= Solution
The <Young–Jucys–Murphy elements> are
$$
X_1=0,\qquad X_r=\sum_{a=1}^{r-1}(a\ r)\quad(r\geq2),
$$
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
$$
\boxed{[X_i,X_j]=0\qquad(1\leq i,j\leq n).}
$$
Thus all the <Young–Jucys–Murphy elements> commute.
Back to article page
