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

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