 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-103/1/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

X_{1} = 0, X_{r} = \sum_{a = 1}^{r - 1} (a r) (r \geq 2),

with products of permutations composed from right to left. Fix

i < j

. In

[X_{i}, X_{j}]

, all pairs of disjoint transpositions commute. The only terms left, for each

a < i

, are

[(a i), (i j)] + [(a i), (a j)] = 0.

Indeed

(a i) (i j) = (a j) (a i)

and

(i j) (a i) = (a i) (a j)

. Summing directly proves

[X_{i}, X_{j}] = 0 (1 \leq i, j \leq n) .

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