On right conjugacy closed loops of twice prime order

TL;DR

This paper classifies right conjugacy closed loops of order 2p, with p an odd prime, providing a comprehensive understanding of their structure up to isomorphism.

Contribution

It offers the first complete classification of right conjugacy closed loops of order 2p for odd prime p, expanding the algebraic theory of loops.

Findings

Complete classification of right conjugacy closed loops of order 2p

Identification of isomorphism classes for these loops

Structural properties of the classified loops

Abstract

The right conjugacy closed loops of order 2p, where p is an odd prime, are classified up to isomorphism.

Tables

Table 1. Table 1. Primitive groups of degree  n 𝑛 n which might have a non-trivial conjugacy class of length less than  n 𝑛 n

	$\mathrm{soc}\left(G\right)$	$n$	Restrictions
(i)	$A_c$	$\left(\frac{c}{k}\right)$	$3\le k\le c-3$
(ii)	$A_{2a}$	$\frac{1}{2}\left(\frac{2a}{a}\right)$	$a\in \{6,9,10,12,36\}$
(iii)	$\mathrm{PSL}(m,r)$	$\frac{{\prod }_{i=0}^{k-1}\left(r^{m-i}-1\right)}{{\prod }_{i=1}^k\left(r^i-1\right)}$	$2\le k\le m-2$,
			$(m,r)\notin \{(4,2),(5,2)\}$
(iv)	$\mathrm{PSL}(m,r)$	$\frac{{\prod }_{i=0}^{2k-1}\left(r^{m-i}-1\right)}{{\left({\prod }_{i=1}^k\left(r^i-1\right)\right)}^2}$	, $m\ge 3$,
			$(m,r)\ne (3,2)$
(v)	$\mathrm{PSL}(2,r)$	$\sqrt{r}\left(r+1\right)/2$	$\sqrt{r}$ an odd prime,
			, $r>9$
(vi)	$\mathrm{PSL}(2,r)$	$r\left(r^2-1\right)/24$	$r$ a prime, $r\equiv \pm 3\left(\mathrm{mod}8\right)$,
			$r\notin \{5,11\}$
(vii)	$\mathrm{PSL}(2,r)$	$r\left(r^2-1\right)/48$	$r$ a prime, $r\equiv \pm 1\left(\mathrm{mod}8\right)$,
			$r\notin \{7,17,23\}$
(viii)	$\mathrm{PSL}(2,r)$	$r\left(r^2-1\right)/120$	$r$ a prime, $r\equiv \pm 1\left(\mathrm{mod}10\right)$,
			$r\notin \{11,19,29,31,41,59\}$
(ix)	$\mathrm{PSU}(4,r)$	$\left(r^2+1\right)\left(r^3+1\right)$
(x)	$\mathrm{PSp}(2m,2)$	$4^m-1$	$m\ge 3$
(xi)	$\mathrm{PSp}(2m,r)$	$\frac{\left(r^{2m}-1\right)\left(r^{2m-2}-1\right)}{\left(r^2-1\right)\left(r-1\right)}$	$m\ge 3$
(xii)	$\Omega (2m+1,r)$	$\frac{\left(r^{2m}-1\right)\left(r^{2m-2}-1\right)}{\left(r^2-1\right)\left(r-1\right)}$	$m\ge 3$
(xiii)	$P{\Omega }^{-}(2m,r)$	$\frac{\left(r^m+1\right)\left(r^{2m-2}-1\right)\left(r^{m-2}-1\right)}{\left(r^2-1\right)\left(r-1\right)}$	$m\ge 3$, $r$ even
(xiv)	$P{\Omega }^{-}(2m,r)$	$\frac{\left(r^m+1\right)\left(r^{2m-2}-1\right)\left(r^{2m-4}-1\right)\left(r^{m-3}-1\right)}{\left(r^3-1\right)\left(r^2-1\right)\left(r-1\right)}$	$m\equiv 0\left(\mathrm{mod}4\right)$, $r$ even
(xv)	$P{\Omega }^{+}(2m,2)$	$\left(2^m-1\right)\left(2^{m-1}+1\right)$	$m\ge 5$ odd
(xvi)	$P{\Omega }^{+}(2m,r)$	$\frac{\left(r^m-1\right)\left(r^{2m-2}-1\right)\left(r^{m-2}+1\right)}{\left(r^2-1\right)\left(r-1\right)}$	$m\ge 3$, $r$ even
(xvii)	$P{\Omega }^{+}(2m,r)$	$\frac{\left(r^m-1\right)\left(r^{2m-2}-1\right)\left(r^{2m-4}-1\right)\left(r^{m-3}+1\right)}{\left(r^3-1\right)\left(r^2-1\right)\left(r-1\right)}$	$m\equiv 3\left(\mathrm{mod}4\right)$, $r$ even
(xviii)	$E_7\left(r\right)$	$\frac{\left(r^{18}-1\right)\left(r^{14}-1\right)\left(r^4-r^2+1\right)}{\left(r^2-1\right)\left(r-1\right)}$

Table 2. Table 2. Smallest size of non-trivial conjugacy classes

Equations

$$|G:C_G(g)|$$

$$n=\prod _{i=0}^{k-1}(r^{m-i}-1)/\prod _{i=1}^k(r^i-1)$$

$$n=\frac{(r^m-(-1{)}^m)(r^{m-1}-(-1{)}^{m-1})}{r^2-1}.$$

$$n=\frac{(r^m-1)(r^{m-1}+1)}{r-1}.$$

$$n=c\cdot \frac{(c-1)(c-2)}{6}.$$

$$n=\prod _{i=0}^{k-1}(r^{m-i}-1)/\prod _{i=1}^k(r^i-1),$$

$$n=\frac{(r^m-1)(r^{m-1}-1)}{(r^2-1)(r-1)},$$

$$n=\prod _{i=0}^{2k-1}(r^{m-i}-1)/(\prod _{i=1}^k(r^i-1){)}^2,$$

$$n=\frac{(r^m-1)(r^{m-1}-1)}{(r-1)(r-1)}$$

$$|G:C_G(g)|=a\cdot |S:C_S(g)|$$

$$Ht=gH=gHh=Hth=H{h}^{-1}th.$$

$$|G:C_G(t)|\le |K:H|\text{for all }t\in T$$

$$N:=\bigcap _{g\in G}{H}^g$$

$$|G|\ge |C_G(s)K_1|\ge \frac{|C_G(s)||K_1|}{|Z|}\ge \frac{|G|}{p}\cdot \frac{p^2}{|Z|}\ge |G|,$$

$$I_{n,d}:=|\{\tau \in S_n\mid {\tau }^2=1,\tau {\zeta }^d={\zeta }^d\tau \}|$$

$$I_n:=I_{n,n}.$$

$$I_{n,d}=\sum _{k=0}^{\lfloor d/2\rfloor }\frac{d!(n/d{)}^k(2-(n/d\text{ mod }2){)}^{d-2k}}{2^{k}k!(d-2k)!}.$$

$$I_n=\sum _{k=0}^{\lfloor n/2\rfloor }\frac{n!}{2^{k}k!(n-2k)!}.$$

$$(\mu ;c_1,\dots ,c_d{)}^2=({\mu }^2;c_1{c}_{1{\mu }^{-1}},c_2{c}_{2{\mu }^{-1}},\dots ,c_d{c}_{d{\mu }^{-1}}).$$

$$I_{p-1}-1+\frac{1}{p-1}\sum _{d=1}^{p-1}{I}_{p-1,d}$$

$$T_1=\{{\pi }_2^{j\tau }{\pi }_1^j\mid 1\le j\le p-1\}\cup \{1\}.$$

$$\mathcal{T}=\{T_1\cup G^{\prime }t\mid T_1\in {\mathcal{T}}_1,t\in G\setminus K\},$$

$$|\mathcal{T}|=p{I}_{p-1}.$$

$${\mathcal{T}}^{\prime }:=\{T_1\cup G^{\prime }\alpha \mid T_1\in {\mathcal{T}}_1\}\subseteq \mathcal{T}.$$

$$\frac{1}{p-1}\sum _{d=1}^{p-1}{\chi }_d,$$

$$p-2+I_{p-1}+\frac{1}{p-1}\sum _{d=1}^{p-1}{I}_{p-1,d}.$$

$$\begin{array}[]{c|cccccccc}p&2&3&5&7&11&13&17&19\\ \hline\cr{\rm(\ref{FormulaForNumberOfAllLoops})}&2&5&18&99&10\,489&151\,973&49\,096\,721&1\,052\,729\,657\end{array}$$

$$H:=\left\{\left(\begin{array}[]{cc}\alpha&0\\ \beta&1\end{array}\right)\mid\alpha\in\mathbb{F}_{q}^{*},\beta\in\mathbb{F}_{q}\right\}.$$

$$T:=C\cup Z.$$

$$t^{-1}h{h}^{\prime }=h{t}^{-1}.$$

Full text

1. Introduction

A quasigroup $\mathcal{L}$ is a set with a binary operation $*:\mathcal{L}\times\mathcal{L}\rightarrow\mathcal{L}$, such that every equation $x*a=b$ or $a*x=b$ with $a,b\in\mathcal{L}$ has a unique solution $x$. In this case, for every $a\in\mathcal{L}$, the right multiplication $R_{a}:\mathcal{L}\rightarrow\mathcal{L},x\mapsto x*a$ is a permutation of $\mathcal{L}$ (and of course so is every left multiplication). A quasigroup is a loop, if it contains an identity element. Thus a group is just a loop, in which the operation is associative, and we will indeed view groups as loops.

In the following, we will only consider finite loops. Let $\mathcal{L}$ be a (finite) loop, whose identity element we denote by $e$. The right multiplication group of $\mathcal{L}$ is the group $G:=\langle R_{a}\mid a\in\mathcal{L}\rangle$, a subgroup of the symmetric group on $\mathcal{L}$. Clearly, $G$ acts faithfully and transitively on $\mathcal{L}$ and $R_{e}$ is the identity element of $G$, which we denote by $1$. Let $H\leq G$ denote the stabiliser in $G$ of $e\in\mathcal{L}$, and let $T:=\{R_{a}\mid a\in\mathcal{L}\}$. Then $T$ is a transversal for $H^{g}\backslash G:=\{H^{g}x\mid x\in G\}$ for every $g\in G$, the identity element of $G$ is contained in $T$, and $\langle T\rangle=G$. The triple $(G,H,T)$ is called the envelope of $\mathcal{L}$, a group theoretic object..

Conversely, starting from group theory, one defines a loop folder to be a triple $(G,H,T)$ of a finite group $G$, a subgroup $H\leq G$ and a subset $T\subseteq G$ with $1\in T$, such that $T$ is a transversal for $H^{g}\backslash G$ for every $g\in G$. Given a loop folder $(G,H,T)$ one can construct a loop $(\mathcal{L},*)$ on the set $H\backslash G$ of right cosets of $H$ in $G$. However, the envelope of $\mathcal{L}$ need not be equal to $(G,H,T)$. In contrast to the right multiplication group of $\mathcal{L}$, in general the group $G$ will not act faithfully on $\mathcal{L}$, and the transversal $T$ will not generate $G$. On the other hand, it is not difficult to construct the envelope of $\mathcal{L}$ from $(G,H,T)$.

These results, as well as the notion of loop folder and envelope of a loop are contained in [2, Section $1$]. However, the connection between loops and their envelopes goes back to Baer [3].

Let $\mathcal{L}$ be a loop with envelop $(G,H,T)$. We say that $\mathcal{L}$ is right conjugacy closed, or an RCC  loop, if $T=\{R_{a}\mid a\in\mathcal{L}\}$ is closed under conjugation by itself. Clearly, this is the case if and only if $T$ is invariant under conjugation in $G=\langle T\rangle$; in other words, if $T$ is a union of conjugacy classes of $G$. We shortly say that $T$ is $G$-invariant in the following. Thus an RCC  loop gives rise to a $G$-invariant transversal of $H$, the stabilizer of $e$ in $G$. (A $G$-invariant transversal of a subgroup $H$ of a group $G$ is sometimes called a distinguished transversal in the literature.) On the group theoretic side, this leads to the notion of an RCC  loop folder. This is a loop folder $(G,H,T)$, where $T$ is $G$-invariant. More definitions regarding loop folders are given at the beginning of Section 3.

