# Decompose the irreducible representation of $S_4$ in terms of the irreducible representations of $S_3$

how would one go about doing this? I know the character tables of both groups and that the multiplicity of the irreducible representation $$V$$ in $$W$$ is

$$\frac{1}{|G|}\sum_{a\in G}\chi_V(a)^*\chi_W(a)$$ My initial thought was to divide the elements of $$S_3$$ into the various conjugacy classes of $$S_4$$, for example $$(12)(3)$$ is in the conjugacy class of $$(12)(3)(4)$$. The following classes of $$S_4$$ contain elements of $$S_3$$

• $$[(1)(2)(3)(4)]$$

• $$[(123)(4)]$$

• $$[(12)(3)(4)]$$

How should I proceed from here? I know in any case that all the elements in the conjugacy classes of $$S_4$$ that I listed above belong to some $$S_3\subset S_4$$, so I tried to use the formula by counting for $$V$$ an irreducible representation of $$S_4$$

$$\chi_V^{S_4}(a)= \chi_V^{S_3}(a)$$ if $$a\in [(1)(2)(3)(4)],[(12)(3)(4)]\textrm{ or }[(123)(4)]$$ and where by $$\chi_V^{S_3}(a)$$ I mean the character of the corresponding conjugacy class in $$S_3$$ (e.g. $$[(12)(3)]$$ for $$[(12)(3)(4)]$$), but this gives wrong results, for example asking how many times the trivial representation of $$S_3$$ appears in the trivial representation of $$S_4$$ yields

$$\frac{1}{24}(1+8+6)=\frac{15}{24}$$ because there is one element in the class $$[(1)(2)(3)(4)]$$, $$6$$ elements in the class $$[(12)(3)(4)]$$ and $$8$$ in the class $$[(123)(4)]$$, but this clearly doesn't make sense. I think I'm misunderstanding the question, what does it even mean to decompose a representation of a group into representations of another group?

• I don't understand the calculation leading to $15/24$ at all. The character of the trivial representation of $S_4$ is constant, $\chi(x)=1$ for all $x\in S_4$. Therefore $\chi(y)=1$ for all $y\in S_3$ also, implying that the restriction of $\chi$ to $S_3$ has inner product $$\langle\chi\vert_{S_3},\psi\rangle=\frac16(1+1+1+1+1+1)=1$$ with the trivial character $\psi$ of $S_3$. – Jyrki Lahtonen Jun 27 at 13:09
• My reasoning was that since I should find representations of $S_4$, I should use the formula for $S_4$, i.e. $|G|=24$, then each term is character of class of $S_4$ times character of corresponding class of $S_3$ times number of elements in that class in $S_4$, and I "assigned" the character $0$ to every conjugacy class of $S_4$ that does not contain elements of $S_3$. I understand this is wrong, but I don't understand why you should consider the restriction of the characters of $S_4$ to $S_3$ instead of an extension of $S_3$ onto $S_4$, since we are trying to find representations of $S_4$ – user438666 Jun 27 at 13:18
• overall I'm really confused about how the formula I've given translates when we're considering a group and a subgroup – user438666 Jun 27 at 13:19
• The title of your question says to me that you are given an irreducible representation of $S_4$ and your task is to calculate how it decomposes when restricted to the subgroup $S_3$. The way to do that is to calculate the inner products of the given character and all the irreducible characters of $S_3$. In other words, $|G|=|S_3|=6$. – Jyrki Lahtonen Jun 27 at 13:25
• If you wan to get a representation of $S_4$ from a given representation of $S_3$, the tool for that is induction, but it is a more complicated process. If $f:S_4\to GL(V)$ is a homomorphism (i.e. a representation of $S_4$), its restriction to $S_3$ is simply the restriction of $f$ to $S_3$. Therefore the character is a restriction also (the matrix representing an element of $S_3$ is the same as the one representing the same permutation as an element of $S_4$). – Jyrki Lahtonen Jun 27 at 13:29