Over an algebraically closed field, an irreducible finite-dimensional representation has only scalar equivariant endomorphisms.Suggest a slogan
Let \(G\) be a group, \(k\) an algebraically closed field, and \(M\) an irreducible representation of \(G\) over \(k\) with \(\dim_k M < \infty\). Then every \(G\)-equivariant endomorphism of \(M\) is multiplication by a scalar, so the map \[ k \mapsvia{\sim} \Endo_G(M),\qquad \lambda\mapsto \lambda\cdot\id_M, \] is an isomorphism.
For an irreducible representation \(M\) of \(G\) over an arbitrary field \(k\), every nonzero \(G\)-equivariant endomorphism of \(M\) has \(G\)-stable kernel \(0\) and image \(M\), so \(\Endo_G(M)\) is a division algebra over \(k\).
Over an algebraically closed field, an irreducible finite-dimensional representation has only scalar equivariant endomorphisms.