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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.

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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, $K \cap B = 1$.

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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 $G_{(W)} \cap G_{(V)} \subset G_{(U)} \cap G_{(V)}$. $\square$


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 $G_{(V^H)}=H$.

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    $\begingroup$ It seems that the first equality in the lemma of Palcoux is not true in general. The kernel of the tensor product of two modules is not necessarily the intersection of the kernels. For example, look at G the Klein four's group. The tensor product of two distinct nontrivial irreducible representations is the third one, and its kernel is nontrivial. $\endgroup$ – Marty Isaacs Mar 28 '15 at 17:35
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    $\begingroup$ @MartyIsaacs: You're right, for $u,v≠0$, $g(u⊗v)=u⊗v$ iff $gu=αu$ and $gv=βv$ with $αβ=1$. Now, we can replace this equality by an inclusion and the argument still works. I've edited that, thank you! $\endgroup$ – Sebastien Palcoux Mar 29 '15 at 6:50
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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 $H \subseteq G_{(X^H)} \subseteq G$).

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