= Solution
Work over $\mathbb C$, with the standard inclusions $S_r\subseteq S_n$ fixing the letters $r+1,\ldots,n$. In the <group algebra>, the <Young–Jucys–Murphy elements> are
$$
\boxed{X_1=0,\qquad X_i=\sum_{a=1}^{i-1}(a\ i)\quad(2\leq i\leq n)}.
$$
Thus $X_i$ adds all <transpositions> involving the newest letter. Conjugating by $S_{i-1}$ merely permutes the summands, so $X_i$ commutes with $\mathbb C S_{i-1}$. In particular, if $j<i$, then $X_j\in\mathbb C S_{i-1}$ and $X_iX_j=X_jX_i$. Hence the whole family is commuting.
Put $\Omega_r=\sum_{1\leq a<b\leq r}(a\ b)$, the <transposition> <conjugacy-class sum> in $S_r$, with $\Omega_1=0$. Then
$$
X_i=\Omega_i-\Omega_{i-1}.
$$
This identity will connect the <Young–Jucys–Murphy elements> with the centres of the successive <group algebras>.
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