Past exam of the algebra course of the University of Oxford / 2009 / A / 1

For a finite-dimensional algebra, every simple module $S$ is finite-dimensional: choose $s\ne 0$; simplicity makes $A\to S$, $a\mapsto as$, surjective, so $\dim S\le \dim A$.

The general Schur's lemma says that a homomorphism $f:M\to N$ between simple modules over any unital ring is either zero or an isomorphism. Indeed, $\ker f$ is a submodule of $M$ and $\mathrm{im}f$ is a submodule of $N$. If $f\ne 0$, simplicity gives $\ker f=0$ and $\mathrm{im}f=N$. In particular, ${\mathrm{End}}_R(M)$ is a division ring.

For a complex algebra and a finite-dimensional simple module $M$, the stronger conclusion isFor $f\in {\mathrm{End}}_R(M)$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. The endomorphism $f-\lambda I$ has nonzero kernel, so the general Schur's lemma forces it to be zero. Consequently a homomorphism between two such simple modules is zero if they are nonisomorphic, and their homomorphism space is one-dimensional if they are isomorphic.

In the finite-dimensional setting of the preceding part, let $S$ be a simple module. Each action map $s\mapsto as$ is an $A$-module endomorphism because $A$ is commutative. The complex Schur's lemma makes every such action a scalar. Therefore every linear subspace of $S$ is an submodule; simplicity forcesEquivalently, a simple module over a commutative ring is a quotient by a maximal ideal, and a finite-dimensional field extension of $\mathbb{C}$ is just $\mathbb{C}$.

The finite-dimensional context matters. If the clause is read as allowing an arbitrary commutative complex algebra, the unrestricted assertion is false: $A=\mathbb{C}(t)$ is simple as a module over itself and is infinite-dimensional over $\mathbb{C}$. This is simple module of a finite-dimensional commutative complex algebra. Thus the one-dimensional conclusion uses the finite-dimensional algebra context, or an alternative hypothesis such as finite dimensionality of the simple module.

For $h\in G$, right-multiplying by $D_2(h{)}^{-1}$ givesMake the bijective substitution $k=hg$, so $g^{-1}{h}^{-1}=k^{-1}$. The sum becomes ${\sum }_k{D}_1(k)B{D}_2(k^{-1})=T$. HenceThus $T$ is an intertwining operator from the second group representation to the first. This is the Schur averaging of rectangular matrices construction.

The preceding intertwining identity makes $T$ a homomorphism between the corresponding simple modules. By Schur's lemma, a nonzero such homomorphism would be an isomorphism. The representations are inequivalent, so this is impossible. Therefore

For equal irreducible group representations, the intertwining identity and the complex Schur's lemma give $T=c{I}_n$. Conjugation preserves trace, soSince $\mathrm{tr}(c{I}_n)=nc$, we obtain