Decomposing the restrictions to $S_3$ of representations of $S_4$ Let $V$ be the standard reputation of $S_4$ and $W$ the unique 2-dimensional representation. I am told that restricting to $S_3$ we should have
$\text{Res}_{S_3}V = V'\oplus 1,\quad\text{Res}_{S_3}(V\otimes\text{sgn})=V'\oplus\text{sgn},\quad\text{Res}_{S_3}W=V'$
where $V'$ is the standard representation on $S_3$. But how do I compute this?
 A: As suggested in the comments, looking at characters is probably the simplest way to go about it. However here is an alternative more combinatorial approach:
If we think of $V$ is the subspace of $\mathbb{C}^4$ where the sum of the coordinates is zero, the decomposition $Res_{S_3}V = V' \oplus 1$ is just given by the subspace where the last coordinate is zero plus the subspace spanned by $(1,1,1,-3)$.
The tensor products commute with restriction. In particular this is equal to $Res(V) \otimes Res(\text{sgn})$ which is just $(V' \oplus 1) \otimes \text{sgn}$ by the first result. One then needs to check that $V' \otimes \text{sgn}$ is $V'$, but $V'$ is the unique irreducible of dimension $2$ so this must be true.
$S_4$ acts on the set of set-partitions of $\{1,2,3,4\}$ into two sets of size 2. There are three such partitions $\{\{1,2\},\{3,4\}\}, \{\{1,3\},\{2,4\}\}$, and $\{\{1,4\},\{2,3\}\}$. Restricting to $S_3$ this action is equivalent to the standard action on $\{1,2,3\}$, which can be seen by just looking at what gets paired with $4$. $W$ is just the space of formal linear combinations of these partitions where the sum of the coefficients is zero, hence $Res(W) = V'$.
