Repeated use of the induction branching rule gives
$$
S^{(2)}\!\uparrow_{S_2}^{S_4}
\cong S^{(4)}\oplus2S^{(3,1)}\oplus S^{(2,2)}\oplus S^{(2,1,1)},
$$
while
$$
S^{(1^3)}\!\uparrow_{S_3}^{S_4}
\cong S^{(2,1,1)}\oplus S^{(1^4)}.
$$
On the five conjugacy classes $1,(12),(12)(34),(123),(1234)$, their characters are respectively
$$
(12,2,0,0,0)
\quad\text{and}\quad
(4,-2,0,1,0).
$$
The tensor-product character is their pointwise product $(48,-4,0,0,0)$. Taking <character orthogonality>[inner products] with the five irreducible characters of $S_4$ gives multiplicities $1,5,4,7,3$. Therefore the <induced-tensor decomposition for the symmetric group on four points> is
$$
\boxed{V\cong S^{(4)}\oplus5S^{(3,1)}\oplus4S^{(2,2)}\oplus7S^{(2,1,1)}\oplus3S^{(1^4)}}.
$$

Solution

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

For the standard inclusions $S_2,S_3\leq S_4$,
$$
\left(S^{(2)}\!\uparrow_{S_2}^{S_4}\right)
\otimes
\left(S^{(1^3)}\!\uparrow_{S_3}^{S_4}\right)
\cong S^{(4)}\oplus5S^{(3,1)}\oplus4S^{(2,2)}\oplus7S^{(2,1,1)}\oplus3S^{(1^4)}.
$$


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