Hopf-Galois structures on a Galois $S_n$-extension

TL;DR

This paper determines the exact number of Hopf-Galois structures on Galois extensions with symmetric group Galois groups, providing precise enumeration for these algebraic structures.

Contribution

It offers a complete enumeration of Hopf-Galois structures specifically for Galois extensions with symmetric group Galois groups, a case not fully understood before.

Findings

Exact count of Hopf-Galois structures for Galois $S_n$-extensions

Provides formulas or methods for enumeration

Enhances understanding of algebraic structures in Galois theory

Abstract

In this paper, we shall determine the exact number of Hopf-Galois structures on a Galois $S_n$-extension, where $S_n$ denotes the symmetric group on $n$ letters.

Equations

$${\xi }_{\mathcal{N}}:\mathcal{N}⟶G;{\xi }_{\mathcal{N}}(\eta )=\eta (1)$$

$$\{\begin{array}{l}\lambda :G⟶\mathrm{Perm}(G);\lambda (\sigma )=(x\mapsto \sigma x),\\ \rho :G⟶\mathrm{Perm}(G);\rho (\sigma )=(x\mapsto x{\sigma }^{-1}),\end{array}$$

$$\mathcal{E}(G)=\{\mbox{regular subgroups of $\mathrm{Perm}(G)$ normalized by $\lambda(G)$}\}.$$

$$\mathcal{E}(G,N)=\left\{\begin{array}[]{c}\mbox{regular subgroups of $\mathrm{Perm}(G)$ which are}\\ \mbox{isomorphic to $N$ and normalized by $\lambda(G)$}\end{array}\right\},$$

$$\#\mathcal{E}(G,N)=\frac{|\mathrm{Aut}(G)|}{|\mathrm{Aut}(N)|}\cdot\#\left\{\begin{array}[]{c}\mbox{regular subgroups in $\mathrm{Hol}(N)$}\\ \mbox{which are isomorphic to $G$}\end{array}\right\},$$

$$\mathrm{Hol}(N)=\rho (N)⋊\mathrm{Aut}(N).$$

$$S_n$$

$$A_n$$

$$C_n$$

$$\#\mathcal{E}(S_1)$$

$$\#\mathcal{E}(S_2)$$

$$\#\mathcal{E}(S_3)=\#\mathcal{E}(S_3,S_3)+\#\mathcal{E}(S_3,C_6)=2+3=5.$$

$$\#\mathcal{E}(S_{4},N)=\begin{cases}8&\mbox{for $N\simeq S_{4}$},\\ 36&\mbox{for $N\simeq A_{4}\times C_{2}$},\\ 24&\mbox{for $N\simeq S_{3}\times C_{2}\times C_{2}$},\\ 48&\mbox{for $N\simeq C_{6}\times C_{2}\times C_{2}$},\\ 0&\mbox{otherwise}.\end{cases}$$

$$\#\mathcal{E}(S_n,S_n)$$

$$\#\mathcal{E}(S_n,A_n\times C_2)$$

$$\#\mathcal{E}(S_6,M_{10})=72\text{mbox}and\#\mathcal{E}(S_6,\text{mbox}PGL(2,9))=0,$$

$$\#\mathcal{E}(S_{n},N)=0\mbox{ for }N\not\simeq\begin{cases}S_{n},A_{n}\times C_{2}&\mbox{if $n\neq 6$},\\ S_{6},A_{6}\times C_{2},M_{10},\mathrm{PGL}(2,9)&\mbox{if $n=6$}.\end{cases}$$

$$\#\mathcal{E}(S_{n})=\begin{cases}2\cdot\sum\limits_{0\leq k\leq n/2}\dfrac{n!}{(n-2k)!\cdot 2^{k}\cdot k!}&\mbox{if $n\neq 6$},\\ 224&\mbox{if $n=6$}.\end{cases}$$

$$\{\rho (\mathfrak{g}(\sigma ))\cdot \mathfrak{f}(\sigma ):\sigma \in S_n\},\text{mbox}where\{\begin{array}{l}\mathfrak{f}\in \mathrm{Hom}(S_n,\mathrm{Aut}(N))\\ \mathfrak{g}\in \text{mbox}Map(S_n,N)\end{array}$$

