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There are $(k-1)!$ $k$-cycles in $S_k$, and each has the same character value of a representation on $S^{\lambda }$. Taking traces in the cycle-sum identity for Young–Jucys–Murphy elements yieldsOnly tableaux whose first $k$ entries form a hook partition contribute.

Write such a shape as ${\mu }_b=(k-b,1^b)$, for $0\le b\le k-1$. There are $f_{{\mu }_b}=\left(\genfrac{}{}{0ex}{}{k-1}{b}\right)$ ways to fill it: choose which $b$ of the labels $2,\dots ,k$ go below the first box. There are $f^{\lambda /{\mu }_b}$ ways to fill the remaining skew diagram with the labels $k+1,\dots ,n$, in standard order. The eigenvalue from part (ii) is $(-1{)}^b(k-b-1)!b!$, so the factoris $1$. Therefore the general character value isHere $f^{\lambda /\mu }$ counts standard skew Young tableaux, and is zero when $\mu$ is not contained in $\lambda$. In particular, for a full $n$-cycle,For $k=1$, the general formula gives $f^{\lambda /(1)}=f_{\lambda }$, the character value at the identity, as it should.

For $k=1$, the product is empty and equals the identity, the only $1$-cycle in $S_1$. Suppose the cycle-sum identity for Young–Jucys–Murphy elements holds at $k-1$:Multiplication by $X_k={\sum }_{a<k}(ak)$ gives terms $\sigma (ak)$. With permutation products acting right to left, this inserts $k$ immediately after $a$ in the cycle of $\sigma$: the arrows become $a\mapsto k\mapsto \sigma (a)$, with the other arrows unchanged.

Conversely, in any $k$-cycle there is a unique predecessor $a$ of $k$. Delete $k$ and join that predecessor to the successor of $k$; this uniquely recovers $\sigma$ and the factor $(ak)$. Hence every $k$-cycle occurs exactly once, and no other permutation occurs. Induction provesThus ${\xi }_k$ is a conjugacy-class sum in $\mathbb{C}{S}_k$, although it need not be central in $\mathbb{C}{S}_n$ when $k<n$.

For the cycle-sum identity for Young–Jucys–Murphy elements, let $C_m$ be the sum of all $m$-cycles on $\{1,\dots ,m\}$. Multiplying such a cycle by $(jm+1)$ 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 $C_m{X}_{m+1}$. Since $X_2=(12)$, induction givesFor $n=1$ the empty product is the identity, also the unique one-cycle.