*Definitions*: Let $W$ be a representation of a group $G$, $K$ a subgroup of $G$, and $X$ a subspace of $W$.  

 Let the *fixed-point subspace* $W^{K}:=\{w \in W \ \vert \ kw=w \ , \forall k \in K \}$.  

Let the *pointwise stabilizer subgroup* $G\_{(X)}:=\{ g \in G \ \vert \ gx=x \ , \forall x \in X \}$. 


Let $G$ be a finite group, $H$ a [core-free subgroup](http://groupprops.subwiki.org/wiki/Core-free_subgroup), $U$ and $V$ two irred. complex representations of $G$.  

Let the subgroup $L := G\_{(U^H)} \cap G\_{(V^H)}$. 


*Question*: Is there an irr. complex rep. $W$ of $G$ such that $G\_{(W^H)} \cap G\_{(V^H)}$ and $G\_{(U^H)} \cap G\_{(W^H)} \subset L$? 


*Motivation*: This would be very helpful for proving the [dual version of a theorem of Øystein Ore](https://mathoverflow.net/q/195366/34538). 



A problem with pointwise stabilizer subgroups of fixed-point subspaces II
**No**, GAP has found the following counter-example. 


`gap> G:=TransitiveGroup(16,39);`  

`gap> H:=Stabilizer(G,1);` 


$G$ and $H$ are order $32$ and $2$. 


`gap> R:=IrreducibleRepresentations(G);`  

`gap> U:=R[9];`  

`gap> V:=R[11];` 


$U$ and $V$ are degree $2$. 


Now $G\_{(U^H)}$ and $G\_{(V^H)}$ are order $8$ and $4$, and $G\_{(U^H)} \cap G\_{(V^H)} = H$. 


We checked that $\forall W$ irr. then $(G\_{(W^H)} \cap G\_{(V^H)},G\_{(U^H)} \cap G\_{(W^H)}) \neq (H,H)$.  

So we get a counter-example (because in general $H \subseteq G\_{(X^H)} \subseteq G$).


