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?

  • $\begingroup$ 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 $S_3$. $\endgroup$ – Jyrki Lahtonen Jun 27 at 13:09
  • $\begingroup$ 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 $S_4$ $\endgroup$ – user438666 Jun 27 at 13:18
  • $\begingroup$ overall I'm really confused about how the formula I've given translates when we're considering a group and a subgroup $\endgroup$ – user438666 Jun 27 at 13:19
  • $\begingroup$ 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, $|G|=|S_3|=6$. $\endgroup$ – Jyrki Lahtonen Jun 27 at 13:25
  • $\begingroup$ 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 $S_4$). $\endgroup$ – Jyrki Lahtonen Jun 27 at 13:29

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