$$\mathfrak{g}(\sigma \tau )=\mathfrak{g}(\sigma )\cdot \mathfrak{f}(\sigma )(\mathfrak{g}(\tau ))\text{mbox}forall\sigma ,\tau \in S_n.$$

$$\mbox{$\mathfrak{g}$ is bijective}\implies N\simeq\begin{cases}S_{n}\mbox{ or }A_{n}\times C_{2}&\mbox{if $n\neq 6$},\\ S_{n}\mbox{ or }A_{n}\times C_{2}\mbox{ or }M_{10}\mbox{ or }\mathrm{PGL}(2,9)&\mbox{if $n=6$}.\end{cases}$$

$$\mathrm{Inn}(\Gamma )$$

$$\mathrm{Out}(\Gamma )$$

$$Z(\Gamma )$$

$$[\Gamma ,\Gamma ]$$

$$N\simeq\begin{cases}S_{n}\mbox{ or }A_{n}\times C_{2}&\mbox{if $n\neq 6$},\\ S_{6}\mbox{ or }A_{6}\times C_{2}\mbox{ or }M_{10}\mbox{ or }\mathrm{PGL}(2,9)&\mbox{if $n=6$}.\end{cases}$$

$$1$$

$$N/M\simeq T^{m},\mbox{ where $T$ is a simple group and $m\in\mathbb{N}$.}$$

$$\mathrm{Aut}(T^m)\simeq \mathrm{Aut}(T{)}^m⋊S_m,$$

$$\mathrm{Aut}(N)⟶\mathrm{Aut}(N/M);\phi \mapsto (\eta M\mapsto \phi (\eta )M)$$

$$\overline{\mathfrak{f}}:S_n$$

$$A_n$$

$$\overline{\mathfrak{g}}:S_n$$

Full text

1. 1 Introduction

2. 2 Hopf-Galois structures on a Galois $S_{n}$-extension for $n\geq 5$

3. 2.1 Ruling out the groups $N$ which are perfect

4. 2.2 Reducing to the groups $N$ which contain a copy of $A_{n}$

5. 2.3 Classifying group extensions of $C_{2}$ by $A_{n}$

6. 3 Acknowledgments

1. Introduction

Let $L/K$ be a finite Galois extension with Galois group $G$. Write $\mathrm{Perm}(G)$ for the symmetric group of $G$. Recall that a subgroup $\mathcal{N}$ of $\mathrm{Perm}(G)$ is said to be regular if the map

[TABLE]

is bijective, or equivalently, if the $\mathcal{N}$-action on $G$ is both transitive and free. For example, the images of the left and right regular representations

[TABLE]

respectively, are both regular subgroups of $\mathrm{Perm}(G)$. By work of C. Greither and B. Pareigis in [9], there is an explicit one-to-one correspondence between Hopf-Galois structures on $L/K$ and elements in

[TABLE]

In turn, this set may be written as a disjoint union of the subsets

[TABLE]

where $N$ ranges over all groups of order $|G|$, up to isomorphism. Hence, it is of interest to enumerate $\mathcal{E}(G)$ and $\mathcal{E}(G,N)$. A useful tool is the formula

[TABLE]

where $\mathrm{Hol}(N)$ denotes the holomorph of $N$ and is given by

[TABLE]

This was shown by N. P. Byott in [2]; or see [6, Section 7]. We shall refer the reader to [6, Chapter 2] for more background on Hopf-Galois structures.

For each $n\in\mathbb{N}$, we shall use the standard notation:

[TABLE]

The purpose of this paper is to determine the exact size of $\mathcal{E}(S_{n},N)$ for each group $N$ of order $n!$ and hence the size of $\mathcal{E}(S_{n})$. For $n=1,2$, trivially

[TABLE]

For $n=3$, by [3, Corollary 6.5], we already know that

[TABLE]

For $n=4$, using the formula (1.1) and the magma code $1$ in the appendix, we may compute that:

Theorem 1.1**.**

Let $N$ be a group of order $4!$. Then, we have

[TABLE]

In particular, we have $\#\mathcal{E}(S_{4})=116$.