It has been shown by Drápal [9] that an RCC  loop of prime order is a group. In this paper we determine all RCC  loops of order $2p$, where $p$ is an odd prime. In order to achieve this, we first describe the possible envelopes $(G,H,T)$ of such loops. Our approach is group theoretic. In Sections 2 and 3 we show that if $(G,H,T)$ is an RCC  loop folder such that $G$ acts faithfully on $H\backslash G$ and the index $|G\colon H|$ is the product of two distinct primes, then $G$ acts imprimitively on $H\backslash G$ (Theorem 3.1). This result uses the classification of the finite simple groups and is based on the classification of finite primitive permutation groups of squarefree degree by Li and Seress, and on the determination of the minimal degrees of permutation representations of finite groups of Lie type by Patton, Cooperstein and Vasilyev. For the purpose of our further investigation, it would suffice to enumerate the primitive permutation groups of degree $2p$ for odd primes $p$; we are not aware of any result in this direction which does not rely on the classification of the finite simple groups.

In Section 4, we continue with some basic results on permutation groups of degree $p$ and give a new proof of Drápal’s theorem on RCC  loops of prime order (Corollary 4.2).

Let $(G,H,T)$ be the envelope of an RCC  loop of order $2p$, where $p$ is an odd prime. Using Theorem 3.1 mentioned above, we may now assume that there is a subgroup $K\lneq G$ with $H\lneq K$, and also that one of the indices $|G\colon K|$ or $|K\colon H|$ is equal to $2$, and the other index is equal to $p$. This configuration is analysed in Section 5 with elementary group theoretical methods. It turns out that there are three possible types for $G$. Firstly, $G$ can be isomorphic to the wreath product $C_{p}\wr C_{2}$, where $C_{p}$ and $C_{2}$ denote (cyclic) groups of order $p$ and $2$, respectively. Secondly, $G$ can be isomorphic to a subgroup of $\mbox{\rm Aff}(1,p)$, the affine group over $\mathbb{F}_{p}$. Thirdly, $G$ can be isomorphic to a group $K\times\langle a\rangle$, where $K$ is an odd order subgroup of $\mbox{\rm Aff}(1,p)$ and $a$ is an element of order $2$ (Theorem 5.13). In particular, $G$ is soluble. Ultimately, our results rely on the classification of the finite simple groups. One could avoid this by assuming from the outset that $G$ is soluble. This would lead to exactly the same list of RCC  loops of order $2p$, but of course without the guarantee to have found them all.

In Section 6, we determine the number of isomorphism classes of loops of order $2p$ (Theorem 6.5).

Finally, Section 7 introduces a series of examples of RCC  loops of order $q^{2}-1$ and multiplication groups $\mbox{\rm GL}(2,q)$ (Proposition 7.1). For $q=4$, we obtain a loop of order $3\cdot 5$, whose multiplication group is not soluble. These examples indicate that a generalisation of our results to RCC  loops of order $pq$ for distinct primes $p$ and $q$ could be substantially more difficult.

This is a good place to discuss some related results. In [19, Theorem A], Stein shows that if $T$ is a conjugacy class in a finite group and at the same time a tranversal to a subgroup, then $\langle T\rangle$ is soluble. This result uses the classification of the finite simple groups. Without the classification, but with the help of the Odd Order Theorem, Csőrgő and Niemenmaa in [6] obtain the solubility of the full multiplication group of a loop under certain conditions on the stabilizer of a point. Their paper contains further references for results along this line. In [5], Csőrgő and Drápal characterise left conjugacy closed loops inside the class of nilpotent loops of nilpotency class two. In the same paper, these authors also determine the nilpotent left conjugacy closed loops of order $p^{2}$ for primes $p$. In [15, Theorem $4.15$], Kunen shows that for each odd prime $p$ there is exactly one non-assciative conjugacy closed loop of order $2p$, up to isomorphism (a loop is conjugacy closed, if it is both left and right conjugacy closed). Burn shows in [4] that every Bol loop of order $p^{2}$ or $2p$ for a prime $p$ is a group. Finally, in [7, Theorem $7.1$] Daly and Vojtěchovský determine the number of nilpotent loops of order $2p$, where again $p$ is a prime, up to isomorphism.

This paper builds upon the PhD thesis of the first author [1], written under the direction of the second author and Alice Niemeyer. Theorem 3.1 is contained in this thesis, but also a complete classification of all RCC  loops of order at most $30$. These have been incorporated into the GAP package Loops of Nagy and Vojtěchovský [17]. The classification of the RCC  loops of order $2p$ is, to the best of our knowledge, new. The examples computed in [1] were of considerable importance for confirming our theoretical results of Section 6. The example of an RCC  loop of order $15$ and multiplication group $\mbox{\rm GL}(2,4)$ contained in [1], gave rise to the series of examples constructed in Section 7.

Our group theoretical notation is standard. For example, we write $G^{\prime}$ for the commutator subgroup of the group $G$. We do recall the notion of an almost simple group and that of the core of a subgroup in the introductions to Section 2 and Section 3, respectively. As already indicated above, a cyclic group of order $n$ is denoted by $C_{n}$, and the symmetric and alternating groups of degree $n$ are denoted by $S_{n}$ and $A_{n}$, respectively.

2. Primitive permutation groups of squarefree degree

We begin with a remark on the sizes of conjugacy classes in almost simple groups. Recall that a group $G$ is almost simple, if there is a non-abelian finite simple group $S$ such that $S\leq G\leq\mbox{\rm Aut}(S)$ (where $S$ is identified with the group of inner automorphisms of $S$). In this context, $S$ is called the socle of $G$.

Remark 2.1**.**

Let $G$ be an almost simple group with socle $S$. Denote by $l$ the smallest index of any proper subgroup of $S$. Since $S$ is simple, $l$ is a lower bound for the size of those non-trivial conjugacy classes of $G$ lying in $S$. Let $g\in G\setminus S$. Then we have

[TABLE]

Notice that if $C_{S}(g)=S$, the element $g$ acts trivially on $S$ which implies that $g=1$. Hence we have $C_{S}(g)\lneq S$. Thus, $l$ is a lower bound on the size of all non-trivial conjugacy classes of $G$. **

The following theorem combines some major results by Li and Seress on finite primitive permutation groups of squarefree degree, and by Patton, Cooperstein and Vasilyev on the minimal degrees of permutation representations of finite groups of Lie type.

Theorem 2.2**.**

Let $G$ be a finite primitive permutation group of degree $n$ (i.e. $G$ acts faithfully and primitively on a set of $n$ points). Suppose that $n$ is square-free (i.e. $p^{2}\nmid n$ for all primes $p$). Then every non-trivial conjugacy class of $G$ has at least $n$ elements, or one of the following holds:

• (a)

We have $n=p$ is a prime and $G$ is isomorphic to a subgroup of $\mbox{\rm Aff}(1,p)$,

• (b)

We either have have $G=S_{8}$ and $n\in\{35,105\}$, or $G=J_{1}$ and $n=2926$, or $G=\mathrm{PGL}(2,r)$ for an odd prime $r$ and $n=r(r+1)/2$,

• (c)

or $G$ is almost simple and $\mathrm{soc}(G)$ and $n$ occur in Table 1. There, $r$ denotes a prime power.

Proof. By [16, Theorem $1$] we either have that $n$ is a prime and $G\leq\mbox{\rm Aff}(1,n)$ as in Case (a), or $G$ is almost simple and $S:=\mathrm{soc}(G)$ as well as $n$ appear in the paper [16] by Li and Seress.

The cases when $S$ is isomorphic to an alternating group, are listed in [16, Table $1$]. If $S$ is as in [16, Table $1$, Line $1$], then $S=A_{c}$ and $n=\binom{c}{k}$ with $1\leq k\leq c-1$. For reasons of symmetry it suffices to consider the case $k\leq c/2$. Table 2 lists the size $s(G)$ of the smallest non-trivial conjugacy class of $G$ for all $G$ with $S\in\{A_{5},A_{6},A_{7},A_{8}\}$. This table, easily compiled or verified with GAP [10], proves our claim for $5\leq c\leq 8$.

If $c\geq 9$, by [8, Theorems $5.2$A,B], the subgroups of $A_{c}$ or $S_{c}$ which have an index less then $c(c-1)/2$ do not occur as centralizers of non-trivial elements. Hence the non-trivial conjugacy classes of $A_{c}$ or $S_{c}$ have at least $c(c-1)/2$ elements, proving our claim for $k\leq 2$. The case $3\leq k\leq c-3$ appears as Case (c)(i) in our statement.

In the remaining cases of [16, Table $1$], a look at Table 2 shows that all non-trivial conjugacy classes of $G$ have at least $n$ elements except for

• •

$G=S_{8}$ and $n\in\{35,105\}$,

• •

$S=A_{2a}$ and $n=\binom{2a}{a}/2$ with $a\in\{6,9,10,12,36\}$.

These cases appear as Case (b) and Case (c)(ii), respectively, in our statement.

The cases when $S$ is a sporadic simple group are listed in [16, Table $2$]. Using GAP, we only find the one exception listed in Case (b).

In [16, Table $3$], the case where $S$ is a classical group are considered. For some small parameter values, we have verified our claim directly with GAP. These cases are listed in the column headed Restrictions of Table 1, and are not commented on any further below. In the following, we refer to the line numbers of [16, Table $3$]. Suppose that $S$ is as in Line $1$. Then $S=\mathrm{PSL}(m,r)$ and

[TABLE]

with $1\leq k<m$. For $k=1$ or $k=m-1$, we have $n=(r^{m}-1)/(r-1)$. If $(m,r)\in\{(2,5),(2,7),(2,9),(2,11),(4,2)\}$, a computation with GAP shows that the non-trivial conjugacy classes of $G$ have more than $n$ elements. Otherwise, $n$ is the smallest index of any proper subgroup of $S$ by [14, Table $5.2$.A]. Applying Remark 2.1, we see that the non-trivial conjugacy classes of $G$ have at least $n$ elements. The case $2\leq k\leq m-2$ is listed as Case (c)(iii) in our statement.

The case when $S$ is as in Line 2, is listed as Case (c)(iv) in our statement.

Suppose that $S$ is as in Line 3 or 4. Then $S=\mathrm{PSL}(2,r)$ and $n=r(r\pm 1)/2$. Since $n$ is squarefree, we have $r=4$ and $n\in\{6,10\}$ or $r$ is an odd prime. If $r=4$, we have $S\cong A_{5}$, and Table 2 proves our claim. If $r$ is an odd prime, then $\mbox{\rm Aut}(\mathrm{PSL}(2,r))=\mathrm{PGL}(2,r)$ and hence $G=\mathrm{PSL}(2,r)$ or $G=\mathrm{PGL}(2,r)$. The conjugacy classes of these groups are well known. We find that only if $G=\mathrm{PGL}(2,r)$ and $n=r(r+1)/2$, there are non-trivial conjugacy classes of $G$ with less than $n$ elements. This case appears in Case (b) in our statement

Suppose that $S$ is as in Line 5. Then $S=\mathrm{PSL}(2,r)$ and $n=\sqrt{r}(r+1)/2$. Since $n$ is squarefree, $r$ is the square of a prime number. The non-trivial conjugacy classes of $\mathrm{PSL}(2,r)$ have at least $n$ elements. Hence $S\lneq G$. This case is listed as Case (c)(v) in our statement.

Suppose that $S$ is as in one of the Lines 6, 7 or 8. Then $S=\mathrm{PSL}(2,r)$ and $n=r(r^{2}-1)/d$ with $d\in\{24,48,120\}$, and $r\equiv\pm 3\ ({\rm mod}\ 8)$ if $d=24$, respectively $r\equiv\pm 1\ ({\rm mod}\ 8)$ if $d=48$, respectively $r\equiv\pm 1\ ({\rm mod}\ 10)$ if $d=120$. In particular, $r$ is odd. Since $n$ is squarefree, $r=9$ or $r$ is an odd prime. If $r=9$ we have $S\cong A_{6}$, and Table 2 proves our claim. The cases where $r$ is an odd prime, are listed as Case (c)(vi) through Case (c)(viii) in our statement.

