Use the <character formula for an induced representation>. For $g\in G$,
$$
\begin{aligned}
\chi(g)(\phi\!\uparrow_H^G)(g)
&=\frac1{|H|}\sum_{\substack{x\in G\\x^{-1}gx\in H}}
\chi(g)\phi(x^{-1}gx)\\
&=\frac1{|H|}\sum_{\substack{x\in G\\x^{-1}gx\in H}}
\chi(x^{-1}gx)\phi(x^{-1}gx)\\
&=\bigl((\chi\!\downarrow_H)\phi\bigr)\!\uparrow_H^G(g).
\end{aligned}
$$
The middle equality uses that a <character of a representation> is constant on conjugacy classes. This proves the <tensor identity for an induced character>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 2 / c / Solution

The point-permutation character is
$$
\xi^{(n-1,1)}=1\!\uparrow_{S_{n-1}}^{S_n}
=\chi^{(n)}+\chi^{(n-1,1)}.
$$
By the <tensor identity for an induced character> and <Frobenius reciprocity>,
$$
\langle\chi^\lambda\chi^\lambda,\xi^{(n-1,1)}\rangle
=\left\langle
\chi^\lambda\!\downarrow_{S_{n-1}},
\chi^\lambda\!\downarrow_{S_{n-1}}
\right\rangle.
$$
The <restriction branching rule for a symmetric group> is multiplicity-free with one constituent for each member of $\lambda^-$, so the right side is $|\lambda^-|$. Also $\langle\chi^\lambda\chi^\lambda,\chi^{(n)}\rangle=1$ because every symmetric-group character is real and irreducible. Subtracting the trivial constituent proves the <standard-character multiplicity in a Specht self-product> formula
$$
\boxed{\langle\chi^\lambda\chi^\lambda,\chi^{(n-1,1)}\rangle=|\lambda^-|-1}.
$$

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 2 / d / i / Solution

For a subgroup $H\leq G$, a character $\chi$ of $G$ and a character $\phi$ of $H$,
$$
\chi\,\operatorname{Ind}_H^G\phi
=\operatorname{Ind}_H^G\bigl((\operatorname{Res}_H^G\chi)\phi\bigr).
$$
It follows directly from the induced-character formula because $\chi(g)=\chi(x^{-1}gx)$.


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 Incoming links: Tensor identity for an induced character