The endomorphism rings of permutation modules of $\frac{3}{2}$-transitive permutation groups

The concept of $\frac{3}{2}$-transitive permutation group goes back to research of Wielandt [10], related with constructing of many combination objects; the defining property of such a group is a non-regular transitive group such that all the nontrivial orbits of a point stabilizer have equal size. More steps towards the classification of the primitive $\frac{3}{2}$-transitive groups were taken in [1] and [4]. It was proved that primitive $\frac{3}{2}$-transitive groups are either affine or almost simple, and the almost simple examples were determined. Further more, if we do not require primitivity of a $\frac{3}{2}$-transitive permutation groups, a recent classification of $\frac{3}{2}$-transitive permutation group can be found in [6].

Given a field $k$ and a natural number $n$, if we assume that $G$ is a permutation subgroup of the symmetric group on $\Omega$ with $n$ letters, then $G$ can be naturally viewed as a subgroup of permutation matrices in the general linear group $\operatorname{GL}(n,k)$. It is well known that the centralizer ring which is defined to be the centralizer of $G$ in the full matrix ring $M_{n}(k)$ is isomorphic to the endomorphism ring of the permutation $kG$-module $k\Omega$. An important research on the centralizer ring theory of Wielandt [10] is to study the centralizer ring corresponding to the permutation group of $kp$ degree for a prime $p$. However, when $G\leq\rm Sym(\Omega)$ is a $\frac{3}{2}$-transitive permutation group, some classical results on permutation groups can be applied to the centralizer ring (or the endomorphism ring). In this paper, we study the endomorphism ring of the natural permutation modules of a $\frac{3}{2}$-transitive groups. The following is our main theorem:

Theorem 1.1 is proved in Section 3. See Subsection 2.2 for related definitions.

Throughout this paper, groups and modules are always finite groups and right modules. For a finite set $X$, a group $G$ and a subgroup $H\leq G$, we always use $|X|$, $|G|$ and $|G:H|$ to denote the cardinality of $X$, the order of $G$ and the index of $H$ in $G$ respectively. Let $k$ be a field, then we always use $k_{H}$ and $k^{G}_{H}$ to denote the trivial $kH$-module and the induced module $k_{H}\otimes_{kH}kG$ respectively. For two $kG$-modules $M$ and $N$, we use $M|N$ to mean that $M$ is isomorphic to a direct summand of $N$ as a $kG$-module. If moreover $G$ is a permutation group on some set $\Omega$, then the $k$-linear span of elements of $\Omega$ which we denote by $k\Omega$ is a natural $kG$-module via the action

$$(\sum_{\alpha\in\Omega}a_{\alpha}\alpha)\cdot g:=\sum_{\alpha\in\Omega}a_{\alpha}(\alpha\cdot g),\forall\ g\in G.$$

If we denote $\underline{H}=\sum_{h\in H}h$ and $\underline{G}=\sum_{g\in G}g$, then it can be easily checked that $k\Omega\cong\underline{H}kG$ as $kG$-module. If ${\rm char}~{}k\nmid|G:H|$, then

$$\underline{H}kG=\underline{G}kG\oplus(\underline{H}-|G:H|^{-1}\underline{G})kG,$$

which implies that $k_{G}\cong\underline{G}kG$ is isomorphic to a direct summand of $k\Omega$ in this situation.

$\mathbf{Notation.}$

Throughout this paper, $\Omega$ is always a finite set. $A_{|\Omega|}$ and $S_{|\Omega|}$ denote the alternating group and symmetric group on $\Omega$, respectively.

The diagonal of the Cartesian product $\Omega\times\Omega$ is always denoted by $1_{\Omega};$ for $\Delta\subseteq\Omega$, we set $1_{\Delta}:=\{(\alpha,\alpha):\alpha\in\Delta\}$.

For a subset $r\subseteq\Omega\times\Omega$ and any $\gamma\in\Omega$, we set $r^{*}=\{(\beta,\alpha):(\alpha,\beta)\in r\}$, $\gamma r=\{\delta\in\Omega:(\gamma,\delta)\in r\}$ and $\Omega(r):=\{\alpha\in\Omega:\alpha s\neq\emptyset\}$.

Note that a permutation group $G\leq\rm{Sym}(\Omega)$ is said to be half-transitive if all orbits of $G$ are of the same cardinality. Moreover, if $G$ acts transitively on $\Omega$, then the cardinalities of the orbits of $G_{\alpha}$ for any given point $\alpha\in\Omega$ are called the $subdegrees$ of $G$ and $G$ is called $primitive$ if a point stabilizer in $G$ is a maximal subgroup. Before we proceed on to prove the main theorem, we first prove the following lemma.