Suppose that $S$ is as in Line 9. Then $S=\mathrm{PSU}(m,r)$ with

[TABLE]

If $m=2$, we have $S\cong\mathrm{PSL}(2,r)$ (see [20, Theorem $10.9$]) and $n=r+1$, a case we have already considered above. For $m=3$ and $r=5$, our claim can be verified with GAP. The case of $m=4$ is listed as Case (c)(ix) in our statement. If $6\mid m$ and $r=2$, then $n=(2^{m}-1)(2^{m-1}+1)/3$ is not squarefree, as $2^{6}-1$ divides $2^{m}-1$ and $3$ divides $2^{m-1}+1$. In the remaining cases, $n$ is the smallest index of any proper subgroup of $S$ (see [14, Table $5.2$.A]). Thus by Remark 2.1, the non-trivial conjugacy classes of $G$ have at least $n$ elements.

Suppose that $S$ is as in Line 10. Then $S=\mathrm{PSp}(2m,r)$ with $(m,r)\neq(2,2)$ and $n=(r^{2m}-1)/(r-1)$. (The case $(m,r)=(2,2)$ leads to $S=A_{6}$ and $n=15$, which can be excluded by Table 2.) Again, we have already considered the case $m=1$, where $S\cong\mathrm{PSL}(2,r)$ (see [20, Theorem $8.1$]). If $m=2$ and $r=3$, then $n=40$ is not squarefree. The case $m\geq 3$ and $r=2$ is listed as Case (c)(x) in our statement. In the remaining cases, $n$ is the smallest index of any proper subgroup of $S$ (see [14, Table $5.2$.A]) and Remark 2.1 proves our claim.

If $S$ is as in Line $12$ or $13$, then $S\cong A_{6}$, and we are done with Table 2.

Suppose that $S$ is as in Line $14$. Then $S=\Omega(2m+1,r)$ and $n=(r^{2m}-1)/(r-1)$. We may assume that $m\geq 3$ and that $r$ is odd, as otherwise $S\cong\mathrm{PSp}(2m,r)$ (see [20, Theorems $11.6$, $11.9$, Corollary $12.32$]), a case already considered. If $r=3$, then $n=(3^{2m}-1)/2$ is not square free. In the other cases, $n$ is the smallest index of any proper subgroup of $S$ (see [14, Table $5.2$.A]), and we are done as above.

Suppose that $S$ is as in Line $16$. Then, $m$ is even and once more by [14, Table $5.2$.A] and Remark 2.1 we obtain our claim. (This includes the case $m=2$, where $S\cong\mathrm{PSL}(2,q^{2})$ (see [20, Corollary $12.43$]) and $n=q^{2}+1$.)

Suppose that $S$ is as in Line $19$. Then $S=\mathrm{P}\Omega^{+}(2m,r)$ with $m\geq 3$ odd and

[TABLE]

If $m=3$, we have $S\cong\mathrm{PSL}(4,r)$ (see [20, Corollary $12.21$]) and $n=(q^{2}+1)(q^{2}+q+1)$. This case is already contained in Case (c)(iii) of our statement. If $r\neq 2$ and $m\geq 5$, we conclude with [14, Table $5.2$.A] and Remark 2.1. The case of $m\geq 5$ and $r=2$ is included as Case (c)(xv) in our statement.

The remaining cases of [16, Table $3$] are listed as Cases (c)(xi) through (c)(xiv) and (c)(xvi) through (c)(xvii), respectively in our statement.

Suppose that $S$ is as in [16, Table $4$], i.e. an exceptional group of Lie type. In [21], [22] and [23], A. V. Vasilyev lists the smallest index $l$ of any proper subgroup of the exceptional simple groups. By Remark 2.1, the non-trivial conjugacy classes of $G$ have at least $l$ elements. We find $l=n$ except for

• •

$S=G_{2}(4)$ and $n=1365$. We verified our claim for the two almost simple groups with socle $G_{2}(4)$ with GAP.

• •

$S=E_{7}(r)$ with $n=(r^{18}-1)(r^{14}-1)(r^{4}-r^{2}+1)/((r^{2}-1)(r-1))$. This case is listed as Case (c)(xviii) in our statement.

This completes our proof. $\Box$

Remark 2.3**.**

For the purpose of this remark, let us call an example a pair $(G,n)$ of a primitive permutation group $G$ of square-free degree $n$ containing a non-trivial conjugacy class with less than $n$ elements.

Now assume the hypotheses of Theorem 2.2. Clearly, not all the instances $(G,n)$ listed there are examples.

If $G$ is as in (a) of this theorem, then $G$ has a conjugacy class of length $p-1$. The symmetric group $S_{8}$ has a conjugacy class of length $28$, and the sporadic simple group $J_{1}$ has a conjugacy class of length $1463$. Thus the groups in (a) and the first three instances of (b) provide examples. This fact also indicates that in order to enumerate all examples one will have to use the classification of the finite simple groups.

The group $G=\mathrm{PGL}(2,r)$ has a conjugacy class of length $r(r-1)/2$ for every odd prime $r$. The group $\mathrm{PGL}(2,3)$ is isomorphic to the symmetric group $S_{4}$, which does not have a primitive permutation representation of degree $6$. Thus $(\mathrm{PGL}(2,r),r(r+1)/2)$ is an example, if and only if $r\geq 5$ and $(r+1)/2$ is square-free.

We expect that not many examples will arise from the pairs $(G,n)$ listed in Theorem 2.2(c), but it would be a tedious task to enumerate all of them. One approach could be to determine all subgroups of $G$ of index less than $n$ and show that most of such subgroups are not centralizers of elements. Still, one has to decide whether one of the remaining numbers $n$ is indeed square-free. This will most certainly lead to difficult, if not intractible, number theoretical questions. **

In the lemma below we are going to make use of Zsigmondy primes, also known as primitive prime divisors. Let $r$ and $d$ be integers greater than $1$. We call a prime $\ell$ a Zsigmondy prime for $r^{d}-1$, if $\ell$ divides $r^{d}-1$, but not $r^{i}-1$ for $1\leq i<d$. A Zsigmondy prime for $r^{d}-1$ exists whenever $d>2$ and $(r,d)\neq(2,6)$ (see [13, Theorem IX.8.3]).

Lemma 2.4**.**

Suppose that $n=pq$, where $p$ and $q$ are distinct primes, and that $G$ is a finite primitive permutation group of degree $n$ such that $G$ has a non-trivial conjugacy class with less than $n$ elements. Then one of the following holds:

• (a)

We have $G\in\{A_{7},S_{7},S_{8}\}$ and $n=35$.

• (b)

We have $G=\mathrm{PGL}(2,r)$ for an odd prime $r$ and $n=r(r+1)/2$.

• (c)

The group $G$ is almost simple with $\mathrm{PSL}(2,r)=\mathrm{soc}(G)\lneq G$, where $\sqrt{r}$ is an odd prime, $r>9$, and $n=\sqrt{r}(r+1)/2$.

• (d)

We have $G\in\{\mathrm{PSL}(2,13),\mathrm{PGL}(2,13)\}$ and $n=91$.

• (e)

We have $G\in\{\mathrm{PSL}(2,61),\mathrm{PGL}(2,61)\}$ and $n=1891$.

• (f)

We have $S=\mathrm{P}\Omega^{+}(2m,2)$, for $m\geq 3$ and $n=(2^{m}-1)(2^{m-1}+1)$.

Proof. We have to exclude those integers $n$ in Theorem 2.2 which are not the product of two different primes. From Cases (a) and (b) of Theorem 2.2, we obtain (part of) Case (a) and Case (b) of our lemma.

So suppose that $G$ is almost simple and that $S:=\mathrm{soc}(G)$ occurs in Table 1. In Case (i) we have $n=\binom{c}{k}$ with $k\geq 3$. For reasons of symmetry it suffices to consider $3\leq k\leq c/2$. By [18, Theorem $7$], the total number (counting multiplicities) of prime factors of the binomial coefficient $\binom{c}{k}$ is greater than or equal to the total number of prime factors of $c$, with equality only if $(c,k)=(8,4)$. Thus $\binom{c}{k}$ is the product of two different primes only if $c$ is a prime. Consider the case $k=3$ first. Then

[TABLE]

If $c=7$ we have $n=35$. This is recorded in Case (a) of our lemma. So assume that $c\geq 11$. Then $n=\binom{c}{3}$ has at least three different prime factors. But then by [18, Theorem $3$], the binomial coefficient $\binom{c}{k}$ with $k>3$ also has at least three different prime factors.

In Cases (ii), (ix) through (xiv) and (xvi) through (xviii) of Table 1, the degree $n$ is clearly not the product of two different primes.

In Case (iii) of Table 1 we have $\mathrm{soc}(G)=\mathrm{PSL}(m,r)$ and

[TABLE]

with $2\leq k<m$. For reasons of symmetry it suffices to consider the integers $k$ with $1\leq k\leq m/2$. Suppose first that $m\geq 5$ and $k\geq 3$. Consider the terms $(r^{m}-1)$, $(r^{m-1}-1)$ and $(r^{m-2}-1)$. They occur only in the numerator and not in the denominator of $n$ and, by Zsigmondy’s theorem, have pairwise distinct primitive prime divisors $r_{1}$, $r_{2}$ and $r_{3}$ (which divide $n$) unless one of the pairs $(m,r)$, $(m-1,r)$ or $(m-2,r)$ is equal to $(6,2)$. But in these cases, i.e. if $m\in\{6,7,8\}$ and $r=2$, we just compute that $n$ is not the product of two different primes for all $3\leq k\leq m/2$. If $m\geq 4$ and $k=2$, then

[TABLE]

which is the product of two different primes if and only if $m\in\{4,5\}$ and $r=2$. But these cases have already been excluded.

In Case (iv) of Table 1 we have $S=\mathrm{PSL}(m,r)$ and

[TABLE]

with $m\geq 3$ and $1\leq k<m/2$. Suppose first we have $m\geq 5$ and $k\geq 2$. Then, again, the terms $(r^{m}-1)$, $(r^{m-1}-1)$ and $(r^{m-2}-1)$ occur only in the numerator and not in the denominator of $n$ and have pairwise distinct primitive prime divisors $r_{1}$, $r_{2}$ and $r_{3}$ (which divides $n$) unless one of the pairs $(m,r)$, $(m-1,r)$ or $(m-2,r)$ is equal to $(6,2)$. But in these cases, i.e. if $m\in\{6,7,8\}$ and $r=2$, we just compute that $n$ is not the product of two different primes for all $2\leq k<m/2$. If $m$ is arbitrary and $k=1$ then

[TABLE]

which is the product of two different primes if and only if $(m,r)=(3,2)$. But this case has already been excluded.

Case (v) of Table 1 is listed as Case (c) in our lemma.

In Cases (vi) through (viii) of Table 1, we have $n=r(r^{2}-1)/d$ with $d\in\{24,48,120\}$. Clearly, $n$ is not the product of two different primes if $r>d+1$. For $r\leq d+1$ and $r$ not equal to one of the primes excluded in Table 1, we have that $n$ is the product of two different primes if and only if $(r,d)=(13,24)$ or $(r,d)=(61,120)$. We have $S=\mathrm{PSL}(2,r)$, and as $r$ is a prime, $G=\mathrm{PSL}(2,r)$ or $G=\mathrm{PGL}(2,r)$. These cases are listed as Case (d) and Case (e) of our lemma.

Finally, Case (xv) of Table 1 is Case (f) of the lemma. $\Box$

3. The action of the right multiplication groups of rcc loops of

twice prime order

The results of the previous section are now applied to RCC  loops whose order is the product of two distinct primes. Recall the notion of the envelope of a loop as introduced in the second paragraph of Section 1 (which follows [2, p. $100$]). Recall also that a loop folder is a triple $(G,H,T)$, where $G$ is a finite group, $H$ is a subgroup of $G$ and $T$ is a transversal, with $1\in T$, for all coset spaces $H^{g}\backslash G$ with $g\in G$ (see the third paragraph of Section 1 and [2, p. $101$]). A loop folder $(G,H,T)$ is an RCC  loop folder, if $T$ is invariant under conjugation by $G$ (see also Section 1). We will now introduce further notation, although only needed in later sections.

