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 $\operatorname{End}\_k(V)$.
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> 
> 


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



For centralizer subgroups, is the endomorphism ring of a restriction generated by endomorphisms and the centralized element?
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.


