For the <cycle-sum identity for Young–Jucys–Murphy elements>, let $C_m$ be the sum of all $m$-cycles on $\{1,\ldots,m\}$. Multiplying such a cycle by $(j\ m+1)$ inserts $m+1$ immediately after $j$ in that cycle, with our composition convention. Every $(m+1)$-cycle has a unique predecessor $j$ of $m+1$, so deleting $m+1$ recovers a unique term of $C_mX_{m+1}$. Since $X_2=(1\ 2)$, induction gives
$$
\boxed{X_2\cdots X_n=C_n.}
$$
For $n=1$ the empty product is the identity, also the unique one-cycle.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 103 / 3 / c / i / Solution

{title2=$X_2\cdots X_n=\sum_{\pi\text{ an }n\text{-cycle}}\pi$}

In the <group algebra> of $S_n$, the product of the <Young–Jucys–Murphy elements> $X_2,\ldots,X_n$ is the sum of all $n$-cycles, each with coefficient one. To prove it, multiply the sum of all $m$-cycles by $X_{m+1}$. Right multiplication by $(j\ m+1)$ inserts $m+1$ immediately after $j$ in the cycle. Every $(m+1)$-cycle has a unique predecessor of $m+1$, so deletion inverts this insertion bijectively. Induction starts at $X_2=(1\ 2)$. The identity turns a product of cell contents into a <central character value of a conjugacy-class sum>.

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