In some recent doodlings, I got myself to the point where what I was trying to understand would work out if the following claim were true:

Let $G$ be a group, $g\in G$, and $\rho:G \to \operatorname{End}_k(V)$ a $G$-module over some commutative ring $k$. Let $H < G$ denote the subgroup centralizing $g$ (i.e. $h\in H$ iff $hg=gh$), and consider restricting $V$ to $H$. Then $\operatorname{End}_H(V)$, the ring of $k$-linear maps commuting with all actions of $\rho(h)$ for all $h\in H$, obviously contains both $\rho(g)$ and $\operatorname{End}_G(V)$. The claim is that in fact $\rho(g)$ together with $\operatorname{End}_G(V)$ generate $\operatorname{End}_H(V)$ as a subring of $\operatorname{End}_k(V)$.

This is presumably a classical fact that I have since forgotten from a first course in representation theory. Does it have an easy proof?

  • $\begingroup$ Of course, since I don't know how to prove the claim, I also entertain the possibility that it's false, in which case I need to go doodle some more. $\endgroup$ – Theo Johnson-Freyd Dec 14 '13 at 18:33
  • $\begingroup$ If $H$ is trivial and $\rho$ is irreducible, we have $End_k =End_H$ and $End_G$=multiples of the identity. Does $\rho(G)$ generate in $End_k$, if it is irreducible? $\endgroup$ – Marc Palm Dec 14 '13 at 19:00
  • $\begingroup$ If $H$ is trivial and $\rho$ is a sum of two non-isomorphic reps, you can't actually map from one component to the other then? $\endgroup$ – Marc Palm Dec 14 '13 at 19:03
  • 2
    $\begingroup$ Maybe we should ask: what is a necessary and sufficient condition for $\mathrm{End}_H(V)$ to be generated by $\rho(g)$ and $\mathrm{End}_G(V)$? $\endgroup$ – Mark Wildon Dec 14 '13 at 19:07
  • $\begingroup$ @MarcPalm: Of course, centralizer subgroups are never trivial, since $g\in H$, and the centralizer of the identity is all of $G$. $\endgroup$ – Theo Johnson-Freyd Dec 14 '13 at 21:53

Here is a counterexample. Let $G = \langle g, k : g^2 = x^5 = 1, x^g = x^{-1} \rangle$ be the dihedral group of order $10$. Take $k = \mathbb{C}$ and let $U$ and $U'$ be the two distinct $2$-dimensional irreducible representations of $G$. Let $\rho : G \rightarrow \mathrm{End}(U \oplus U')$ be the corresponding representation. Let $H = \mathrm{Cent}_G(g) = \langle g \rangle$.

By Schur's Lemma, $\mathrm{End}_G(U \oplus U') \cong \mathbb{C} \oplus \mathbb{C}$. In particular, $\mathrm{End}_G(U \oplus U')$ is commutative. Hence the algebra generated by $\mathrm{End}_G(U \oplus U')$ and $\rho(g)$ is commutative. On the other hand,

$$ U \!\downarrow_H \, \cong U'\!\downarrow_H \, \cong \mathbb{C} \oplus W$$

where $W$ is the unique non-trivial irreducible representation of $H$ and so $$\mathrm{End}_H(U \oplus U') = \mathrm{End}_H(\mathbb{C} \oplus \mathbb{C} \oplus W \oplus W)$$ is the sum of two copies of the matrix algebra $\mathrm{Mat}_2(\mathbb{C})$.

Edit: Julian Rosen has already posted a simpler counterexample. The example here shows that $\rho$ can be faithful.

  • $\begingroup$ Very nice. This also answers (in the negative) the version of my question with the restrictions to finite groups and $\mathbb C$-representations. C'est la vie. $\endgroup$ – Theo Johnson-Freyd Dec 14 '13 at 22:01

I don't think this is true.

Let $G=F_2=\langle a,b\rangle$, the free group on two generators. Take $V=k^2$ (say $k$ a field of characteristic not 2), with $a$ acting trivially and $b$ acting by $(x,y)\mapsto (2x,y)$. Take $g=a$, so that $H=\langle a\rangle$.

We have $\mathrm{End}_H(V)=\mathrm{End}_k(V)$, $\rho(g)=\mathrm{Id}_V$, and $\mathrm{End}_G(V)=\{\text{diagonal matrices}\}$.


Assume $H$ is trivial, and $\rho$ a sum of two irreducible, non-isomorphic representations $V=V_1+V_2$ then $End_G(V) = \mathbb{C}^2$ and you have $End_k(V) = End_H(V)$. Because $\rho(G)$ will preserve the $V_i$'s, your conjecture will fail in that case. You can't construct an operator mapping $V_1 \rightarrow V_2$ with elements of type $\rho(g)$.

If $\rho$ is irreducible and $H$ is trivial, you claim that $\rho(G)$ generates $End_k(V)$. You have that $\{ \rho(g) v : g \in G\}$ spans $V$, but I would guess that isn't sufficient as well.


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