Suppose $\Psi:G\rar\UU(E)$ is irreducible and $\dim E=n$. 1. Show that for all $g\in G$ and all $x,y\in E$: $$ \frac{l}{|G|}\Big(\sum_{g\in G}\Psi(g)\otimes\Psi(g)^*\Big)x\otimes y=y\otimes x~. $$ 2. Let $e_1,\ldots,e_n$ be an orthonormal basis for $E$ and put $E^{jk}x\colon=\la x,e_k\ra e_j$ and define the exchange or swap operator $P_{ex}\in\Hom(E\otimes E)$ by $$ P_{ex}=\sum_{j,k}E^{kj}\otimes E^{jk}=\sum_{j,k}E^{kj}\otimes E^{kj*}~. $$ Verify that $P_{ex}(x\otimes y)=y\otimes x$, that the eigen-values of the unitary operator $P_{ex}$ are $\pm1$ and that for all $A\in\Hom(E)$: $[A\otimes A,P_{ex}]=0$. 3. Cf. exam and prove that for all $A\in\Hom(E)$: $\tr_1((A\otimes1)P_{ex})=A$.

1. This follows immediately from the Great Orthogonality Theorem. 2. Now $E^{jk}e_l=\d_{kl}e_j$, i.e. the matrix of $E^{jk}$ with respect to the basis $e_1,\ldots,e_n$ has just one non-zero entry: the $(j,k)$ entry, which is $1$. Hence we get $$ P_{ex}(e_l\otimes e_m) =\sum_{j,k}E^{kj}e_l\otimes E^{jk}e_m =\sum_{j,k}\d_{jl}\d_{km}e_k\otimes e_j =e_m\otimes e_l $$ and therefore $P_{ex}(x\otimes y)=y\otimes x$. Obviously $P_{ex}^2=1$ and $P_{ex}^*=P_{ex}$, for $$ \la P_{ex}(x\otimes y),u\otimes v\ra =\la y,u\ra\la x,v\ra =\la x\otimes y,P_{ex}(u\otimes v)\ra, $$ i.e. $P_{ex}$ is unitary and the only possible eigen-values are $\pm1$. For $A\in\Hom(E)$ and $x,y\in E$ we have: $$ (A\otimes A)P_{ex}(x\otimes y) =Ay\otimes Ax =P_{ex}(A\otimes A)(x,y) $$ 3. Finally for any $A\in\Hom(E)$ we have: \begin{eqnarray*} \tr_1((A\otimes1)P_{ex}) &=&\sum_{j,k}\tr(AE^{kj})\otimes E^{jk}\\ &=&\sum_{j,k,l,m}(a_{lm}\d_{mk}\d_{jl})E^{jk} =\sum_{j,k}a_{jk}E^{jk} =A~. \end{eqnarray*}