Let $(G,H,T)$ be a loop folder. By definition, the order of $(G,H,T)$ is the size of $T$. We say that $(G,H,T)$ is faithful, if $G$ acts faithfully on $H\backslash G$. This is the case if and only if the core of $H$ in $G$ is trivial. Recall that the smallest normal subgroup $C$ of $G$ containd in $H$ is called the core of $H$ in $G$. Thus $C$ is the intersection of all the $G$-conjugates of $H$ in $G$, i.e. $C:=\cap_{g\in G}H^{g}$. The core of $H$ in $G$ is equal to the kernel of the permutation representation of $G$ on the (right or left) cosets of $H$. Clearly, the envelope of a loop is a faithful loop folder.

Here is the main result of this section. It is only used in the setup of Subsection 5.1, and nowhere else in this paper.

Theorem 3.1**.**

Let $(G,H,T)$ denote the envelope of an RCC  loop of order $n=pq$, where $p$ and $q$ are distinct primes. Then $G$ acts imprimitively on $H\backslash G$.

Proof. Let $n=pq=|H\backslash G|$. Suppose that $G$ acts primitively on $H\backslash G$. Since $|T|=n$ and $T$ is a union of conjugacy classes one of which is the trivial class, $G$ has a non-trivial conjugacy class with less than $n$ elements. Hence $G$ is one of the groups of Lemma 2.4.

In Cases (a), (b), (d) and (e) of Lemma 2.4, the concerned groups have at most two non-trivial conjugacy classes with less than $n$ elements. Elementary combinatorics shows that in these cases there is no union of conjugacy classes $T$ with $|T|=n$.

Suppose that $G$ is as in Case (c) of Lemma 2.4. Then $G$ is almost simple with $S:=\mathrm{soc}(G)=\mathrm{PSL}(2,r)$, where $\sqrt{r}$ an odd prime, $r>9$, and $n=\sqrt{r}(r+1)/2$. Moreover, $S\lneq G$. The subgroups of $S=\mathrm{PSL}(2,r)$, are classified in Dickson’s Theorem; see [11, Hauptsatz II.8.27]. This shows that if $r\geq 17$, then only the maximal subgroups of index $r+1$ have an index less then $n=\sqrt{r}(r+1)/2$. Consider the non-trivial conjugacy classes of $G$. Those which contains elements of $S$ have at least $n$ elements as we already mentioned in the proof of Theorem 2.2.

Let $g\in G\setminus S$. By Remark 2.1 we have

[TABLE]

for some positive integer $a$. Hence $|G\colon C_{G}(g)|\geq n$ except if $C_{S}(g)\leq M$, where $M$ is a maximal subgroup of $S$ with $|S\colon M|=r+1$. In this case $r+1\mid|S\colon C_{S}(g)|$. Hence, if $|G\colon C_{G}(g)|$ is less than $n$, it is a multiple of $r+1$. Thus, a union $T$ of conjugacy classes of sizes less than $n$ with $1\in T$ has a size congruent to $1$ modulo $r+1$ and is therefore not divisible by the prime $(r+1)/2$. Therefore, it is not possible to have $|T|=n$.

Finally, assume that $G$ is as in Case (f) of Lemma 2.4. Then $G$ is almost simple with $S:=\mathrm{soc}(G)=\mathrm{P}\Omega^{+}(2m,2)$, where $m\geq 3$ and $n=(2^{m}-1)(2^{m-1}+1)$. In order for $n$ to be the product of two distinct primes, it is necessary that $p:=2^{m-1}+1$ is a Fermat prime and $q:=2^{m}-1$ is a Mersenne prime. In particular, $m$ is a Fermat prime. By [14, Table $5.2$.A], the smallest index of any proper subgroup of $S$ equals $2^{m-1}(2^{m}-1)$. Remark 2.1 shows that any nontrivial conjugacy class of $G$ has at least $2^{m-1}(2^{m}-1)$ elements. As twice this number is greater than $n$, we conclude that $T$ is the union of two conjugacy classes of $G$, one of which has length $2(2^{2m-2}+2^{m-2}-1)$. We have $m\geq 5$ and $S=\mbox{\rm SO}^{+}(2m,2)$ (in the notation of [20, p. $160$]). In particular, $G\in\{\mbox{\rm SO}^{+}(2m,2),O^{+}(2m,2)\}$. For the order of $G$ see [20, p. $141$, $165$].

Let $\ell$ be a Zsigmondy prime for $2^{2m-2}-1$. As $\ell$ does not divide $2^{m-1}-1$, and as $p=2^{m-1}+1$, we conlcude that $\ell=p$. Let $t$ be a nontrivial element in $T$. Now $p$ does not divide $2(2^{2m-2}+2^{m-2}-1)=|G\colon C_{G}(t)|$, and so there is $g\in C_{G}(t)$ with $|g|=p$. Let $V$ denote the natural $2m$-dimensional $\mathbb{F}_{2}$-vector space of $G$, equipped with the quadratic form $Q$ defining $G$. Since $p$ is a Zsigmondy prime for $2^{2m-2}-1$, it follows that $g$ acts irreducibly on some $(2m-2)$-dimensional subspace $V_{0}$ of $V$. As the dimension of $V_{0}$ is larger than $1$, either $V_{0}$ is totally singular or non-degenerate with respect to $Q$ (for these notions see [20, p. $56$]). The maximal dimension of a totally singular subspace of $V$ equals $m$, and thus $V_{0}$ is in fact non-degenerate. It follows that $g$ fixes $V_{1}:=V_{0}^{\perp}$. In particular, $g\in O(V_{0})\times O(V_{1})$, where $O(V_{i})$ denotes the orthogonal group with respect to the restriction $Q_{i}$ of $Q$ to $V_{i}$, $i=0,1$. We may thus write $g=g_{0}\oplus g_{1}$, whith $g_{i}$ the restriction of $g$ to $V_{i}$, $i=0,1$. Now $O(V_{0})$ contains a cyclic, irreducible subgroup, and thus the Witt index of $Q_{0}$ equals $m-2$ by [12, Satz 3c)]. Hence $V_{1}$ does not contain any non-trivial singular vector with respect to $Q_{1}$, and thus $O(V_{1})\cong S_{3}$, the symmetric group on three letters (see [20, Theorem $11.4$]). As $|g|=p\geq 17$, we conclude that $g$ acts trivially on $V_{1}$, i.e. $g_{1}=1$ and $V_{1}$ is the fixed space of $g$. It follows that $C_{G}(g)$ fixes $V_{1}$ and $V_{0}=V_{1}^{\perp}$, and thus $C_{G}(g)\leq C_{O(V_{0})}(g_{0})\times O(V_{1})$. As $g_{0}$ acts irreducibly on $V_{0}$, its centralizer in $O(V_{0})$ is cyclic and irreducible, and thus equals $\langle g_{0}\rangle$, again by [12, Satz 3c)]. In particular, $t\in C_{G}(g)\leq\langle g_{0}\rangle\times O(V_{1})$. If $p\mid|t|$ or $3\mid|t|$, then $C_{G}(t)\leq O(V_{0})\times O(V_{1})$. Otherwise, $|t|=2$ and $t$ has a $(2m-1)$-dimensional fixed space $V^{\prime}$ on $V$ and $C_{G}(t)\leq O(V^{\prime})$. In any case, the $2$-part of $|C_{G}(t)|$ is less than $2^{(m-1)^{2}}$, whereas the $2$-part of $|G|$ equals $2^{m(m-1)+1}$ (see [20, p. $141$]). In particular, the $2$-part of $|G\colon C_{G}(t)|$ is larger than $2$, a contradiction. $\Box$

We end this section with two general results on loop folders with certain invariance properties. The first will be used in an extension of a theorem of Drápal [9].

Lemma 3.2**.**

Let $(G,H,T)$ be a loop folder such that $T$ is invariant under conjugation by $H$. Suppose the $t\in T$ is such that $Ht$ is also a left $H$-coset in $G$, i.e. there exist $g\in G$ with $Ht=gH$ (this is the case in particular if $t$ normalizes $H$). Then $[t,H]=1$.

Proof. Let $h\in H$. Then

[TABLE]

This implies $t=h^{-1}th$, as $t,h^{-1}th\in T$. $\Box$

Lemma 3.3**.**

Let $(G,H,T)$ denote an RCC  loop folder. Let $K\leq G$ such that $HG^{\prime}\leq K$. Then

[TABLE]

(i.e. the length of the conjugacy classes of the elements in $T$ are bounded above by $|K\colon H|$).

Proof. Let $g\in G$. Then the right coset $Kg$ is a union of exactly $|K\colon H|$ right cosets of $H$ in $G$. Thus $|Kg\cap T|=|K\colon H|$.

Now let $t\in T$ and $x\in G$. Then $t^{x}t^{-1}\in G^{\prime}\leq K$. It follows that $Kt^{x}=Kt$ and thus $t^{x}\in Kt$ for all $x\in G$. As $t^{x}\in T$ for all $x\in G$ by assumption, we conclude that $|G\colon C_{G}(t)|=|\{t^{x}\mid x\in G\}|\leq|K\colon H|$. $\Box$

4. Right conjugacy closed loops of prime order

In this section we give a new proof of a theorem of Drápal [9] which states that left conjugacy closed loops of prime order are groups. We prove the analogue for RCC  loops, but as the opposite loop of a left conjugacy closed loop is an RCC loop, our version is equivalent to Drápal’s result. Recall the notions related to loop folders summarized at the beginning of Section 3.

We begin with an easy lemma.

Lemma 4.1**.**

Let $p$ be a prime and let $G\leq S_{p}$ with $p\mid|G|$. Then the following statements hold for every $1\neq g\in G$.

(a)* If $p\nmid|g|$, then $p\mid|G\colon C_{G}(g)|$.*

(b)* If $p\mid|g|$, then $|C_{G}(g)|=p$.*

(c)* Suppose that $|G\colon C_{G}(g)|<p$. Then $G$ has a unique Sylow $p$-subgroup $P$ and $g\in P$. Moreover, if $G\neq P$, then $G$ is a Frobenius group with kernel $P$ and a Frobenius complement of order $r$ dividing $p-1$. In this case, $P$ is the Frattini subgroup of $G$.*

(d)* We have $|G\colon C_{G}(g)|\neq 2(p-1)$.*

Proof. In view of the cycle decomposition of $g$, the first two parts are trivial. So let us assume that $|G\colon C_{G}(g)|<p$ or that $|G\colon C_{G}(g)|=2(p-1)$. By (a) and (b) we have $|C_{G}(g)|=p$, and in particular $P:=\langle g\rangle$ is a Sylow $p$-subgroup of $G$. Under the hypothesis of (c) we get $|G\colon N_{G}(P)|<p$, and under the hypothesis of (d) we get $|G\colon N_{G}(P)|\mid 2(p-1)$. In each case Sylow’s theorems imply $P\unlhd G$. Now (d) and the last two statements of (c) follow from the fact that $G$ embeds into $N_{S_{p}}(P)$, which is isomorphic to $\mbox{\rm Aff}(1,p)$. $\Box$

Corollary 4.2** (Drápal [9]).**

Let $p$ be a prime and let $(G,H,T)$ denote the envelope of an RCC  loop $\mathcal{L}$ of order $p$. Then $H=1$, i.e. $\mathcal{L}$ is a group (isomorphic to $G$).

Proof. We may assume that $G\leq S_{p}$ and we have $p\mid|G|$. Now $T=\{1\}\cup T^{\prime}$ with $T^{\prime}:=T\setminus\{1\}$. By assumption, $T^{\prime}$ is a union of conjugacy classes of $G$ of lengths at most $p-1$. It follows from Lemma 4.1(c) that $G$ has a unique Sylow $p$-subgroup $P$ and that $T\subseteq P$. Hence $G=\langle T\rangle=P$, i.e. $H=1$. $\Box$

We will also need the following generalization of Corollary 4.2.