Remark 1.2*.*

The first and last cases in Theorem 1.1 were known previously, by [5, Theorem 7] and [7, Proposition 4], respectively.

For $n\geq 5$, it was computed in [5, Corollaries 6 and 10], respectively, that

[TABLE]

Also, the case when $n=6$ is slightly different because $S_{6}$ has an exceptional outer automorphism. In [5, p. 91], it was shown that

[TABLE]

where $M_{10}$ denotes the Mathieu group of degree $10$. Our main result is that $\#\mathcal{E}(S_{n},N)=0$ for all other choices of $N$. More precisely, we shall prove:

Theorem 1.3**.**

For $n\geq 5$, let $N$ be a group of order $n!$. Then, we have

[TABLE]

In particular, we have

[TABLE]

2. Hopf-Galois structures on a Galois $S_{n}$-extension for $n\geq 5$

In this section, assume that $n\geq 5$, in which case $A_{n}$ is non-abelian simple, and let $N$ be a group of order $n!$. In [10, Theorem 1.7], the author gave some necessary conditions on $N$ in order for $\#\mathcal{E}(S_{n},N)$ to be non-zero. Below, we shall refine the arguments there to prove Theorem 1.3.

As already noted in [10, Proposition 2.1], which is a consequence of (1.2), a subgroup of $\mathrm{Hol}(N)$ isomorphic to $S_{n}$ is of the shape

[TABLE]

satisfy the relation

[TABLE]

Moreover, the subgroup (2.1) is regular precisely when $\mathfrak{g}$ is bijective. Hence, by (1.1), if $\#\mathcal{E}(S_{n},N)$ is non-zero, then we may choose $\mathfrak{g}$ to be bijective. In view of (1.3) and (1.4), to prove Theorem 1.3, it then suffices to show that

[TABLE]

To that end, given any group $\Gamma$, we shall use the following notation:

[TABLE]

Recall that $\Gamma$ is said to be perfect if $\Gamma=[\Gamma,\Gamma]$. The implication (2.3), and in particular Theorem 1.3, is a direct consequence of the propositions below.

Proposition 2.1**.**

Suppose that $\mathfrak{g}$ is bijective. Then, we have:

1. (a)

The group $N$ cannot be perfect. 2. (b)

The group $N$ contains a copy of $A_{n}$.

Proposition 2.2**.**

Suppose that $N$ contains a copy of $A_{n}$. Then, we have:

[TABLE]

Hence, it remains to prove Propositions 2.1 and 2.2, which we shall do in the subsequent subsections.

2.1. Ruling out the groups $N$ which are perfect

In what follows, suppose that $\mathfrak{g}$ is bijective, and we shall prove Proposition 2.1 (a). Let us remark that this is the hardest step to the proof of Theorem 1.3.

Suppose for contradiction that $N$ is perfect. Note that then $N$ cannot have a subgroup of index two, because such a subgroup is normal, and its quotient is abelian. Also, it shall be helpful to recall that $A_{n}$ is the unique non-trivial proper normal subgroup of $S_{n}$, since $n\geq 5$.

Lemma 2.3**.**

The homomorphism $\mathfrak{f}$ is injective.

Proof.

Notice that $\mathfrak{g}$ restricts to a homomorphism $\ker(\mathfrak{f})\longrightarrow N$ by (2.2). This implies that $\ker(\mathfrak{f})$ cannot contain $A_{n}$, since $N$ has no subgroup of index two, and hence $\ker(\mathfrak{f})$ must be trivial. ∎

Recall that the double cover of $A_{n}$ is the unique group $2A_{n}$, up to isomorphism, fitting into a short exact sequence

[TABLE]

such that the image of $C_{2}$ in $2A_{n}$ lies in $Z(2A_{n})$ and $[2A_{n},2A_{n}]$. It is known that $2A_{n}$ is a perfect group whose center has order two.

