A problem with pointwise stabilizer subgroups of fixed-point subspaces II

Question

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, $U$ 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.

Answer 1

Here is a proof of a special case of the problem: when $H$ is the trivial subgroup of $G$.

In this case, we have two irreducible characters $\alpha$ and $\beta$ of $G$ with kernels $A$ and $B$, respectively, and $L = A \cap B$. We want to show that there is an irreducible character $\chi$ of $G$ whose kernel $K$ satisfies $K \cap A \subseteq L$ and $K \cap B \subseteq L$. We can replace $G$ by $G/L$, so we can assume $L = 1$. Let $M = AB$, so $M$ is the direct product of $A$ and $B$.

Let $\mu$ and $\nu$ be irreducible constituents of $\beta_A$ and $\alpha_B$, respectively, and let $\gamma = \mu \times \nu \in$ Irr($M$). Let $\chi$ be any irreducible character of $G$ lying over $\gamma$, and let $K$ be the kernel of $\chi$. I claim that $K \cap A = 1$ and $K \cap B = 1$

Suppose $x \in K \cap A$. Now $K =\,$ker($\chi$)$\,\subseteq\,$ker($\gamma$), so $\mu(1)\nu(1) = \gamma(1) = \gamma(x) = \mu(x)\nu(1)$, and thus $x \in\,$ker($\mu$). This shows that $K \cap A \subseteq\,$ker($\mu$). But $K \cap A$ is normal in $G$ and $\mu$ lies under $\beta$, and it follows that $K \cap A \subseteq\,$ker($\beta$)$\,=B$. Thus $K \cap A \subseteq A \cap B = 1$. as wanted, and similarly, $K \cap B = 1$.

Answer 2

Here is an other proof for the case $H$ trivial.

Definition: Let $\langle U,V \rangle$ be the direct sum of all the irreducible representations generated by the irreducible complex representations $U$ and $V$ of the finite group $G$, for $\otimes$.

Lemma: $G_{(U \otimes V)} \supset G_{(U)} \cap G_{(V)} = G_{(\langle U,V \rangle)}$.
proof: The first inclusion is clear. Next, if $G_{(U)} \cap G_{(V)} \subset G_{(W_i)}$, then $G_{(U)} \cap G_{(V)} \subset G_{(W_1)} \cap G_{(W_2)}$ $ \subset G_{(W_1 \otimes W_2)}$, so $G_{(U)} \cap G_{(V)} \subset G_{(\langle U,V \rangle)}$. Now, $U,V \le \langle U,V \rangle$, so $G_{(\langle U,V \rangle)} \subset G_{(U)}, G_{(V)}$. $\square$

Next by Frobenius reciprocity, if $W \le U \otimes V$ then $U \le W \otimes \overline{V}$ and $V \le \overline{U} \otimes W$.
So $G_{(W \otimes \overline{V})} \subset G_{(U)}$ and $G_{(\overline{U} \otimes W)} \subset G_{(V)}$. Finally, $\langle W,\overline{V} \rangle = \langle W,V \rangle$ and $\langle \overline{U}, W \rangle = \langle U, W \rangle$ (because $G$ is finite), so by the previous lemma: $G_{(U)} \cap G_{(W)}$ and $G_{(W)} \cap G_{(V)} \subset G_{(U)} \cap G_{(V)}$. $\square$

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Remark on the general case

If $U^H = 0$ then $W^H = 0$, $\forall W \le U \otimes V$. So if $G_{(V^H)} \neq G$ then $G_{(U)} \cap G_{(W)} \not\subset G_{(U)} \cap G_{(V)}$. Ex: $G = S_4$, $H=G_1 = S_3$, $U$ the non-trivial $1$-dim. irr. rep., $V$ the $3$-dim. irr. rep. with $G_{(V^H)}=H$.

Answer 3

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 $H \subseteq G_{(X^H)} \subseteq G$).