 Solution

ID: past-exam-of-the-algebra-course-of-the-university-of-oxford/2009/a/1/b/iii/solution

For equal irreducible group representations, the intertwining identity and the complex Schur's lemma give

T = c I_{n}

. Conjugation preserves trace, so

tr T = \sum_{g \in G} tr (D_{1} (g) B D_{1} (g)^{- 1}) = ∣ G ∣ tr B .

Since

tr (c I_{n}) = n c

, we obtain

T = \frac{∣ G ∣ tr B}{n} I_{n} .

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