Lemma 2.4**.**

*We have $N\simeq 2A_{n}$ and $\mathfrak{f}(A_{n})\subset\mathrm{Inn}(N)$. *

Proof.

Let $M$ be any proper and maximal characteristic subgroup of $N$. The quotient group $N/M$ is then characteristically simple, and so we have

[TABLE]

Since $N$ is perfect, necessarily $T$ is non-abelian. Also, it is known that

[TABLE]

by [4, Lemma 3.2], for example. Note that there is a natural homomorphism

[TABLE]

because $M$ is characteristic. We then have a homomorphism

[TABLE]

induced by $\mathfrak{f}$. Replacing $2A_{n}$ by $A_{n}$ in the proof of [10, Lemma 4.8], the exact same argument shows that the homomorphism

[TABLE]

must be trivial, and in turn the homomorphism

[TABLE]

must be trivial as well. Hence, we conclude that $\overline{\mathfrak{f}}(A_{n})$ lies in $\mathrm{Inn}(T)^{m}$. Let us remark that the argument in [10, Lemma 4.8] uses [10, Proposition 3.4], which is a consequence of the classification of finite simple groups.

Next, consider the surjective map

[TABLE]

induced by $\mathfrak{g}$. As shown in [10, Lemma 4.1], the relation (2.2) implies that:

• •

$\overline{\mathfrak{g}}$ restricts to a homomorphism $\ker(\overline{\mathfrak{f}})\longrightarrow T^{m}$,

• •

$\mathfrak{g}^{-1}(M)$ is a subgroup of $S_{n}$ of index $[N:M]$.

The latter in turn implies that $A_{n}\not\subset\mathfrak{g}^{-1}(M)$, for otherwise

[TABLE]

which is impossible because $T$ is non-abelian. We consider two possibilities:

1. (1)

If $\ker(\overline{\mathfrak{f}})\supset A_{n}$, then $\overline{\mathfrak{g}}$ restricts to a homomorphism $A_{n}\longrightarrow T^{m}$, which cannot be trivial because $A_{n}\not\subset\mathfrak{g}^{-1}(M)$, and so must be an embedding. 2. (2)

If $\ker(\overline{\mathfrak{f}})=1$, the $\overline{\mathfrak{f}}$ restricts to an embedding $A_{n}\longrightarrow\mathrm{Inn}(T)^{m}\simeq T^{m}$.

In both cases, we have $|A_{n}|\leq|T|^{m}$ and hence $|M|\leq 2$. Necessarily $|M|=2$, for otherwise $N\simeq T^{m}$, and $A_{n}$ would embed into $N$ as a subgroup of index two. It follows that $|A_{n}|=|T|^{m}$, whence

[TABLE]

Since any normal subgroup of order two lies in the center, by the uniqueness of the double cover of $A_{n}$, we see that $N\simeq 2A_{n}$ and $M=Z(N)$.

Since $M=Z(N)$, the map (2.4) is injective, by [10, Proposition 3.5 (c)], for example. It follows that $\mathrm{Out}(N)$ embeds into $\mathrm{Out}(N/M)\simeq\mathrm{Out}(A_{n})$ and so is abelian. Thus, the homomorphism

[TABLE]

must be trivial, whence $\mathfrak{f}(A_{n})$ lies in $\mathrm{Inn}(N)$. ∎

From Lemmas 2.3 and 2.4, we deduce that $\mathfrak{f}$ induces an isomorphism

[TABLE]

where $\widetilde{f}:A_{n}\longrightarrow N$ is any injective map satisfying

[TABLE]

Also, let $\zeta\in S_{n}$ be the element such that $\mathfrak{g}(\zeta)$ is the generator of $Z(N)$.

Lemma 2.5**.**

We have $\zeta\in A_{n}$.

Proof.