Proposition 4.3**.**

Let $p$ be a prime and let $(G,H,T)$ be an RCC  loop folder of order $p$ with $\langle T\rangle=G$. Then $G$ is abelian.

Proof. Let

[TABLE]

denote the kernel of the action of $G$ on $H\backslash G$ and put $\bar{G}:=G/N$, $\bar{H}:=H/N$ and $\bar{T}:=\{Nt\mid t\in T\}\subseteq\bar{G}$. Then $(\bar{G},\bar{H},\bar{T})$ is a faithful RCC  loop folder of order $p$ with $\langle\bar{T}\rangle=\bar{G}$. Thus $(\bar{G},\bar{H},\bar{T})$ is the envelope of an RCC  loop of order $p$ (see [2, $1.7.(4)$]). By the result of Drápal (see Corollary 4.2), such a loop is a group. It follows that $\bar{T}=\bar{G}$ and $\bar{H}=1$. In particular, $H=N$ is a normal subgroup of $G$ of index $p$.

To show that $G$ is abelian, let $t\in T$. By Lemma 3.2 we have $[t,H]=1$, as $H\unlhd G$. It follows that $\langle H,t\rangle\leq C_{G}(t)$. If $t\neq 1$, we have $\langle H,t\rangle=G$, as $H$ is of index $p$ in $G$. Hence $t\in Z(G)$ for all $t\in T$. The claim follows from $\langle T\rangle=G$. $\Box$

Notice that the above proposition is a generalization of Drápal’s theorem (see Corollary 4.2); indeed, if $(G,H,T)$ is the envelope of a loop, then $G=\langle T\rangle$, and if, moreover, $G$ is abelian, then $H=1$, as the core of $H$ in $G$ is trivial.

5. The right multiplication groups of rcc loops

of twice prime order

We refer the reader to the introduction of Section 3 for the notions related to loop folders.

5.1. Generalities

Let $p$ and $q$ be distinct primes and let $(G,H,T)$ denote the envelope of an RCC  loop of order $pq$. This implies in particular that $G$ acts faithfully on $H\backslash G$, i.e. the core of $H$ in $G$ is trivial. It is at this stage, and only here, where we impose an important consequence of Theorem 3.1. This states that $G$ acts imprimitively on $H\backslash G$, and hence $H$ is not a maximal subgroup of $G$. We let $K\lneq G$ such that $H\lneq K$. We choose notation such that $|G\colon K|=q$ and $|K\colon H|=p$.

Put $T_{1}:=T\cap K$, $K_{1}:=\langle T_{1}\rangle\leq K$ and $H_{1}:=H\cap K_{1}\leq H$.

We collect first properties.

Lemma 5.1**.**

Let the notation be as above. Then $(K_{1},H_{1},T_{1})$ is an RCC  loop folder of order $p$ with $K_{1}$ abelian. Also, $K_{1}\unlhd K$ and $K=HK_{1}$. Finally, $H_{1}\unlhd K$.

Proof. Clearly, $K$ is the disjoint union of the cosets $Ht$ for $t\in T_{1}$. Thus $|T_{1}|=p$ and $K_{1}$ is the disjoint union of the cosets $K_{1}t$ for $t\in T_{1}$. As $T_{1}$ is invariant under conjugation in $K$, the first statement follows. The second statement follows from Proposition 4.3, and the next two are obvious. The last statement follows from $K=HK_{1}$ and the fact that $H_{1}\unlhd H$ and that $K_{1}$ is abelian. $\Box$

5.2. The case $q=2$.

Let us assume throughout this subsection that $q=2$. Then $K\unlhd G$. Moreover, $K_{1}\unlhd G$, as $K_{1}^{g}=\langle(T\cap K)^{g}\rangle=\langle T\cap K\rangle=K_{1}$ for all $g\in G$.

Lemma 5.2**.**

Let $L\leq H$ with $L\unlhd K$. Then $L\cap L^{a}=1$ and $LL^{a}=L\times L^{a}\unlhd G$ for all $a\in G\setminus K$.

Proof. Let $a\in G\setminus K$. Clearly, $L\cap L^{a}$ and $LL^{a}$ are normal subgroups of $G$, as $a^{2}\in K$, and $L\unlhd K$. Thus $L\cap L^{a}=1$ since $L\cap L^{a}\leq H$ and the core of $H$ in $G$ is trivial. As $L^{a}\unlhd K$, the product $LL^{a}$ is direct. $\Box$

We record two consequences which will be used later on.

Corollary 5.3**.**

We have $H_{1}\in\{1,p\}$ and $K_{1}$ is elementary abelian of order $p$ or $p^{2}$.

Proof. If $H_{1}$ is trivial, $K_{1}$ has order $p$ by Lemma 5.1. Suppose that $H_{1}$ is nontrivial and let $a\in T\setminus K$. Then $H_{1}\cap H_{1}^{a}=1$ by Lemma 5.2. As $K_{1}\unlhd G$, we have $H_{1}H_{1}^{a}\leq K_{1}$. It follows that $|H_{1}|^{2}$ divides $|K_{1}|=p|H_{1}|$. This implies $|H_{1}|=p$ and $K_{1}=H_{1}H_{1}^{a}$, yielding our claim. $\Box$

Corollary 5.4**.**

Suppose that $1\neq H\unlhd K$. Then $|H|=p$ and there is an involution $a\in G\setminus K$ such that $K=H\times H^{a}$. In particular, $G$ is isomorphic to the wreath product $C_{p}\wr C_{2}$.

Proof. By Lemma 5.2 we have $H\cap H^{a}=1$ and $HH^{a}\leq K$ for every $a\in G\setminus K$. As $|K|=p|H|$, this implies that $|H|=p$ and $K=H\times H^{a}$. It also follows that the involutions in $G$ are contained in $G\setminus K$ and thus $G\cong C_{p}\wr C_{2}$. $\Box$

We now distinguish two cases.

5.2.1. Case $1$

Assume that $H\neq 1$ and that $[s,K_{1}]=1$ for all $s\in T\setminus K$.

Proposition 5.5**.**

Under the assumptions of 5.2.1, we have $H_{1}=1$ and $K_{1}\leq Z(G)$. Moreover, $G$ is isomorphic to the wreath product $C_{p}\wr C_{2}$.

Proof. The fact that $K_{1}$ is abelian and our hypothesis imply that $T\subseteq C_{G}(K_{1})$, and hence $G=\langle T\rangle\leq C_{G}(K_{1})$, i.e. $K_{1}\leq Z(G)$. Thus $H_{1}\unlhd G$, which implies $H_{1}=1$, as $H_{1}\leq H$ and the core of $H$ in $G$ is trivial. Now $K=HK_{1}$ by Lemma 5.1, and thus $H\unlhd K$. The claim follows from Corollary 5.4. $\Box$

5.2.2. Case $2$

Assume that there is $s\in T\setminus K$ with $[s,K_{1}]\neq 1$. In this case we put $Z:=K_{1}\cap C_{G}(s)$. Also, we let $C:=\cap_{k\in K}H^{k}\unlhd K$ denote the kernel of the action of $K$ on the cosets of $H$ in $K$.

Lemma 5.6**.**

Under the assumptions and with the notation of 5.2.2, the following statements hold.

(a)* We have $|G\colon C_{G}(s)|=p$ and $|Z|=|H_{1}|$.*

(b)* We have $G=C_{G}(s)K_{1}$ and $Z\leq Z(G)$.*

(c)* The centralizer $C_{G}(s)$ is abelian and $C_{G}(s)/Z$ is cyclic.*

Proof. By assumption, $K_{1}\not\leq C_{G}(s)$. Corollary 5.3 implies that $|Z|\in\{1,p\}$. By Lemma 5.1 we have $|T_{1}|=p$, which implies that $|T\setminus K|=p$ (recall that $q=2$ and thus $|T|=2p$). As $T\setminus K$ is a union of conjugacy classes of $G$, we have $|G\colon C_{G}(s)|\leq p$, i.e. $|C_{G}(s)|\geq|G|/p$.

Now if $H_{1}=1$, i.e. $|K_{1}|=p$, we also have $|Z|=1$ and $G=C_{G}(s)K_{1}$. Thus all statements of (a) and (b) hold in this case.

Now assume that $|H_{1}|=p$, i.e. $K_{1}$ is elementary abelian of order $p^{2}$. Then $|Z|\leq p$ and

[TABLE]

and we must have equality everywhere in the above chain of inequalities. This implies $|Z|=p$ and $|C_{G}(s)K_{1}|=|G|$, again yielding all the claims of (a) and the first claim of (b).

In any case, the set $T\setminus K$ is a conjugacy class of $G$, consisting of the elements $\{s^{k}\mid k\in K_{1}\}$. Write $\bar{\ }:G\rightarrow\bar{G}:=G/K_{1}$ for the canonical epimorphism. We have $\bar{G}=\langle\bar{T}\rangle=\langle\bar{s}\rangle$ as $T_{1}\subseteq K_{1}$. The natural isomorphism $\bar{G}\rightarrow C_{G}(s)/Z$ maps $\bar{s}$ to $Zs\in C_{G}(s)/Z$. Thus $C_{G}(s)/Z=\langle Zs\rangle$ and $C_{G}(s)=\langle Z,s\rangle$. In particular, $C_{G}(s)/Z$ is cyclic and $C_{G}(s)$ is abelian, as $Z\leq C_{G}(s)$. Now $G=C_{G}(s)K_{1}$ and $Z=K_{1}\cap C_{G}(s)$ imply that $Z\leq Z(G)$. $\Box$

The previous result implies in particular that $G$ is soluble. Indeed, $K_{1}$ is a normal subgroup of $G$, as we have remarked at the beginning of 5.2.2. Now $K_{1}$ is abelian by Corollary 5.3, and $G/K_{1}=C_{G}(s)K_{1}/K_{1}\cong C_{G}(s)/Z$ by Part (b) of the lemma above and by the definition of $Z$. By Part (c) of the lemma, $C_{G}(s)/Z$ is cyclic, hence $G$ is soluble.

Corollary 5.7**.**

Let the assumptions and notation be as in 5.2.2. If $H_{1}\neq 1$, then $H\unlhd K$.

Proof. Suppose that $H_{1}\neq 1$. Then $|Z|=p$ by Corollary 5.3 and Lemma 5.6(a). Now $Z\leq Z(G)$ by Lemma 5.6(b), and thus $Z\cap H=1$, since the core of $H$ in $G$ is trivial. It follows that $K=H\times Z$, as $|K\colon H|=p$ and $Z\leq K_{1}\leq K$.

Now $K/C$ is a soluble permutation group on $p$ points. It follows from a theorem of Galois (see [11, Satz II.3.6]) that $K/C$ is isomorphic to a subgroup of the affine group $\mbox{\rm Aff}(1,p)$.

We have $K/C=(H\times Z)/C=H/C\times ZC/C\cong H/C\times Z$. This implies that $H/C$ is trivial, i.e. $H=C\unlhd K$. $\Box$

Lemma 5.8**.**

Let the assumptions and notation be as in 5.2.2. If $H_{1}=1$, then $C=1$ or $H$ is a $p$-group.

Proof. Put $L:=C_{G}(s)$. Then $L$ is cyclic, $G=L\ltimes K_{1}$ and $K=H\ltimes K_{1}$, with $|K_{1}|=p$ (see Corollary 5.3 and Lemma 5.6). In particular, $H$ is abelian, as it is isomorphic to a subgroup of $L$.

Let $\ell$ be a prime different from $p$ and let $S\leq H$ denote a Sylow $\ell$-subgroup. As $|G\colon L|=p$, there is $g\in G$ such that $S^{g}\leq L$. As $L$ is abelian, we have $L\leq C_{G}(S^{g})$. Suppose that $C_{G}(S^{g})=G$, i.e. $S^{g}\leq Z(G)$. Then $S\leq Z(G)$ wich implies $S=1$, as the core of $H$ in $G$ is trivial.