Proof. Note that $M_{n}(k)^{G}={\rm End}_{kG}(k\Omega)$ when regarding $G$ as permutation matrices, thus we can use matrices to describe ${\rm End}_{kG}(k\Omega)$. We take $\Omega$ as the set $\{1,2,\cdots,n\}$ for convenience. Now we denote the orbits under the action of $G$ on $\Omega\times\Omega$ by

$$S=\{s_{1},\cdots,s_{t},s_{t+1},\cdots,s_{m}\},$$

where $\{s_{1},\cdots,s_{t}\}$ are precisely the reflexive orbits. For any orbit $s_{l}\in S$, we define a matrix as follows:

$$(A_{s_{l}})_{ij}=\left\{\begin{array}[]{rcl}1&&{(i,j)\in s_{l}}\\ 0&&{(i,j)\notin s_{l}}\\ \end{array}\right.$$

called the adjacency matrix of $s_{l}$. It is known from [2, Theorem 2.3.5] that

$$M_{n}(k)^{G}=\mathop{\oplus}\limits_{i=1}^{m}kA_{s_{i}}$$

as vector spaces. We fix $\mathcal{B}=\{A_{s_{i}},1\leq i\leq m\}$ as a basis of $M_{n}(k)^{G}$. For $s_{l}\in S$, we conclude from [2, Section 1.1] that $\Omega(s_{l})$ is an orbit of the action of $G$ on $\Omega$. We now define a map

$$\varphi:M_{n}(k)^{G}\times M_{n}(k)^{G}\rightarrow k,~{}(A_{s_{l}},A_{s_{l^{\prime}}})\mapsto C^{1_{\Omega(s_{l})}}_{s_{l}s_{l^{\prime}}}1_{k}$$

for all $A_{s_{l}},A_{s_{l^{\prime}}}\in\mathcal{B}$, where $1_{k}$ is the identity of $k$. From [2, Chapter 1], $\varphi$ is clearly bilinear, while $\varphi(A_{s_{l}},A_{s_{l^{\prime}}}A_{s_{l^{\prime\prime}}})$ and $\varphi(A_{s_{l}}A_{s_{l^{\prime}}},A_{s_{l^{\prime\prime}}})$ are just the coefficient of $A_{1_{\Omega(s_{l})}}$ in the expression of $A_{s_{l}}A_{s_{l^{\prime}}}A_{s_{l^{\prime\prime}}}$ with respect to the basis $\mathcal{B}$, they must be equal and therefore $\varphi$ is associative. Then we only need to show that $\varphi$ is symmetric and non-degenerate. To show it is symmetric, it suffices to show that for all $s_{l},s_{l^{\prime}}\in S$, the following equality holds:

$$C^{1_{\Omega(s_{l})}}_{s_{l}s_{l^{\prime}}}1_{k}=C^{1_{\Omega(s_{l^{\prime}})}}_{s_{l^{\prime}}s_{l}}1_{k}.$$

Since $G$ is half-transitive on $\Omega$, ie., all orbits of $G$ on $\Omega$ have the same size, but from [2, Chapter 1], we have

$$C^{1_{\Omega(s_{l^{\prime}})}}_{s_{l^{\prime}}s_{l}}|\Omega(s_{l^{\prime}})|1_{k}=C^{1_{\Omega(s_{l})}}_{s_{l}s_{l^{\prime}}}|\Omega(s_{l})|1_{k},$$

thus we obtain the required equation.

Now it remains to show that $\varphi$ is non-degenerate, denote

$$\mathrm{Rad}\varphi:=\{x\in M_{n}(k)^{G}:\varphi(x,y)=0~{}\text{for~{}all}~{}y\in M_{n}(k)^{G}\},$$

we have to show that Rad$\varphi=0$. Now choose arbitrary $x\in$ Rad$\varphi$, we can write $x=\sum^{m}_{i=1}a_{s_{i}}A_{s_{i}}$, where $a_{s_{i}}\in k$ for all $1\leq i\leq m$. For any $s_{l}\in S$, it is clear that

$$C^{1_{\Omega(s_{l})}}_{s_{l}s_{l^{\prime}}}1_{k}\neq 0\Leftrightarrow s_{l}=s^{*}_{l^{\prime}}.$$

