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Let $\rho : G \to GL(V)$ be an irreducible representation of a finite group. Schur's lemma says if $\pi:GL(V) \to GL(V)$ intertwines with $\rho$, that is, $\pi \rho(g) = \rho(g) \pi$ for every $g\in G$, then $\pi = \lambda I$ for some $\lambda \in \mathbb{C}$.

Is there a similar lemma for $\rho = m_1 \rho_1 \oplus \ldots \oplus m_k \rho_k$, where $\rho_1, \ldots, \rho_k$ are different irreducible representations? The question is, given such $\rho$, if $\pi \rho = \rho \pi$, then what is the structure of $\pi$?

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up vote 10 down vote accepted

This is an easy exercise. If $\rho _i$ are "different" (i.e. inequivalent) then by Schur's lemma, $Hom _G(\rho _i,\rho _i)={\mathbb C}I$ and $Hom _G(\rho _i, \rho _j)=0$. Hence the commutant of $G$ in $End(\rho)$ is easily seen to be the product

$$M_{m_1}({\mathbb C})\times \cdots \times M_{m_k}({\mathbb C}),$$ where $M_n({\mathbb C})$ is the algebra of $n\times n$ matrices with complex entries.

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Schur's lemma has a different generalization when the coefficient field $F$ is not algebraically closed. Then you get $M_{m_1}(D_1)\times\cdots\times M_{m_k}(D_k)$ where $D_i:={\rm Hom}_{FG}(\rho_i,\rho_i)$ is a division algebra over $F$ by Schur's lemma. If $F$ is infinite, noncommutative $D_i$ can arise. For example, the quaternion group of order 8 has a 4-dimensional irreducible representation $\rho$ over the rationals, where ${\rm Hom}_{\mathbb{Q}G}(\rho,\rho)$ is the rational quaternions.

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