Using (2.2), it is easy to check that the map

[TABLE]

is a homomorphism. It cannot be trivial, for otherwise by picking a different $\widetilde{f}$ if necessary, we may assume that $\mathfrak{g}(\sigma)=\widetilde{f}(\sigma)^{-1}$ for all $\sigma\in A_{n}$. But then again by (2.2), for all $\sigma,\tau\in A_{n}$, we have

[TABLE]

which means that $\widetilde{f}$ is in fact a homomorphism. This is impossible since $\widetilde{f}$ is injective and $N$ cannot have any subgroup of index two. We have thus shown that $g$ must be an isomorphism. Put $\varphi=f^{-1}\circ g$, which is an automorphism on $A_{n}$. If $\zeta\notin A_{n}$, then for any $\sigma\in A_{n}$, we would have

[TABLE]

namely $\varphi$ is fixed point free. But $A_{n}$ has no such automorphism since $n\geq 5$. Hence, we must have $\zeta\in A_{n}$, as claimed. ∎

Lemma 2.6**.**

For any $\sigma\in S_{n}$, we have

[TABLE]

Moreover, the element $\zeta$ has order two.

Proof.

By (2.2), we have

[TABLE]

Since $Z(N)$ has order two, its generator $\mathfrak{g}(\zeta)$ has to be fixed by all automorphisms on $N$. We then see that $\mathfrak{g}(\zeta^{2})=\mathfrak{g}(\zeta)^{2}=1$ as well as

[TABLE]

The claim then follows since $\zeta\in A_{n}$ by Lemma 2.5 and $\mathfrak{g}$ is bijective. ∎

By Lemmas 2.5 and 2.6, we know that $\zeta$ has order two and lies in $A_{n}$. We may then assume, without loss of generality, that

[TABLE]

Also, recall from Lemma 2.4 that $N\simeq 2A_{n}$, and write

[TABLE]

for some lift of $\zeta$ in $2A_{n}$, where the notation is as in [11, Chapter 2.7.2]. Given any element $\gamma$ in a group $\Gamma$, let $\mathrm{Cent}_{\Gamma}(\gamma)$ denote the centralizer of $\gamma$ in $\Gamma$.

Lemma 2.7**.**

We have

[TABLE]

Proof.

The first equality is obvious because $(1\hskip 5.69054pt2)\notin A_{n}$ commutes with $\zeta$. As for the second equality, note that the quotient map induces a homomorphism

[TABLE]

which is $2$-to-$1$ and whose image has index at most two. But

[TABLE]

in the notation of [11, Chapter 2.7.2], where $z\in Z(2A_{n})$ is non-trivial. This means that the index is in fact equal to two, from which the claim follows. ∎

We are now ready to prove Proposition 2.1 (a). From Lemma 2.6 and the fact that $f$ is an isomorphism, we deduce that

[TABLE]

But this contradicts Lemma 2.7, whence $N$ cannot be perfect, as desired.

2.2. Reducing to the groups $N$ which contain a copy of $A_{n}$

In what follows, suppose that $\mathfrak{g}$ is bijective, and we shall prove Proposition 2.1 (b)

By Proposition 2.1 (a), the group $N$ is not perfect, so it has a proper and maximal characteristic subgroup $M$ containing $[N,N]$. The quotient $N/M$ is then abelian, and by the proof of the second statement of [10, Theorem 1.7], we know that $A_{n}=\mathfrak{g}^{-1}(M)$. This means that $\mathfrak{g}$ restricts to a bijective map

[TABLE]

Also, since $M$ is characteristic, the map $\mathfrak{f}$ induces a homomorphism

[TABLE]

Clearly, it follows directly from (2.2) that

[TABLE]

Analogous to (2.1), this implies that $\mathrm{Hol}(M)$ has a regular subgroup isomorphic to $A_{n}$. Since $A_{n}$ is non-abelian simple, we deduce from [4] that $M\simeq A_{n}$. This proves that $N$ contains a copy of $A_{n}$, as desired.

2.3. Classifying group extensions of $C_{2}$ by $A_{n}$

In what follows, suppose that $N$ contains a copy of $A_{n}$, and we shall prove Proposition 2.2. Since any subgroup of index two is necessarily normal, the hypothesis implies that $N$ fits in a short exact sequence

[TABLE]

We shall write $A$ for the image of $A_{n}$ in $N$ and $\epsilon$ for the non-trivial element in $C_{2}$. Let us make two observations.