Now assume that $\ell\mid|H|$. By the above, we must have $C_{G}(S^{g})=L$. Then $C_{G}(S)$ is a cyclic complement of $K_{1}$ in $G$ containing $H$. As $C\leq H\leq C_{G}(S)$ and $C\unlhd K$, it follows that $C\unlhd G$ und thus $C=1$. $\Box$

Corollary 5.9**.**

Let the assumptions and notation be as in 5.2.2. If $H_{1}=1$, then $C=1$ or $H\unlhd K$.

Proof. Suppose that $C\neq 1$. Then $K=HK_{1}$ is a $p$-group by Lemma 5.8. As $K/C$ is isomorphic to a subgroup of the affine group $\mbox{\rm Aff}(1,p)$ by Galois’ theorem (see [11, Satz II.3.6]), it follows that $|K/C|=p$. Now $C\cap C^{s}=1$ and $CC^{s}=C\times C^{s}\unlhd G$ by Lemma 5.2. Thus $C^{s}\cong(C\times C^{s})/C\leq K/C$, and hence $C$ has order $p$. Therefore, $|K|=p^{2}$ and hence $K$ is abelian, proving our claim. $\Box$

5.3. The case $p=2$.

Let us assume throughout this subsection that $p=2$. Then $H\unlhd K$. Here, we put $D:=\cap_{g\in G}K^{g}$, the kernel of the action of $G$ on the set of right cosets of $K$. Then $D$ is an elementary abelian $2$-group, as $G$ acts faithfully on the set of right cosets of $H$ and as $H$ has index $2$ in $K$. We write $\bar{\ }:G\rightarrow\bar{G}:=G/D$ for the canonical epimorphism. Then $\bar{G}$ is a faithful permutation group on $q$ letters, i.e. $\bar{G}$ is isomorphic to a subgroup of $S_{q}$.

Lemma 5.10**.**

We have $|K\cap T|=2$, and writing $K\cap T=\{1,z\}$, we have $z\in Z(K)$. In particular, $C_{G}(z)\in\{K,G\}$.

Proof. The first assertion follows from $|K\colon H|=2$, and the second from the fact that all $K$-conjugates of $z$ again are in $K\cap T$. $\Box$

5.3.1. Case 1

Here, we consider the case that $K$ is normal in $G$. Let us keep the notation of Lemma 5.10 in the following.

Lemma 5.11**.**

Suppose that $K\unlhd G$. Then $G$ is abelian and $H=1$.

Proof. In this case $z^{g}\in K\cap T$ for all $g\in G$, and thus $\langle z\rangle\leq Z(G)$. Now $\langle z\rangle\cap H=1$, as the core of $H$ in $G$ is trivial. It follows that $|z|=2$ and $K=H\times\langle z\rangle$. This implies that $H^{\prime}=K^{\prime}\unlhd G$, and thus $K^{\prime}=1$, i.e. $K$ is abelian. In turn, $K$ is a $2$-group, as $O_{2^{\prime}}(H)=O_{2^{\prime}}(K)\unlhd G$ (recall that $O_{2^{\prime}}(H)$ denotes the largest normal subgroup of $H$ of odd order). Thus $K$ is the unique Sylow $2$-subgroup of $G$.

Now let $t\in T\setminus K$. Then $t$ is not a $2$-element as otherwise $t\in K$. Let $r$ be an integer such that $Q:=\langle t^{r}\rangle$ is a Sylow $q$-subgroup of $G$. As $G/K$ is cyclic, we have $G^{\prime}\leq K$ and we may thus apply Lemma 3.3. This yields $|G\colon C_{G}(t)|\leq 2$. Hence $|G\colon C_{G}(Q)|\leq 2$ and thus $C_{G}(Q)\unlhd G$. Now $Q$ is abelian and hence $Q\leq C_{G}(Q)$. It follows that $Q\unlhd C_{G}(Q)$ and thus $Q=O_{q}(C_{G}(Q))$. In particular, $Q\unlhd G$, implying that $G=Q\times K$ is abelian. $\Box$

5.3.2. Case 2

Here, we consider the case that $K$ is not normal in $G$. Again, we use the notation of Lemma 5.10.

Proposition 5.12**.**

Suppose that $N_{G}(K)=K$. Then there is $M\unlhd G$ such that $|G\colon HM|=2$.

Proof. Let $Z$ denote the $G$-conjugacy class of $z$ and put $T^{\prime}:=T\setminus(Z\cup\{1\})$. By Lemma 5.10, we have $C_{G}(z)\in\{K,G\}$. Suppose first that $C_{G}(z)=K$. Then $|Z|=|G\colon K|=q$, and thus $|T^{\prime}|=q-1$. If $C_{G}(z)=G$, then $|T^{\prime}|=2(q-1)$. In particular, $q-1\mid|T^{\prime}|$. Let $X_{1},\ldots,X_{m}$ denote the $G$-conjugacy classes contained in $T^{\prime}$, numbered in such a way that $|X_{1}|\leq\cdots\leq|X_{m}|$. Thus $|X_{1}|\leq q-1$, unless $m=1$ and $C_{G}(z)=G$, in which case $|X_{1}|=2(q-1)$.

Let $t\in T^{\prime}$. Then $t\not\in K$, as $t\not\in\{1,z\}=K\cap T$. In particular, $\bar{t}\neq 1$, since $D\leq K$. Let $X$ denote the $G$-conjugacy class of $t$. Then $\bar{X}$ is the $\bar{G}$-conjugacy class of $\bar{t}$ and $|\bar{X}|$ divides $|X|$. Consider the case that $X=X_{1}$. If $|X|\leq q-1$, then $|\bar{X}|\leq q-1$. If $|X|=2(q-1)$, then $|\bar{X}|$ is a proper divisor of $2(q-1)$ by Lemma 4.1(d), and thus, again, $|\bar{X}|\leq q-1$. Lemma 4.1(c) implies that $\bar{G}$ has a normal Sylow $q$-subgroup $Q$. Moreover, $|\bar{G}\colon\bar{K}|=|G\colon K|=q$, and $\bar{K}\neq 1$, as otherwise $K=D$ would be normal in $G$. Thus $\bar{G}$ is a Frobenius group of order $qr$ with $r\mid(q-1)$, again by Lemma 4.1(c).

Since $\bar{G}$ is a Frobenius group, every non-trivial conjugacy class of $\bar{G}$ has length $q$ or $r$, and the conjugacy classes of length $r$ lie in $Q$. Suppose that there is some $1\leq j\leq m$ such that $|\bar{X}_{j}|=q$. Then $|X_{j}|=q$ as $|\bar{X}_{j}|$ divides $|X_{j}|$ and $|X_{j}|\leq 2(q-1)$. Also, $X_{j}$ is the unique conjugacy class of length $q$ contained in $T^{\prime}$. If $1\leq i\neq j\leq m$, then $|X_{i}|\leq q-1$, and thus $|\bar{X}_{i}|=r$. In particular, $r\mid|X_{i}|$. Now $q-1$ divides $|T^{\prime}|$, as we have already observed above. It follows that $r$ divides $q$, a contradiction. This shows that $\bar{T}^{\prime}\subseteq Q$.

We have $z\in Z(K)$ and $D\leq K$, and thus $Z\subseteq C_{G}(D)$ as $D\unlhd G$. Moreover, $D$ is abelian and hence $\langle Z,D\rangle\leq C_{G}(D)$. Now $\bar{G}=\langle\bar{T}\rangle=\langle\bar{Z}\cup\bar{T}^{\prime}\rangle=\langle\bar{Z}\rangle$, as $\bar{T}^{\prime}\subseteq Q$, the Frattini subgroup of $\bar{G}$. Thus $G=\langle Z,D\rangle\leq Z_{G}(D)$, i.e. $D\leq Z(G)$. This implies that $D\cap H=1$, as the core of $H$ in $G$ is trivial. Hence $|H||D|=|HD|\leq|K|=2|H|$, and so $|D|\in\{1,2\}$.

Let $M$ denote the inverse image of $Q$ in $G$. If $D=1$, then $|M|=q$ an thus $G=K\ltimes M$ and $|G\colon HM|=2$ as claimed. Now suppose that $D=\langle d\rangle$ with $|d|=2$. Then $HM\neq G$ as otherwise $d\in HM$ and, in turn, $d\in H$. Now $G=KM$ as $\bar{G}=\bar{K}\bar{M}$, and thus $G=KM=HM\cup HMd$, i.e. $|G\colon HM|=2$, as claimed. $\Box$

5.4. The main result

We can now summarize our results for envelopes of RCC  loop folders of orders $2p$ for odd primes $p$.

Theorem 5.13**.**

Let $(G,H,T)$ be the envelope of an RCC  loop of order $2p$, where $p$ is an odd prime. Then there is a subgroup $K\leq G$ with $H\leq K$ and $|G\colon K|=2$ and one of the following occurs.

(a)* The group $G$ is isomorphic to the wreath product $C_{p}\wr C_{2}$.*

(b)* The group $G$ is isomorphic to a subgroup of the affine group $\mbox{\rm Aff}(1,p)$.*

(c)* We have $G=K\times\langle a\rangle$, and $K$ has odd order and is isomorphic to a subgroup of the affine group $\mbox{\rm Aff}(1,p)$.*

In Cases (b) and (c), $\langle T\cap K\rangle$ is a normal subgroup of $G$ of order $p$. The Cases (a), (b) and (c) are disjoint.

Proof. The first statement follows from Lemmas 5.11 and Proposition 5.12 (with $q$ replaced by $p$). In particular, we are in the situation of Subsection 5.2.

In the following, we resume to the notation introduced at the beginning of Subsection 5.1. Suppose that $G$ is not isomorphic to the wreath product $C_{p}\wr C_{2}$. By Proposition 5.5, we may assume that we are in the situation of 5.2.2. Corollary 5.4 implies that $H$ is not a normal subgroup of $K$. Hence $H_{1}=1$ and $C=1$ by Corollaries 5.7 and 5.9. In particular, $|K_{1}|=p$ by Corollary 5.3.

If $C_{G}(K_{1})=K_{1}$, then $G/K_{1}$ injects into the automorphism group of $K_{1}$, and thus $G$ is as in (b). Assume now that $K_{1}\lneq C_{G}(K_{1})$. As $K=HK_{1}$ by Lemma 5.1, we have $C_{H}(K_{1})\leq C$, and thus $H\cap C_{G}(K_{1})=C_{H}(K_{1})=1$. Hence $|HC_{G}(K_{1})|=|H||C_{G}(K_{1})|>|H||K_{1}|=|K|$, and thus $HC_{G}(K_{1})=G$ and $|C_{G}(K_{1})|=2|K_{1}|$. It follows that $C_{G}(K_{1})=K_{1}\times\langle a\rangle$ for some $a\in G\setminus K$ of order $2$. As $K\unlhd G$, and $\langle a\rangle=O_{2}(C_{G}(K_{1}))\unlhd G$, we have $G=K\times\langle a\rangle$. Now $K$ acts faithfully on the set of $H$-cosets in $K$, and thus $K=HK_{1}$ is isomorphic to a subgroup of the

affine group $\mbox{\rm Aff}(1,p)$. Finally, $K$ has odd order since $G/K_{1}$ is cyclic by Lemma 5.6. $\Box$

6. The rcc loops of twice prime order

Let $p$ be an odd prime. In this section we determine the number of isomorphism classes of RCC loops of order $2p$. Let $\mathcal{L}$ denote such a loop and let $(G,H,T)$ be its envelope. By numbering the elements of $\mathcal{L}$ by the integers $1,\ldots,2p$, where $1$ numbers the identity element of $\mathcal{L}$, we may and will view $G$ as a subgroup of $S_{2p}$, and $H$ as the stabilizer in $G$ of $1$. If $\mathcal{L}_{1}$ and $(G_{1},H_{1},T_{1})$ is another such configuration, then $\mathcal{L}$ and $\mathcal{L}_{1}$ are isomorphic as loops, if and only if there is an element of $S_{2p}$, conjugating $(G,H,T)$ to $(G_{1},H_{1},T_{1})$. The isomorphism types of the right multiplication groups arising in RCC loops of order $2p$ have been described in Theorem 5.13. For each of these groups $G$ we have to determine their embeddings into $S_{2p}$ up to conjugation. This will yield the possible pairs $(G,H)$ to be considered. For each of these pairs we have to determine the normalizer $N$ in $S_{2p}$ of $G$ and $H$, and then find the distinct $N$-orbits of $G$-invariant transversals $T$ for $H\backslash G$ such that $1\in T$ and $\langle T\rangle=G$. We will refer to the three different types of $G$ in Theorem 5.13(a), (b), and (c) as Case (a), (b) and (c), respectively.