Choose an element $(a,b)\in s_{l}$ and denote $G_{a}$ the stabilizer for $a$ in $G$, then we deduce that

$$C^{1_{\Omega(s_{l})}}_{s_{l}s_{l^{\prime}}}1_{k}=|b^{G_{a}}|1_{k}$$

which is an orbit size under the action of $G_{a}$ on $\Omega$. Therefore we have the following equations

$$\varphi(x,s^{*}_{l})=a_{s_{l}}C^{1_{\Omega(s_{l})}}_{s_{l}s^{*}_{l}}=0,$$

which implies that $a_{s_{l}}=0$ . Hence we derive that $a_{s_{i}}=0$ when $s^{*}_{i}$ runs over the orbits of $G$ on $\Omega\times\Omega$, therefore $a=0$, as required. $\square$

Proof of Theorem 1.1.  We divide our proof into four parts according to the four situations in the classification of $\frac{3}{2}$-transitive permutation groups.

(1) $G$ is 2-transitive.

Denote $|\Omega|=n$, then

$${\rm End}_{kG}(k\Omega)\cong k[X]/(X^{2})~{}\text{or}~{}{\rm End}_{kG}(k\Omega)\cong k\times k$$

depending whether ${\rm char}~{}k$ divides $n$ or not, ${\rm End}_{kG}(k\Omega)$ is always a symmetric algebra in both cases.

(2) $G$ is a Frobenius group.

Denote $|\Omega|=n$ and ${\rm char}~{}k=p$, let $\alpha\in\Omega$ and $H=G_{\alpha}$ be the point stabilizer of $\alpha$ in $G$. We may assume that $p\mid|G|$, otherwise ${\rm End}_{kG}(k\Omega)$ would be semisimple by Maschke’s theorem. That $G$ is a Frobenius group implies that $|H|$ and $n$ are coprime, thus either $p\mid|H|$ or $p\mid n$.

We first suppose that $p\mid|H|$. Let $P$ be a Sylow $p$-subgroup of $H$, then $N_{G}(P)\leq H$. Now $k\Omega\cong k^{G}_{H}$ as $kG$-module and $P$ is a vertex of $k_{H}$, hence the Scott module $k_{G}$ is the Green correspondence of $k_{H}$ and the vertice of any indecomposable summands of $k^{G}_{H}$ except $k_{G}$ is of the form $P\cap P^{g}$ for some $g\in G-H$. But that $G$ is a Frobenius group yields that $H$ has trivial intersection property, therefore the vertices of all indecomposable summands of $k^{G}_{H}$ except $k_{G}$ are the identity and hence those summands are projective. Whence we can write $k\Omega=k_{G}\oplus M$ for some projective $kG$-module and

$${\rm End}_{kG}(k\Omega)\cong k\times{\rm End}_{kG}(P).$$

It is well-known that $kG$ is a symmetric algebra and thus so is ${\rm End}_{kG}(P)$.

If $p\mid n$, then $|H|$ is invertible in $k$ and we set

$$e:=\frac{1}{|H|}\sum_{h\in H}h.$$

Then direct computation shows that $k\Omega\cong ekG$ as $kG$-module, and then

$${\rm End}_{kG}(k\Omega)\cong ekGe$$

is a symmetric algebra.

In fact, we have the following more general result using the concept of Schur rings:

Proof. By Lemma 3.1, it suffices to assume that ${\rm char}\ k$ divides some subdegrees of $G$. Fixed some $\alpha\in\Omega$, for any $t\in T,g\in G$, we define $t^{g}$ to be the unique element in $T$ such that $\alpha\cdot t^{g}=\alpha\cdot(tg)$. It can be directly checked that this define a group action of $G$ on $T$ which is permutation isomorphic to the group action of $G$ on $\Omega$. As a consequence of normality of $T$, we have $x^{h}=h^{-1}xh$ for all $x\in T,h\in H$. Denote $O_{1},\cdots,O_{s}$ by the orbits under the action of $H$ on $T$. Then by Lemma 2.1,

$$\mathcal{A}:={\rm span}_{k}\{\underline{O_{i}}:1\leq i\leq s\}$$

a Schur ring $\mathcal{A}$ over $T$. Invoking Schur’s method, we know that ${\rm End}_{kG}(k\Omega)$ is isomorphic to $\mathcal{A}$. We have ${\rm char}\ k\nmid|T|$ by assumption, which yields that $kT$ is a semisimple algebra by Maschke’s theorem. Let $t\in T$ be the element such that $\alpha\cdot t=\beta$ and denote $O_{t}$ by the orbit containing $t$, then

$$\sum_{h\in H}t^{h}=|H_{\beta}|\underline{O_{t}}$$

is nonzero in $kT$, where $H_{\beta}$ is the stabilizer of $\beta$ in $H$. Note that the conjugation of $H$ on $T$ can be extended to an automorphism of the group algebra $kT$ just via the natural way. Invoking [7, Corollary 5.4], we deduce that

$$(kT)^{H}:=\{x\in kT:x^{h}=x~{}\text{for~{}all}~{}h\in H\}$$

is also a semisimple algebra. But $(kT)^{H}$ is exactly this Schur ring $\mathcal{A}$, we are done. $\square$

