The <deletion projection for a symmetric group> removes $n$ from the cycle containing it. A cycle $(a_1\ \cdots\ a_r\ n)$ becomes $(a_1\ \cdots\ a_r)$, with a singleton understood as a fixed point; all other cycles remain unchanged. This describes a genuine <permutation> of $\{1,\ldots,n-1\}$ and immediately proves the two identity assertions.

Put $H=S_{n-1}$. For $\sigma,\theta\in H$ and $k<n$, if $\pi(\theta(k))\ne n$, deletion gives $\sigma\pi(\theta(k))$. If $\pi(\theta(k))=n$, it gives $(\sigma\pi\theta)(n)=\sigma\pi(n)$. These are exactly the two cases in $\sigma\pi_n\theta(k)$. Thus
$$
\boxed{(\sigma\pi\theta)_n=\sigma\pi_n\theta.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 103 / 1 / c / v / Solution

{title2=$\Pi_n$}

Delete $n$ from the cycle notation of a <permutation> in $S_n$ to obtain a permutation in $S_{n-1}$. Extend this set map linearly to $\Pi_n:\mathbb C S_n\to\mathbb C S_{n-1}$. It is an $S_{n-1}$-bimodule map and the identity on $\mathbb C S_{n-1}$, but is not in general an algebra homomorphism. In particular,
$$
\Pi_n(X_n)=(n-1)1,\qquad \Pi_n(X_n^2)=(n-1)1+2\sum_{i<j<n}(i\ j).
$$
For $n\geq5$ it is the unique bimodule-equivariant set map $S_n\to S_{n-1}$; at $n=4$ another such map exists because $C_{S_3}(S_2)=S_2$.

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 Incoming links: Deletion projection for a symmetric group