1
$\begingroup$

$G$ is a finite group with a subgroup $H$. Let $\rho_1:G \to GL(V)$ and $\rho_2:H \to GL(U)$ be irreducible representations. $Z=\mathbb{C}[G]^H$, i.e., $Z$ is the centralizer of $H$ in $\mathbb{C}[G]$.

Show that the following defines an action of $Z$ on the space $\def\Hom{\operatorname{Hom}}\Hom_H(U,V )$ of functions: $f\mapsto g.f := \rho_1(g)\circ f$. Further, Show that this action is a transitive action (or equivalently $\Hom_H(U,V)$ is an irreducible $C[G]^H$ module).

It's easy to see that the given map is an action of $Z$ on $\Hom_H(U,V)$. But I am unable to prove that the action is transitive i.e. if $f_1,f_2$ (nonzero) belongs to $\Hom_H(U,V) $ then there exists $g$ in $Z$ such that $g.f_1=f_2$. Any Hints/ideas ?

$\endgroup$
  • $\begingroup$ The action on $\text{Hom}_H(U,V)$ by $Z$ is the action of an algebra (thus $\text{Hom}_H(U,V)$ becomes a $Z$-module), not of a group, so what do you mean when you say the action of $Z$ is supposed to be transitive? $\endgroup$ – Matthias Klupsch Apr 17 '15 at 8:46
  • $\begingroup$ Dear @MatthiasKlupsch sorry for confusion. I have added the details. $\endgroup$ – Arpit Kansal Apr 17 '15 at 8:52
  • 4
    $\begingroup$ By your definition, transitivity is impossible as far as I can tell, because for example the zero map $U \rightarrow V$ is obviously invariant under the action of $Z$. $\endgroup$ – Matthias Klupsch Apr 17 '15 at 12:26
  • 1
    $\begingroup$ Is this related to this other question, which also is false as stated? Note that a $Z$-module $W$ is irreducible iff $Zw=W$ for all $w\in W\setminus0$, which is equivalent to $\forall w_1,w_2\in W\setminus0,~\exists z\in Z:zw_1=w_2$, which is almost the definition of a transitive action (although $Z$ is not a group). Please provide a source for the question so we can get to the bottom of this. $\endgroup$ – whacka Apr 17 '15 at 13:18
  • 1
    $\begingroup$ This is still false, as can be seen by slightly augmenting the reasoning in my answer in the other question I linked to. If $H=1$ and $m=\dim U>1$ then $\hom_H(U,V)\cong V^{\oplus m}$ as a representation of $Z=\Bbb C[G]$, which cannot be irreducible. $\endgroup$ – whacka Apr 19 '15 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.