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 $\pi$. 



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> Is is true that $\langle \mathbb{1}\_K, \phi\_1, \ldots, \phi\_\ell \rangle\_\mathbb{C}$ is a subalgebra of the character ring of $K$?
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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. 



Permutation groups having a regular cyclic subgroup and a conjectured algebra of characters
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*](https://link.springer.com/article/10.1007/BF01162390), 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*](https://ac.els-cdn.com/S0021869385712069/1-s2.0-S0021869385712069-main.pdf?_tid=4a44dd19-a39a-4917-9865-05344aaca29c&acdnat=1546089437_64d5d1713a1e8e24e618e0b753116619), 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 $\mathcal{C}(K)$. $\Box$