(3) $G$ is an affine primitive permutation group

That is, $G={\rm T}(V)H\leq\rm AGL(V)$ for some vector space $V$, where $H\leq\rm GL(V)$ is a $\frac{1}{2}$-transitive irreducible linear group, ${\rm T}(V)$ is the group of translations and $V$ is of dimension $d$ over a finite field $\mathbb{F}_{p}$ for a prime $p$. We denote $T$ by the normal regular subgroup of $G$. This case will follow from the following easy but useful lemma.

Proof. By Lemma 3.1, it suffices to consider the situation when the characteristic of $k$ divides some nontrivial subdegree of $G$. We can similarly define an action of $G$ on $T$ as that in the proof of Theorem 3.1. Denote $H$ for some point stabilizer and $O_{1},\cdots,O_{s}$ the orbits under the action of $H$ on $T$. Therefore we know that ${\rm End}_{kG}(k\Omega)$ is isomorphic to a Schur ring $\mathcal{A}$ over $T$, where

$$\mathcal{A}:={\rm span}_{k}\{\underline{O_{i}}:1\leq i\leq s\}.$$

Since ${\rm char}\ k$ does not divide the order of $T$, we deduce that $kT$ is a semisimple algebra thanks to Maschke’s theorem. As $kT$ is abelian, it has no nonzero nilpotent elements. Hence $\mathcal{A}$ as a subalgebra of $kT$ also contains no nonzero nilpotent elements, forcing $\mathcal{A}$ to be semisimple, we are done. $\square$

Note that if we omit the condition that any subdegree of $G$ is coprime to the order of $T$ in Lemma 3.2, the conclusion may be false, we will give an example in the following.

(4) $G$ is almost simple

For this case, we need the following two lemmas.

Proof. First, we will show that for a given field $k$ and any permutation group $G$ on some set $\Omega$ with $|\Omega|=n$, the endomorphism algebra of $k\Omega$ is isomorphic to the adjacent algebra of the so-called coherent configuration. Since the natural permutation module $k\Omega$ is in some sense equivalent to the permutation representation

$$\rho:kG\rightarrow M_{n}(k),$$

and to compute the endomorphism ring of $k\Omega$ is to calculate the centralizer ring for permutation matrices corresponding to $G$ in $M_{n}(k)$, this is

$${\rm End}_{kG}(k\Omega)=M_{n}(k)^{G}.$$

Now by hypothesis, we have $\rho(G_{1})\leq\rho(G_{2})$, which implies that

$${\rm End}_{kG_{2}}(k\Omega)=M_{n}(k)^{G_{2}}\subseteq M_{n}(k)^{G_{1}}={\rm End}_{kG_{1}}(k\Omega).$$

By comparing their rank, we get the equality, as required. $\square$

Proof. Write $n$ for $|\Omega|$. All the nontrivial subdegrees of $G$ are the same since $G$ is $\frac{3}{2}$-transitive. If $N_{G}(P)\nleqslant H$, then

$$P\leq H\cap H^{x}$$

for some $x\in N_{G}(P)-H$ and hence $p\nmid|H:H\cap H^{x}|$. But $|H:H\cap H^{x}|$ equals to some subdegree, a contradiction, as required. $\square$

Now we come back to our analysis. In this case, then by [6, Corollary 2.3], we can obtain the following two cases:

(a) $n=21$, $G=A_{7}$ or $S_{7}$ acting on the set of pairs in $\{1,\cdots,7\}$,

(b) $n=\frac{1}{2}q(q-1)$ where $q=2^{m}\geq 8$, and either $G=\operatorname{PSL}(2,q)$, or $G=$ $\mathrm{PGL}(2,\mathrm{q})$ with $m$ prime.

