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Schur's lemma says that for finite group representations, this space between non-equivalent irreps has dimension 0, and that the morphisms between identical irreps are homothety. Yet I forgot how to prove that the space of morphisms between equivalent irreps has dimension 1 using these two results. Could anybody give me a hint or answer?

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Let $\rho\colon G \to \mathrm{GL}(V)$ be representation and $k$ the field of scalars.

Morphisms from the representation $\rho$ to itself are linear maps $V \to V$ that commute with the action of $\rho$. The important bit hear is that the map $V \to V$ is linear and a linear homothety is just multiplication by a scalar. So if we define $\phi_\lambda$ by $v \mapsto \lambda v$ then $\lambda \mapsto \phi_\lambda$ gives the isomorphism between $k$ and the space of endomorphisms of the representation $\rho$.

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  • $\begingroup$ Now I can understand the mechanism. Thanks a lot! $\endgroup$ – Math.StackExchange Mar 10 '15 at 4:14

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