Let \(G\) be a finite group, \(k\) an algebraically closed field, and \(V\) an irreducible \(k\)-linear representation of \(G\).
Show that \(\hom_{kG}(V,V)\) is a division algebra with \(k\) in its center.
Show that \(V\) is finite-dimensional over \(k\), and conclude that \(\hom_{kG}(V,V)\) is also finite dimensional.
Show the inclusion \(k\hookrightarrow\hom_{kG}(V,V)\) found in (a) is an isomorphism. (For \(f\in\hom_{kG}(V,V)\), view \(f\) as a linear transformation and consider \(f-\alpha I\), where \(\alpha\) is an eigenvalue of \(f\)).
Solution
1. Every nonzero element of \(\operatorname{Hom}_{kG}(V,V)\) is invertible, so \(\operatorname{Hom}_{kG}(V,V)\) is a division algebra.
Let \(0\ne f\in\operatorname{Hom}_{kG}(V,V)\). Both \(\ker f\) and \(\operatorname{im}f\) are \(kG\)-submodules of \(V\). Since \(V\) is irreducible, \[ \ker f\in\{0,V\},\qquad \operatorname{im}f\in\{0,V\}. \] Because \(f\ne0\), one cannot have \(\ker f=V\) or \(\operatorname{im}f=0\). Hence \(\ker f=0\) and \(\operatorname{im}f=V\), so \(f\) is bijective. Its inverse is again \(kG\)-linear: if \(v=f(u)\), then for \(g\in G\), \[ f^{-1}(gv)=f^{-1}(gf(u))=f^{-1}(f(gu))=gu=g f^{-1}(v). \] Thus every nonzero endomorphism is a unit.
2. The scalar maps embed \(k\) into the center of \(\operatorname{Hom}_{kG}(V,V)\).
For \(\lambda\in k\), the map \(\lambda I_V\) is \(kG\)-linear, and \[ (\lambda I_V)f=f(\lambda I_V) \] for every \(k\)-linear endomorphism \(f\). The map \(\lambda\mapsto\lambda I_V\) is injective because \(V\ne0\).
3. The representation \(V\) is finite-dimensional over \(k\).
Choose \(0\ne v\in V\). The \(k\)-span \[ W=\operatorname{span}_k\{gv:g\in G\} \] is a nonzero \(kG\)-submodule of \(V\). Since \(V\) is irreducible, \(W=V\). Because \(G\) is finite, \(W\) is spanned by at most \(|G|\) vectors, hence \[ \dim_kV\le |G|<\infty. \]
4. The algebra \(\operatorname{Hom}_{kG}(V,V)\) is finite-dimensional over \(k\).
It is a \(k\)-linear subspace of the finite-dimensional vector space \(\operatorname{End}_k(V)\), whose dimension is \((\dim_kV)^2\).
5. Every \(kG\)-endomorphism of \(V\) is scalar. Therefore\[ k\xrightarrow{\ \sim\ }\operatorname{Hom}_{kG}(V,V), \qquad \lambda\longmapsto\lambda I_V. \]
Let \(f\in\operatorname{Hom}_{kG}(V,V)\). Since \(V\) is finite-dimensional and \(k\) is algebraically closed, the characteristic polynomial of \(f\) has a root \(\alpha\in k\). Hence \(f-\alpha I_V\) has nonzero kernel and is therefore not invertible. But by 1 every nonzero element of \(\operatorname{Hom}_{kG}(V,V)\) is invertible. Consequently \[ f-\alpha I_V=0, \] so \(f=\alpha I_V\). Together with 2, this proves the scalar inclusion is an isomorphism.