Let $G$ be a finite group with $|G|=n$, let $S_G=S_n$ be the group of $n!$ permutations of the set $G$. Then $G$ is a subgroup of $S_G$ via left-translation (i.e. $g\in G$ corresponds to the permutation $h \mapsto gh$).

My question: What can be said about the decomposition into irreducibles of the $S_G$-representation $\mathrm{Ind}_G^{S_G} 1$, where $1$ is the trivial $G$-representation? Is there some combinatorial way to identify the partitions $\lambda$ of $n$ corresponding to the irreducible $S_G$ representations $V_\lambda$ that appear in this induced representation?

Using Frobenius reciprocity, the representation $V_\lambda$ occurs iff restricted to $G$ it contains a copy of the trivial representation. The question can also be expressed by asking for which $\lambda$ we have $$\frac{1}{n} \sum_{g \in G} \chi_\lambda(h \mapsto gh)\neq 0,$$ where $\chi_\lambda$ is the character of $V_\lambda$. As the cycle-type of $h \mapsto gh$ is uniquely determined by $\mathrm{ord}(g)$, this shows that the answer of the question above only depends on the multiset $\{\mathrm{ord}(g): g \in G\}$.

  • I think it is very hard in general, otherwise one would immediately be able to solve the Foulkes conjecture where $G$ is the wreath product of two symmetric groups. – Abdelmalek Abdesselam Apr 10 '17 at 13:31
  • If I understand the question right, it's asking about permutation actions of symmetric groups on the cosets of a regular subgroup. So it's not general enough to include Foulkes' Conjecture. – Mark Wildon Apr 10 '17 at 15:06
  • The easiest non-trivial case is when $G = \langle g \rangle$ is cyclic of prime order $p$. By the formula in the question, the multiplicity of $\chi^\lambda$ is $\frac{1}{p}\chi^\lambda(1) + \frac{p-1}{p} \chi^\lambda(g)$. Now $\chi^\lambda(g) = 0$ unless $\lambda$ is a hook, of the form $(p-r,1^r)$ with $0 \le r < p$, in which case $\chi^\lambda(g) = (-1)^r$. So, except for a tiny correction term in the hook case, the multiplicity of $\chi^\lambda$ is $1/p$ times the number of standard Young tableaux of shape $\lambda$. – Mark Wildon Apr 10 '17 at 16:00
  • I just found that for $G$ cyclic of order $n$, by Reutenauer, Free Lie Algebras, Thm. 8.9 and 8.8, the multiplicity of $\chi^\lambda$ is the number of standard Young tableaux of shape $\lambda$ and major index divisible by $n$. Assuming for $n=p$ prime that the major index is random mod $p$ this fits very well with the heuristic by @MarkWildon . – JoS Apr 10 '17 at 16:37
  • @JoS: this result is for the $n$th Lie power and concerns the character $\mathrm{Ind}_{\langle g \rangle}^{S_n} \theta$ where $\theta$ is a faithful character of $\langle g \rangle$. – Mark Wildon Apr 10 '17 at 17:21

Let me collect two partial results that I learned since asking the question. I would still be very happy to have a more general answer of course.

1) For $G=\mathbb{Z}/n\mathbb{Z}$ the question can be explicitly answered (as described in the comments above). Indeed, combining Theorem 8.9 (case $i=0$) and Theorem 8.8 of the book Free Lie Algebras by Reutenauer, the multiplicity with which the representation $V_\lambda$ appears in $\mathrm{Ind}_{G}^{S_n} 1$ is the number of standard tableaux of shape $\lambda$ and major index congruent to $0$ mod $n$. For convenience, let me recall that in a standard tableau $T$, a descent is an index $i \in \{1, \ldots, n-1\}$ such that $i+1$ is in a lower row than $i$ in $T$ and the major index is the sum of all descents for $T$.

2) For general finite groups $G$, the only useful result I know is that the restriction of $\mathrm{Ind}_{G}^{S_n} 1$ to $S_{n-1}$ is isomorphic to the regular representation of $S_{n-1}$ by Mackey's formula. Indeed, note that since $G$ acts simply transitively on itself by left-translation, we have $S_n=G \cdot S_{n-1}$. Since any nontrivial element of $G$ has no fixed point as a permutation of $G$, we have $G \cap S_{n-1}=\{()\}$. Then by Mackey's formula $$\left(\mathrm{Ind}_{G}^{S_n} 1 \right)|_{S_{n-1}} = \mathrm{Ind}_{\{()\}}^{S_{n-1}}1 = \mathrm{Reg}_{S_{n-1}}.$$ Translated to partitions, this means the following: take all partitions $\lambda$ with the multiplicity in which $V_\lambda$ appears in $\mathrm{Ind}_{G}^{S_n} 1$. From this form the multiset of all partitions $\mu$ obtainable by removing a block of the partitions $\lambda$. Then the collection of $\mu$ obtained this way is the set of all partitions of $n-1$ with the multiplicity given by the dimension of $V_{\mu}$. In a certain sense this says that a lot of partitions $\lambda$ appear in $\mathrm{Ind}_{G}^{S_n} 1$. Indeed, any partition $\mu$ of $n-1$ must be obtainable from such a $\lambda$ by removing one block.

  • Suppose $n = 2^m$ and $G \cong C_2^m$. For instance take $G = \langle (12)(34), (13)(24)\rangle$ if $m=2$. All non-identity elements in $G$ have cycle type $(2^m)$. If $g$ has this cycle type then $\chi^\lambda(g)$ is, up to a sign, the number of domino tableaux of shape $\lambda$. This number is typically far smaller than $\chi^\lambda(1)$. (It can be computed using the $2$-quotient of $\lambda$.) So again the multiplicity of $\chi^\lambda$ is $\chi^\lambda(1) / n$, up to a small correction factor. This generalizes to $C_a^m$ in the obvious way. – Mark Wildon Apr 13 '17 at 20:28

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