I came across [this preprint](https://arxiv.org/pdf/1210.7440.pdf) that claims in Lemma 1.1 that Gelfand's trick (also known as Gelfand's lemma) only works in characteristic zero:



> 
> Let $H < G$ be finite groups. Suppose we have an anti-involution $\sigma : G \rightarrow G$ that preserves all H double-cosets. Then over algebraically closed fields **of characteristic zero** (G, H) is a Gelfand pair.
> 
> 
> 


It is not obvious to me where the characteristic of the ground field being zero is used in the proofs from Lang's $SL\_2(\mathbb{R})$ book (Theorem 1 and Theorem 3 in Chapter IV), the introduction in [this preprint](https://arxiv.org/pdf/1810.12888.pdf), the last slides in [these slides](http://www.wisdom.weizmann.ac.il/~dimagur/FunRepGel.pdf), or anything else that I've seen.


What causes Gelfand's trick to fail in positive characteristic? In this setting, the groups are finite but I would also like to know the answer for the more general versions of Gelfand's trick (i.e. also for locally compact groups with compact subgroups or even reductive groups over local fields with closed subgroups).



Gelfand's trick (Gelfand's lemma) in positive characteristic?
Let me try to muddle my way through. In positive characteristic the Gelfand trick will still tell you that the Hecke algebra of $H$-biinvariant functions $A=Func(H\backslash G/H,k)$ is commutative.


Then, by general nonsense, the Hecke algebra is isomorphic to $End\_G(P)$, where $P=k(G/H)=Ind\_H^G (k\_{triv})$ is the permutation representation.


Since $P$ is not semisimple, its structure can still be complicated. The sort of information you need to conclude that it is a Gelfand pair is that $Soc(P)$ or $P/Rad(P)$ is multiplicity-free. There is no way you can conclude that about $Soc(P)$.


But, here, Doc, is a happy ending for your opera: it follows from Nakayama Lemma that $$End(P/Rad(P))\cong A/Rad(A),$$
forcing the semisimple module $P/Rad(P)$ to be multiplicity free.


Let $V$ be an irreducible $G$-module. By Frobenius-Nakayama-Fudd Reciprocity,
$$
Hom\_H(k\_{triv},V)=Hom\_G(P,V)=Hom\_G(P/Rad(P),V)
$$
must be at most one-dimensional! And so is
$$
Hom\_H(V,k\_{triv})=Hom\_H(k\_{triv},V^\ast).
$$


