of the pair $(G\rtimes H,H)$ is in fact equal to the fixed point algebra $\mathbb C[G]^H$. This explains your observation when $G$ is abelian.

  • $\begingroup$ The fact that $\mathbb{C}[H \backslash (G\rtimes H) / H] \simeq \mathbb C[G]^H$ explains also your first paragraph because (for the action by conjugation) $\mathbb C[G]^G = Z(\mathbb C[G])$ which is commutative by definition. $\endgroup$ – Sebastien Palcoux Jun 15 '16 at 18:28
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    $\begingroup$ The equality holds because $X:=G\rtimes H/H$ equals $G$ with $H$ acting by automorphisms. Thus $H\backslash X$ is the set of $H$-orbits on $G$. But those also give a basis of $\mathbb C[G]^H$ (take the sum over each orbit). $\endgroup$ – Friedrich Knop Jun 15 '16 at 18:29
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    $\begingroup$ "$X:=G\rtimes H/H$ equals $G$" means that they can be naturally identified as set: $X=\{ (g,H) \ | \ g \in G \}$. Then $H\backslash X = \{ (\sigma_H(g),H) \ | \ g \in G \}$, and $\{ \sum_{h \in H}\sigma_h(g) \ | \ g \in G \}$ generates $\mathbb C[G]^H$. $\endgroup$ – Sebastien Palcoux Jun 15 '16 at 18:45

It is perhaps worth pointing out the special case that if the semidirect product $GH$ is a Frobenius group with kernel $G$ and complement $H$, then $(GH,H)$ is a Gelfand pair if and only if $G$ is Abelian.

In general, we have a Gelfand pair from such a semidirect product if and only if the permutation character $\theta = {\rm Ind}_{H}^{GH}(1)$ is multiplicity free. One explanation why we always get a Gelfand pair when $G$ is Abelian is that even the restriction ${\rm Res}^{GH}_{G}(\theta)$ is the regular character of $G$, which is multiplicity free when $G$ is Abelian.

Suppose now that $GH$ is such a Frobenius group, but that $G$ is non-Abelian. Then $G$ has a non-linear irreducible character $\mu$, which occurs with multiplicity $\mu(1)$ in ${\rm Res}^{GH}_{G}(\theta)$. On the the other hand, there is just one irreducible character $\chi$ of $G$ which contains $\mu$ with non-zero multiplicity on restriction to $G$, and that is $\chi = {\rm Ind}^{GH}_{G}(\mu)$. Hence $\chi$ must occur with multiplicity $\mu(1) > 1$ in $\theta $, so that $(GH,H)$ is not a Gelfand pair.

More generally, when $H$ is cyclic, or when $H$ is Abelian of order relatively prime to $|G|$, it is possible to use Clifford's theorem to check that $(GH,H)$ is not a Gelfand pair if there is an irreducible character $\mu$ of $G$ such that $\mu(1) > |I_{H}(\mu)|$, where $I_{H}(\mu) = \{h \in H : \mu^{h} = \mu \}$, the Frobenius group case being an extreme example of this where $I_{H}(\mu)$ is the trivial group.


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