Let $G$ be a transitive permutation group of degree $d$ having a cyclic regular subgroup $K = \langle k \rangle \cong C_d$. Let $\pi(g) = |\mathrm{Fix}(g)|$ be the permutation character of $G$ and let

$$\pi = \mathbb{1} + \pi_1 + \cdots + \pi_\ell $$

be the decomposition of $\pi$ into irreducible characters of $G$. Let $\theta$ be a character of $K$ of order $d$. Since $\pi\!\downarrow_K = \mathbb{1}\!\!\uparrow_{1}^K = 1 + \theta + \cdots + \theta^{d-1}$, on restriction to $K$, each $\pi_i$ splits as a sum of distinct powers of $\theta$. Let $\phi_i = \pi_i\!\downarrow_K$. Let $\langle \mathbb{1}_K, \phi_1, \ldots, \phi_\ell \rangle_\mathbb{C}$ be the vector subspace of the character ring of $K$ spanned by the restrictions of all the constituents of $\pi$.

Is is true that $\langle \mathbb{1}_K, \phi_1, \ldots, \phi_\ell \rangle_\mathbb{C}$ is a subalgebra of the character ring of $K$?

For $d \le 14$, I have checked that this is the case by computer algebra.

Example. Take $G = \langle (0,2,4), (1,3,5) \rangle \rtimes \langle (0,1)(2,3)(4,5) \rangle \cong C_3 \wr C_2$ and $k = (0,1,2,3,4,5)$. The point stabiliser of $0$ is $\langle (1,3,5) \rangle$ and so

$$\pi = 1_{\langle (1,3,5) \rangle}\!\!\uparrow^G = \mathbb{1}_G + \mathrm{sgn} + (1 \times \alpha)\!\!\uparrow_{\langle (0,2,4), (1,3,5) \rangle}^G + (1 \times \alpha^2)\!\!\uparrow_{\langle (0,2,4), (1,3,5) \rangle}^G$$

where $\alpha$ is a non-trivial character of $\langle (1,3,5)\rangle$ and $\mathrm{sgn}$ is the non-trivial character of $G$ having the base group $\langle (0,2,4), (1,3,5) \rangle$ in its kernel. Restricting to $K$ one finds that the $\phi_i$ are, in the order above, $\theta^3$, $\theta + \theta^4$ and $\theta^2 + \theta^5$. As predicted, $\langle \mathbb{1}_K, \theta^3, \theta+\theta^4, \theta^2 + \theta^5 \rangle_\mathbb{Q}$ is closed under multiplication. For instance, $(\theta+\theta^4)^2 = 2(\theta^2 + \theta^5)$.

Motivation. If true, this conjecture gives an algebra of characters analogous to the Schur ring of a permutation group.

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    $\begingroup$ Do you know counterexamples if you allow $K$ to be abelian but not necessarily cyclic? $\endgroup$ Dec 29, 2018 at 11:07

1 Answer 1


The conjecture is true and holds, in a generalized version, whenever $K$ is an abelian regular subgroup. In fact by Theorem 1.9(d) in O. Tamaschke, Zur Theorie der Permutationsgruppen mit regulärer Untergruppe. I, Math. Zeit. (1963) 80 328–354, there is an even more general result that holds when $K$ is not necessarily abelian. A special case of this theorem is that $\langle \mathbb{1}_K, \phi_1, \ldots, \phi_\ell \rangle_\mathbb{C}$ has the structure of a unitary central $S$-ring. In particular, it is closed under the natural product.

Below I'll give a proof in the cyclic case, using some ideas from Wolfgang Knapp On Burnside's Method, J. Alg. (1995) 175 644–660. In outline, what we show is that $\langle \mathbb{1}_K, \phi_1, \ldots, \phi_\ell \rangle$ under multiplication is isomorphic to the Schur ring for $G$ with respect to the product that sends two $H$-invariant subsets to their intersection.