For case (a), it is well-known that the ranks of both groups are $3$ and the subdegrees for them are $\{1,10,10\}$. We may assume that ${\rm char}~{}k=2$ or $5$ by lemma 3.1. Under the assumption,

$$k^{G}_{H}=k_{G}\oplus M$$

for some $kG$-module $M$ and

$${\rm End}_{kG}(k\Omega)\simeq k\times{\rm End}_{kG}(M),$$

which implies that ${\rm End}_{kG}(M)$ is 2-dimensional and hence must be a symmetric algebra, thus the same is true for ${\rm End}_{kG}(k\Omega)$.

For case (b), it suffices to analyse $G={\rm PSL}(2,q)$ by Lemma 3.3 while the stabilizer $H$ is equal to $D_{2(q+1)}$. Since it suffices to consider the case when ${\rm char}~{}k$ divides the subdegree $q+1$ of $G$, in particular, ${\rm char}~{}k$ is an odd prime $p$. Now

$$|H\cap H^{x}|=2$$

for any $x\in G-H$ implies that any Sylow $p$-subgroup $P$ of $H$ has the property that $P\cap P^{x}=1$ for all $x\in G-H$. Since Lemma 3.4 guarantees that Green’s correspondence theorem can be applied, we are done by the same analysis as in situation 2. $\square$

$\mathbf{Remark}$: Note that by O’Nan-Scott theorem, a primitive permutation group $G$ on $\Omega$ with degree a prime power is either almost simple, or of affine type or of product type. If we suppose that $G$ is almost simple, then by [5, Corollary 2], we know that ${\rm soc}(G)$ which is defined to be the subgroup generated by all minimal normal subgroups of $G$, must act $2$-transitive on $\Omega$ except for the case ${\rm soc}(G)\cong\rm PSU(4,2)$ and $|\Omega|=27$. For the exceptional case, the subdegree for ${\rm soc}(G)$ are $\{1,10,16\}$, but it is not hard to check that ${\rm End}_{k{{\rm soc}(G)}}(k\Omega)$ is always a symmetric algebra for arbitrary field $k$. Invoking Lemma 3.3, we deduce that ${\rm End}_{kG}(k\Omega)$ is always a symmetric algebra when $G$ is an almost simple primitive permutation group with prime power degree.

To end up, we mention the following result as a consequence of the methods used in this paper. Recall that a block for a transitive permutation group $G$ on $\Omega$ is a subset $\Delta\subset\Omega$ such that either $\Delta^{g}=\Delta$ or $\Delta^{g}\cap\Delta=\emptyset$ for all $g\in G$.

Proof. First we suppose that $q<p$. Then it suffices to consider the case $e=2$ by Maschke’s theorem. If $G$ is not primitive, then by [11, Theorem 16.1], $G$ contains an intransitive normal subgroup $N$ inducing $q$ blocks such that $G/N$ acts faithfully on these blocks. In particular, the assumption $q<p$ yields that $p\nmid|G/N|$ . Denote by $H$ a point stabilizer for some $\alpha\in\Omega$. Write $K=N\cap H$ and denote

$$C={\rm Core}_{N}(K):=\cap_{n\in N}K^{n}.$$

Since $N/C$ acts faithfully on the block containing $\alpha$, we deduce that $p\nmid|N/C|$ and hence $p\nmid|H:C|$. Thus $k_{H}|k^{H}_{C}$, which means that $k^{G}_{H}$ is a direct summand of $k^{G}_{C}=(k^{H}_{C})^{G}_{H}$. That $p\nmid|N/C||G/N|$ implies that $k^{G}_{C}=(k^{N}_{C})^{G}_{N}$ is a semisimple $kG$-module, forcing $k^{G}_{H}$ to be a semisimple module too. Therefore

$${\rm End}_{kG}(k\Omega)\cong{\rm End}_{kG}(k^{G}_{H})$$

is a semisimple algebra, we are done in this case. Thus we may assume that $G$ is primitive. If $G$ is $2$-transitive, then we are in Situation 1 and we are done. Now we are left with the case $G$ is uniprimitive, ie., a primitive group which is not $2$-transitive. By [11, Theorem 16.3], a Sylow $q$-subgroup of $G$ is of order $q^{2}$ which is clearly an abelian regular subgroup. Invoking Lemma 3.2, we obtain the conclusion.

Next we suppose that $q>p$. Since a Sylow $q$-subgroup of $G$ is not normal, the result in [3, Theorem 3] implies that either $G$ is $2$-transitive on $\Omega$ or $G$ has a regular subgroup of order $q^{e}$. If $G$ is $2$-transitive, then analysis in Situation 1 yields the conclusion. If $G$ has a regular subgroup of order $q^{e}$, then the same reasoning as in the proof of Lemma 3.2 can be applied, as required. $\square$