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 $\{\mathrm{ord}(g): g \in G\}$.

  • $\begingroup$ 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. $\endgroup$ – Abdelmalek Abdesselam Apr 10 '17 at 13:31
  • $\begingroup$ 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. $\endgroup$ – Mark Wildon Apr 10 '17 at 15:06
  • $\begingroup$ 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 $\lambda$. $\endgroup$ – Mark Wildon Apr 10 '17 at 16:00
  • $\begingroup$ 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 . $\endgroup$ – JoS Apr 10 '17 at 16:37
  • $\begingroup$ @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 $\langle g \rangle$. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Mark Wildon Apr 13 '17 at 20:28
  • $\begingroup$ In case (1), a couple of years ago I was motivated by an equivalent question of Sheila Sundaram to explicitly determine which irreps appear (for any character). See this paper, Thm. 3. My idea was basically the same as Mark's, but getting provable bounds to squash all the cases is a little delicate. The theorem Reutenauer mentions is due to Kraskiewicz--Weyman. At the end of Schocker's Embeddings of higher Lie modules, Thm. 3.4, an answer is given for the Lie module of type (d, d) but the proof is omitted. Those cases are surely interesting and hard. $\endgroup$ – Joshua P. Swanson Sep 25 '19 at 5:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.