We begin with some preliminary results. As usual, the largest normal $p$-subgroup of a finite group $U$ is denoted by $O_{p}(U)$, and its largest normal subgroup of odd order by $O_{2^{\prime}}(U)$.

Lemma 6.1**.**

Let $\pi_{1},\alpha\in S_{2p}$ be defined by $\pi_{1}:=(1,2,\ldots,p)$ and $\alpha:=(1,p+1)(2,p+2)\cdots(p,2p)$. Put $\pi_{2}:=\pi_{1}^{\alpha}=(p+1,p+2,\ldots,2p)$. Let $\nu_{1}\in S_{p}$ be an element of order $p-1$ such that $N_{S_{p}}(\langle\pi_{1}\rangle)=\langle\pi_{1},\nu_{1}\rangle$. Put $\nu_{2}:=\nu_{1}^{\alpha}$ and $\nu:=\nu_{1}\nu_{2}$.

(a)* Let $G:=\langle\pi_{2},\alpha\rangle$. Then $Z(G)=\langle\pi_{1}\pi_{2}\rangle$, $G^{\prime}=\langle\pi_{1}^{-1}\pi_{2}\rangle$ and $G=C_{S_{2p}}(Z(G))\cong C_{p}\wr C_{2}$. Put $N:=N_{S_{2p}}(G)$. Then $N=N_{S_{2p}}(Z(G))=\langle\nu\rangle\ltimes G$.*

(b)* Let $U\leq N$ with $O_{p}(U)=Z(G)$ and $U\not\leq G$. Then there is $n\in N$ such that $N_{S_{2p}}(U^{n})=A\times\langle\alpha\rangle$ with $A=\langle\nu\rangle\ltimes Z(G)$.*

Proof. (a) The statements about $G$ are trivially verified. From $G=C_{S_{2p}}(Z(G))$ we conclude that $G\unlhd N_{S_{2p}}(Z(G))$, and thus $N=N_{S_{2p}}(Z(G))$. Moreover, $N/G$ is isomorphic to a subgroup of $\mbox{\rm Aut}(Z(G))$, which is a cyclic group of order $p-1$. Now $\langle\nu\rangle\cap G=1$, as the elements in $G$ fixing the set $\{1,\ldots,p\}$ have order divisible by $p$. Also, $\nu$ normalizes $G$, and thus $N=\langle\nu\rangle\ltimes G$.

(b) From $O_{p}(U)=Z(G)$ we conclude that $N_{S_{2p}}(U)\leq N_{S_{2p}}(Z(G))=N$, and thus $N_{S_{2p}}(U)=N_{N}(U)$. Let $V$ denote a complement to $Z(G)$ in $U$, and let $W$ be a Hall $p^{\prime}$-group of $N$ containing $V$ (see [11, Hauptsatz VI.1.7]). Then $W$ is a complement to $O_{p}(N)$ in $N$. As $\alpha$ centralizes $\nu$, we have $\langle\nu,\alpha\rangle=\langle\nu\rangle\times\langle\alpha\rangle$, and thus $\langle\nu\rangle\times\langle\alpha\rangle$ is another complement to $O_{p}(N)$ in $N$. As all such complements are conjugate in $N$, there is $n\in N$ such that $W^{n}=\langle\nu\rangle\times\langle\alpha\rangle$ and $U^{n}=Z(G)V^{n}$. By replacing $U$ with $U^{n}$, we may assume that $V\leq W=\langle\nu\rangle\times\langle\alpha\rangle$. In particular, $W$ is abelian. It follows that $WZ(G)$ normalizes $U=VZ(G)$. As $U\not\leq G$, there is an element $\nu^{i}\alpha^{j}\in V$ such that $\nu^{i}\neq 1$. Then $[\nu^{i}\alpha^{j},\pi_{1}]\not\in Z(G)$. In particular, $U$ is not normal in $N$. As $WZ(G)$ has index $p$ in $N$, we conclude that $N_{N}(U)=WZ(G)$, which proves our claim. $\Box$

Let $n,d$ be positive integers, and let $\zeta\in S_{n}$ denote an $n$-cycle. Let us put

[TABLE]

and

[TABLE]

Thus $I_{n,d}$ is one more than the number of involutions in $C_{S_{n}}(\zeta^{d})$ and  $I_{n}$ is one more than the number of involutions in $S_{n}$. Notice that the definition of $I_{n,d}$ does not depend on the chosen $n$-cycle $\zeta$, as all $n$-cycles are conjugate in $S_{n}$.

It is not difficult to derive a formula for $I_{n,d}$, where the formula for $I_{n}$ is certainly well known. In the following result, $n\text{\rm\ mod\ }2\in\{0,1\}$ denotes the remainder of the division of $n$ by $2$.

Lemma 6.2**.**

Let $n,d,e$ and $f$ be positive integers such that $d\mid n$ and $\mbox{\rm gcd}(e,n/d)=1$. Then $I_{n,de}=I_{n,d}$ and $I_{n,f}=I_{n,\gcd(n,f)}$. Moreover, we have

[TABLE]

In particular,

[TABLE]

Proof. Let $\zeta\in S_{n}$ be an $n$-cycle. As $e$ is relatively prime to $n/d$, we have that $\zeta^{de}=(\zeta^{d})^{e}$ is the product of $d$ cycles of length $n/d$. In particular, $\zeta^{d}$ and $\zeta^{de}$ are conjugate in $S_{n}$ and thus $I_{n,de}=I_{n,d}$. Writing $f=de$ with $d=\gcd(n,f)$ and $e=f/\gcd(n,f)$, we obtain $I_{n,f}=I_{n,\gcd(n,f)}$, as $f/\gcd(n,f)$ and $n/\gcd(n,f)$ are relatively prime.

By definition, $I_{n,d}$ equals the number of elements $\tau\in C_{S_{n}}(\zeta^{d})$ with $\tau^{2}=1$. The structure of $C_{S_{n}}(\zeta^{d})$ is well known; it is a wreath product isomorphic to $C_{n/d}\wr S_{d}$, where $C_{n/d}$ denotes a cyclic group of order $n/d$. We view the elements of $C_{S_{n}}(\zeta^{d})$ as $(d+1)$-tuples $(\mu;c_{1},\ldots,c_{d})$, where each $c_{i}$ lies in one of the $d$ cycles of $\zeta^{d}$, and where $\mu\in S_{d}$ permutes the numbers $\{1,\ldots,d\}$. We have

[TABLE]

Let $\tau:=(\mu;c_{1},\ldots,c_{d})\in C_{S_{n}}(\zeta^{d})$ satisfy $\tau^{2}=1$. Then $\mu^{2}=1$ and $c_{i\mu}=c_{i}^{-1}$ for all $1\leq i\leq d$. Suppose that $\mu$ is a product of exactly $k$ transpositions for some $0\leq k\leq\lfloor d/2\rfloor$. Then $c_{j}=c_{i}^{-1}$, if $(i,j)$ is a transposition of $\mu$, and $c_{i}^{2}=1$ if $i$ is a fixed point of $\mu$. This way, a fixed $\mu$ gives rise to $(n/d)^{k}(2-(n/d\text{\rm\ mod\ }2))^{d-2k}$ elements $\tau\in C_{S_{n}}(\zeta^{d})$ with $\tau^{2}=1$. The centraliser of $\mu$ in $S_{d}$ has order $2^{k}k!(d-2k)!$, yielding our formula for $I_{n,d}$. The one for $I_{n}$ follows from this by putting $d=n$. $\Box$

Proposition 6.3**.**

There are exactly

[TABLE]

distinct isomorphism types of RCC  loops with multiplication group $G$ as in Case (a).

Proof. Let $(G,H,T)$ denote the envelope of an RCC  loop of order $2p$ with $G$ as in Case (a), i.e. $G$ is isomorphic to the wreath product $C_{p}\wr C_{2}$. In this case, $H$ is cyclic of order $p$. By numbering the right cosets of $H$ in $G$ from $1$ to $2p$, we obtain an embedding $G\rightarrow S_{2p}$, and we identify $G$ with its image in $S_{2p}$ from now on. Let $\pi_{1}$, $\alpha$ and $\pi_{2}$ be defined as in Lemma 6.1. We may choose the numbering of the cosets of $H$ in $G$ in such a way that $H=\langle\pi_{2}\rangle$ and $G=\langle\pi_{2},\alpha\rangle$. From Lemma 6.1(a) we obtain $Z(G)=\langle\pi_{1}\pi_{2}\rangle$, $G^{\prime}=\langle\pi_{1}^{-1}\pi_{2}\rangle$ and $G=C_{S_{2p}}(Z(G))$. Also, $N:=N_{S_{2p}}(G)$ equals $\langle\nu\rangle\ltimes G$ with $\nu$ as in Lemma 6.1. Observe that $N$ normalises $H$.

Let $\mathcal{T}$ denote the set of $G$-invariant transversals for $H\backslash G$ containing $1$. Put $K:=\langle\pi_{1},\pi_{2}\rangle=O_{p}(G)$ and let  $\mathcal{T}_{1}$ denote the set of $G$-invariant transversals for $H\backslash K$ containing $1$. Let $t\in G\setminus K$. Then $|C_{G}(t)|=2p$ and thus $t$ lies in a conjugacy class of length $p$. As every conjugacy class of $G$ lies in some coset of $G^{\prime}$, we find that $G^{\prime}t$ is the conjugacy class of $G$ containing $t$. Hence if $T\in\mathcal{T}$, we have $T=(K\cap T)\cup G^{\prime}t$ for some $t\in G\setminus K$, and $K\cap T\in\mathcal{T}_{1}$. Conversely, if $T_{1}\in\mathcal{T}_{1}$, and if $t$ is any element of $G\setminus K$, then $T_{1}\cup G^{\prime}t\in\mathcal{T}$.