Setup. Let $\zeta \in \mathbb{C}$ be a primitive $d$-th root of unity. For $j \in \{0,\ldots, d-1\}$ let

$$v_j = \sum_{i=0}^{d-1} \zeta^{-ij} k^i \in \mathbb{C}K.$$

The $v_j/d$ are the orthogonal primitive idempotents in the commutative group algebra $\mathbb{C}K$. Since $v_j k = \zeta^j v_j$, the subspace $\langle v_j \rangle$ affords the character $\theta^j$. Let $\odot$ denote the product on $\mathbb{C}K$ defined by multiplying coefficients of each $k^j$: thus if $d=3$ then

$$(a_0+a_1k+a_2k^2) \odot(b_0+b_1k+b_2k^2) = a_0b_0 + a_1b_1 k + a_2b_2k^2.$$

Let $\mathcal{C}(K)$ denote the character ring of $K$. Consider the linear map $\mathcal{C}(K) \rightarrow \mathbb{C}K$

$$\chi \mapsto \sum_{i=0}^{d-1} \overline{\chi(k^j)}k^j.$$

Since the image of $\theta^j$ is $v_j$, this map is a linear isomorphism. Observe that

$$v_j \odot v_{j'} = v_{j + j' \,\mathrm{mod}\; d}.$$

Therefore $\mathcal{C}(K)$, with its usual product, corresponding to the tensor product of modules, is isomorphic to $(\mathbb{C}K, \odot)$.

Proof of conjecture. Let $G$ act on $\{0,1,\ldots,d-1\}$. Let $H \le G$ be the point stabiliser of $0$. Let $\mathcal{O}_0, \mathcal{O}_1, \ldots, \mathcal{O}_\ell$ be the orbits of $H$ on $\{0,1,\ldots,d-1\}$, ordered so that $\mathcal{O}_0 = \{0\}$. (Note that the number of orbits is the rank of the $G$-action, which is the number of irreducible constituents of $\pi$, namely $\ell+1$.) Identify the $G$-set $\{0,1,\ldots,d-1\}$ with the regular subgroup $K$ by $j \mapsto k^j$. The orbit sums are then

$$e_m = \sum_{j \in \mathcal{O}_m} k^j \in \mathbb{C}K$$

for $m \in \{0,1,\ldots,\ell\}$. By definition $S = \langle e_0, e_1, \ldots, e_\ell \rangle_\mathbb{C}$ is the Schur ring for $G$. The $e_m$ are Schur's `primitive elements' and are primitive idempotents for the $\odot$ product. (The Schur ring is also a unital ring under the normal product on $\mathbb{C}K$, but, perhaps surprisingly, we do not need this fact here.)

By our identification, $\mathbb{C}K$ affords the natural permutation module for $G$. Let

$$\mathbb{C}K = V_0 \oplus V_1 \oplus \cdots \oplus V_\ell$$

be its direct sum decomposition into irreducible representations, where

$$V_0 = \langle 1_K + k + \cdots + k^{d-1} \rangle$$

affords the trivial representation, and $V_i$ affords the character $\pi_i$ in the question. Note that $V_0 = \langle v_0 \rangle$. More generally, let $B_i$ be the set of $j$ such that $\langle \pi_i\!\!\downarrow_K, \theta^j \rangle = 1$. Then $V_i = \langle v_j : j \in B_i\rangle$. Moreover, by Frobenius reciprocity

$$\langle \pi_i \!\!\downarrow_H, \mathbb{1}_H \rangle = \langle \pi_i, \mathbb{1}_H\!\!\uparrow^G \rangle = \langle \pi_i, \pi \rangle = 1$$

for each $i$, so each $V_i$ has a unique (up to scalars) $H$-invariant vector. Since $e_0 = \frac{1}{d}(v_0 + v_1 + \cdots + v_{d-1})$ is $H$-invariant, taking the projection into $V_i$ we see that a suitable choice for this $H$-invariant vector is $\sum_{j \in B_i} v_j$. Therefore an alternative basis for the Schur ring $S$ is

$$\big\{ \sum_{j \in B_i} v_j : j \in \{0,1,\ldots, d-1\bigr\}.$$

Under the isomorphism in the 'setup', we have

$$\phi_i = \sum_{j \in B_i} \theta^j \mapsto \sum_{j \in B_i} v_j.$$

Therefore, by the inverse of this isomorphism, $(S, \odot)$ is isomorphic to $\langle \mathbb{1}_K, \theta_1, \ldots, \theta_\ell \rangle$. In particular, the linear span of characters is closed under the product in $\mathcal{C}(K)$. $\Box$


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