Let $F$ be a finite extension of $\mathbb Q\_p$, where p is an odd prime. Let $G$ be a connected reductive group defined over $F$. Let $M, H$ be closed $F$-subgroups of $G$ (in particular, I'm interested in the case when $M$ is a Levi subgroup of $G$, and $H$ is the group of fixed points of an $F$-involution of $G$.)


I'm looking for a proof or a reference for the following claim.


Claim: if the Galois cohomology of $M\cap H$ over $F$ is trivial, then $(MH)(F) = M(F)H(F)$, where $X(F)$ denotes the $F$-rational points of an $F$-variety $X$.



F-points of product of closed subgroups vs. product of F-points, F a local field, reference?
I am just posting my comment above as an answer. Let $K/F$ be a finite Galois extension, and let $m \in M(K)$ and $h\in H(K)$ be elements such that the element $g=m\cdot h^{-1}$ is Galois invariant, i.e., $g$ is an element of $G(F)$. Consider the non-Abelian $1$-cocycles in $G(K)$ for $\Gamma = \text{Aut}(K/F)$, $( m^{-1}\cdot \gamma(m) )\_{\gamma\in \Gamma}$ and $(h^{-1}\cdot \gamma(h))\_{\gamma\in \Gamma}$. Since $m\cdot h^{-1}$ is Galois invariant, these $1$-cocycles are equal, and they are $1$-cocycles in $(M\cap H)(K)$. By your hypothesis, this is a coboundary $(r^{-1}\cdot \gamma(r))\_{\gamma\in \Gamma}$ for an element $r\in (M\cap H)(K)$. Then defining $\widetilde{m}=m\cdot r^{-1}$ and $\widetilde{h} = h\cdot r^{-1}$, these satisfy $\widetilde{m}\cdot \widetilde{h}^{-1}$ equals $g$, and both $\widetilde{m}$ and $\widetilde{h}$ are Galois invariant, i.e., they are elements of $M(F)$, resp. $H(F)$.