As $K=H\times H^{\alpha}$, we have $K=\cup_{0\leq j\leq p-1}H\pi_{1}^{j}$. A transversal for $H\backslash K$ contains exactly one element of each coset $H\pi_{1}^{j}$, $0\leq j\leq p-1$. As we insist that our transversals contain the trivial element, a transversal $T_{1}$ for $H\backslash K$ determines a map $\tau:\{1,\ldots,p-1\}\rightarrow\{0,1,\ldots,p-1\}$ such that

[TABLE]

Conjugating the element $\pi_{2}^{j\tau}\pi_{1}^{j}\in T_{1}\setminus\{1\}$ by $\alpha$, we obtain $\pi_{1}^{j\tau}\pi_{2}^{j}=\pi_{2}^{j}\pi_{1}^{j\tau}$. If $T_{1}$ is $G$-invariant, we must have, firstly, that $j\tau\neq 0$ and, secondly, that $\pi_{2}^{j}\pi_{1}^{j\tau}\in T_{1}\setminus\{1\}$ for all $1\leq j\leq p-1$. The latter condition implies that $\pi_{2}^{j\tau^{2}}\pi_{1}^{j\tau}=\pi_{2}^{j}\pi_{1}^{j\tau}$ for all $1\leq j\leq p-1$, and thus $\tau^{2}=1$. In particular, $\tau$ is a permutation of order at most $2$ of the set $\{1,\ldots,p-1\}$. Conversely, if $\tau$ is a permutation of the latter set with $\tau^{2}=1$, then $T_{1}$ defined by (3) lies in $\mathcal{T}_{1}$. In particular, $|\mathcal{T}_{1}|=I_{p-1}$. As the number of conjugacy classes of $G$ in $G\setminus K$ equals $p$, we conclude from

[TABLE]

that

[TABLE]

We next determine the number of $N$-orbits on $\mathcal{T}$. This is the same as the number of $\langle\nu\rangle$-orbits on $\mathcal{T}$. To compute this number, put

[TABLE]

Observe that $\mathcal{T}_{1}$ is $\langle\nu\rangle$-invariant, as $\nu$ normalises $H$. In addition, $\nu$ centralises $\alpha$, and thus $\mathcal{T}^{\prime}$ is $\langle\nu\rangle$-invariant as well. As $Z(G)$ is a set of representatives for the set of right cosets of $G^{\prime}$ in $K$, every conjugacy class of $G$ contained in $G\setminus K$ is of the form $G^{\prime}z\alpha$ for some $z\in Z(G)$. As $\langle\nu\rangle$ acts transitively on $Z(G)\setminus\{1\}$, we conclude that every orbit of $\langle\nu\rangle$ on $\mathcal{T}\setminus\mathcal{T}^{\prime}$ has length $p-1$, and thus there are exactly $I_{p-1}$ such orbits. We are thus left with the determination of the number of $\langle\nu\rangle$-orbits on $\mathcal{T}^{\prime}$, which is the same as the number of $\langle\nu\rangle$-orbits on $\mathcal{T}_{1}$. By the Burnside-Cauchy-Frobenius lemma, the latter number equals

[TABLE]

where $\chi_{d}$ is the number of fixed points of $\nu^{d}$ on $\mathcal{T}_{1}$. The action of $\langle\nu\rangle$ on $K$ determines a $(p-1)$-cycle $\zeta$ on the set $\{1,\ldots,p-1\}$ such that $\nu^{-1}\pi_{i}^{j}\nu=\pi_{i}^{j\zeta}$ for $i=1,2$ and all $1\leq j\leq p-1$. Now let $T_{1}\in\mathcal{T}_{1}$ be given by (3) with respect to $\tau\in S_{p-1}$ with $\tau^{2}=1$. Then $T_{1}$ is fixed by $\nu^{d}$, if and only if $\zeta^{d}$ centralises $\tau$. Thus $\chi_{d}=I_{p-1,d}$.

It remains to determine those $N$-orbits on $\mathcal{T}$ containing transversals that generate $G$. Let $T\in\mathcal{T}$ such that $\langle T\rangle\neq G$. Then $\langle T\rangle$ is a normal subgroup of $G$ of index $p$. Thus $G^{\prime}\leq\langle T\rangle$ and $T=G^{\prime}\cup G^{\prime}z\alpha$ for some $z\in Z(G)$. Since $\langle T\rangle\neq G$, we must have $z=1$, i.e. $\langle T\rangle=T=G^{\prime}\cup G^{\prime}\alpha$. As this is $N$-invariant, our result follows. $\Box$

Proposition 6.4**.**

Write $p-1=2^{n}r$ with positive integers $n$ and $r$ and with $r$ odd. Then there are exactly $p-r-1$ distinct isomorphism types of RCC  loops with multiplication group $G$ as in Case (b), and there are exactly $r$ isomorphism types of RCC  loops with multiplication group $G$ as in Case (c).

Proof. Let $(G,H,T)$ denote the envelope of an RCC  loop of order $2p$ with $G$ as in Case (b) or (c). If $H=1$, then $T=G$ is a group of order $2p$, which is non-abelian in Case (b), and cyclic in Case (c). In each case, we obtain a unique isomorphism class of RCC  loops.

Thus let us assume that $H\neq 1$ in the following. As in the proof of Proposition 6.3, we identify $G$ with its image in $S_{2p}$ through an embedding obtained by numbering the right cosets of $H$ in $G$ from $1$ to $2p$. Put $P:=O_{p}(G)$, the unique Sylow $p$-subgroup of $G$. Let $\pi_{1}$, $\alpha$ and $\pi_{2}$ be defined as in Lemma 6.1. We may choose the numbering of the cosets of $H$ in $G$ in such a way that $P=\langle\pi_{1}\pi_{2}\rangle$, and that $Z(G)=\langle\alpha\rangle$ in Case (c). Put $N:=N_{S_{2p}}(G)$. We now apply Lemma 6.1(b) with our $G$ taking the role of $U$ of that lemma. As $H\neq 1$, we have $G\not\leq\langle\pi_{2},\alpha\rangle$, and thus, replacing $G$ by a suitable conjugate within $N_{S_{2p}}(P)$, we find that $N=A\times\langle\alpha\rangle$, with $A\cong\mbox{\rm Aff}(1,p)$. We have $A=L\ltimes P$, with $L$ cyclic of order $p-1$.

Assume that $G$ is as in Case (b). Then $G\cap L$ is a complement to $P$ in $G$. As all such complements are conjugate in $G$ by Schur’s theorem (see [11, Satz I.17.5]), we may assume that $H\leq L$. In particular, $G\leq A$, and $H$ is $N$-invariant. Let $T$ be a $G$-invariant transversal for $H\backslash G$. Then $P\subseteq T$ by Theorem 5.13. Let $\tau\in T\setminus P$. Then $|C_{G}(\tau)|=2|H|$ and thus $T\setminus P$ consists of the $G$-conjugacy class containing $\tau$. If, moreover, $G=\langle P,\tau\rangle$, we have $2p|H|=|G|=p|\tau|$ and $\tau$ has even order larger than $2$. Every element $\tau^{\prime}$ which is conjugate to $\tau$ in $A$ gives rise to an isomorphic loop with multiplication group $\langle P,\tau^{\prime}\rangle$, as in Case (b). It follows that the isomorphism types of RCC  loops with a multiplication group as in Case (b) equals the number of $A$-conjugacy classes of elements of $A$ of even order larger than $2$. As $A$ has $(p-r-2)p$ such elements, the result follows.

Assuming now that $G$ is as in Case (c), we have $G=K\times\langle\alpha\rangle$, with $K=O_{2^{\prime}}(G)$, and thus $K\unlhd N$. In turn, $K\leq A$ as every Sylow subgroup of $K$ is conjugate to a subgroup of $A$. Again, $H$ is $N$-invariant. Let $T$ be a $G$-invariant transversal for $H\backslash G$. As in Case (b), we have $T=P\cup C$, where $C$ is a $G$-conjugacy class of an element $\tau\in G\setminus K$. Every element $\tau^{\prime}$ in the $A$-conjugacy class containing $\tau$ gives rise to an isomorphic loop with multiplication group $\langle P,\tau^{\prime}\rangle$. Now $\tau=\tau_{1}\alpha$ for some $\tau_{1}\in K$. It follows that the isomorphism types of RCC  loops with a multiplication group as in Case (c) equals the number of $A$-conjugacy classes of elements of $A$ of odd order different from $p$. All these elements lie in the unique subgroup of $A$ of order $pr$, and thus there are $(r-1)p$ non-trivial such elements. As the trivial element yields a group, the result follows. $\Box$

We summarise our results in the following theorem.

Theorem 6.5**.**

Let $p$ be a prime. Then the number of isomorphism types of RCC  loops of order $2p$ (including groups) equals

[TABLE]

Proof. Every loop of order $4$ is a group. As $I_{1,1}=I_{1}=1$, formula (4) holds for $p=2$. For odd $p$ it follows from Propositions 6.3 and 6.4, as the cases in Theorem 5.13 are disjoint. $\Box$

The table below contains the numbers obtained by evaluating formula (4) for small values of $p$. These numbers have also been obtained for $p\leq 13$ in the PhD-thesis of the first author [1] by different methods.

[TABLE]

One of the referees has kindly pointed out that formula (4) evaluates to an integer, even if $p$ is not a prime (and larger than $1$). This follows from the fact that for general positive integers $n,d$, the number $I_{n,d}$ equals the number of fixed points of the element $\zeta^{d}$ on the set $\{\tau\in S_{n}\mid\tau^{2}=1\}$, where the $n$-cycle $\zeta$ acts by conjugation. Thus, by the Burnside-Cauchy-Frobenius lemma, the number of orbits of $\langle\zeta\rangle$ on $\{\tau\in S_{n}\mid\tau^{2}=1\}$ equals $1/n\sum_{d=1}^{n}I_{n,d}$, so that this number is an integer.

7. A series of examples

According to Theorem 5.13, the right multiplication group of an RCC  loop of order $2p$, where $p$ is an odd prime, is soluble. This is no longer the case for right multiplication groups of RCC  loops of order $pq$, where $p$ and $q$ are distinct primes. An example is given in [1, Table B.7] of an RCC  loop of order $15$ with right multiplication group isomorphic to $\mbox{\rm GL}(2,4)$. This fits into an infinite series of examples.

Proposition 7.1**.**

Let $q$ be a power of a prime with $q>2$. Then there is an RCC  loop of of order $q^{2}-1$ and right multiplication group isomorphic to $\mbox{\rm GL}(2,q)$.

Proof. Let $G:=\mbox{\rm GL}(2,q)$, acting from the right on $\mathbb{F}_{q}^{1\times 2}$, and let

[TABLE]

Let $Z:=Z(G)$ denote the set of scalar matrices in $G$ and let $C$ be a $G$-conjugacy class of elements of order $q^{2}-1$, i.e. the elements of $C$ are Singer cycles. Then $|C_{G}(t)|=q^{2}-1$ for all $t\in C$; in particular $|C|=q(q-1)$. Now put

[TABLE]

We claim that $T$ is a $G$-invariant transversal for $H\backslash G$. Clearly, $T$ is $G$-invariant and $|T|=q^{2}-1=|G\colon H|$. Let $t,t^{\prime}\in C$. We have to show that $t^{\prime}t^{-1}\in H$ if and only if $t=t^{\prime}$. To see this, first observe that $|C_{G}(t)||H|=|G|$ and that $C_{G}(t)\cap H=1$, as $|C_{G}(t)\cap H|$ divides $\mbox{\rm gcd}(|C_{G}(t)|,|H|)=q-1$, and the only elements in $C_{G}(t)$ of order dividing $q-1$ are the elements of $Z$. We conclude that $G=C_{G}(t)H$. It follows that there is $h\in H$ with $t^{\prime}=h^{-1}th$. Put $h^{\prime}:=t^{\prime}t^{-1}=h^{-1}tht^{-1}$. Thus

[TABLE]

Now assume that $h^{\prime}\in H$. As $\mbox{\rm det}(h^{\prime})=\mbox{\rm det}(t^{\prime}t^{-1})=1$, we have

[TABLE]

for some $\gamma\in\mathbb{F}_{q}$. Let

[TABLE]

with $\alpha\in\mathbb{F}_{q}^{*}$ and $\beta\in\mathbb{F}_{q}$, and let

[TABLE]

with $a,b,c,d\in\mathbb{F}_{q}$. Then

[TABLE]

and

[TABLE]

where we do not need to specify the entries in the first columns of $t^{-1}hh^{\prime}$ respectively $ht^{-1}$. As $t$ acts irreducibly on the natural vector space $\mathbb{F}_{q}^{1\times 2}$ for $G$, we conclude that $b\neq 0$. Equation (5) yields $\alpha=1$ and $\beta=0$, i.e. $h=1$, and thus $t=t^{\prime}$. If $z,z^{\prime}\in Z$, then $z^{\prime}z^{-1}\in H$ if and only if $z=z^{\prime}$. Now let $z\in Z$ and $t\in C$ and assume that $tz^{-1}\in H$. Then $t\in HZ$; but $|HZ|=q(q-1)^{2}$, whereas $|t|=q^{2}-1\nmid q(q-1)^{2}$, a contradiction.

Finally, it is easy to check that $\langle T\rangle=G$, by a direct computation if $q=3$, and by using the fact that $G/Z$ is almost simple if $q\neq 3$. This completes the proof. $\Box$

Acknowledgements

We thank Alice Niemeyer for her support and her interest in this work. We also thank Barbara Baumeister for introducing us to the fascinating topic of invariant transversals. Finally, we are very much indebted to the anonymous referees for several suggestions which improved the exposition of this paper, and also for drawing our attention to related work.