Lemma 2.8**.**

The short exact sequence (2.5) splits if $n\neq 6$.

Proof.

For each $\eta\in N$, note that we have an automorphism

[TABLE]

Fix an element $\eta_{0}\in N\setminus A$. Suppose now that $n\neq 6$, so then $\mathrm{Aut}(A)\simeq S_{n}$, with $\mathrm{Inn}(A)$ corresponding to $A_{n}$.

1. (1)

If $\varphi_{\eta_{0}}\notin\mathrm{Inn}(A)$, then there exists $a\in A$ such that $\varphi_{\eta_{0}}\varphi_{a}$ has order two, which means that $(\eta_{0}a)^{2}$ centralizes $A$. 2. (2)

If $\varphi_{\eta_{0}}\in\mathrm{Inn}(A)$, then there exists $a\in A$ such that $\varphi_{\eta_{0}}\varphi_{a}$ is the identity, which means that $\eta_{0}a$ centralizes $A$.

In both cases, since $(\eta_{0}a)^{2}\in A$ and $Z(A)=1$, we deduce that $\eta_{0}a$ has order two. Thus, by sending $\epsilon\mapsto\eta_{0}a$, we obtain a homomorphism $C_{2}\longrightarrow N$ which splits the short exact sequence (2.5), as desired. ∎

Lemma 2.9**.**

Let $\psi:C_{2}\longrightarrow\mathrm{Aut}(A_{n})$ be a homomorphism such that

[TABLE]

for some $\sigma_{0}\in S_{n}$. Then, we have

[TABLE]

Proof.

Note that $\sigma_{0}$ must have order dividing two. Also, we have

[TABLE]

for all $\tau_{1},\tau_{2}\in A_{n}$ and $i_{1},i_{2}\in\mathbb{Z}$. It is then straightforward to verify that

[TABLE]

defines an isomorphism if $\sigma_{0}\notin A_{n}$, and

[TABLE]

defines an isomorphism if $\sigma_{0}\in A_{n}$. This proves the claim. ∎

We are now ready to prove Proposition 2.2. For $n=6$, the claim may be verified using the magma code $2$ in the appendix. For $n\neq 6$, the claim is a direct consequence of Lemmas 2.8 and 2.9 because then $\mathrm{Aut}(A_{n})\simeq S_{n}$.

3. Acknowledgments

The author would like to thank the referee for some helpful comments. She would also like to recognize the software magma [1] and gap [8] which were used in the computations to deal with the cases $n=4,6$.

Appendix: magma codes

magma code $1$:

⬇

G:=SymmetricGroup(4);

AutG:=AutomorphismGroup(G);

NN:=SmallGroups(24);

for i in [1..#NN] do

N:=SmallGroup(24,i);

AutN:=AutomorphismGroup(N);

Hol:=Holomorph(N);

RegSub:=RegularSubgroups(Hol);

L:=[0]; // sizes of the conjugacy classes of regular subgroups in Hol(N) isomorphic to G

for R in RegSub do

if IsIsomorphic(R‘subgroup,G) then

Append(~L,R‘length);

end if;

end for;

E:=(#AutG/#AutN)*&+L; // formula (1.1)

if E ne 0 then

print <i,E>;

end if;

end for;

magma code $1$ output:

⬇

<12,8> // SmallGroup(24,12) = S4

<13,36> // SmallGroup(24,13) = A4 x C2

<14,24> // SmallGroup(24,14) = S3 x C2 x C2

<15,48> // SmallGroup(24,15) = C6 x C2 x C2

// The identifications of SmallGroup(24,i) here may be checked using

the StructureDescription() or IdGroup() command in gap.

magma code $2$:

⬇

A6:=AlternatingGroup(6);

NN:=SmallGroups(720);

for i in [1..#NN] do

N:=SmallGroup(720,i);

if not IsSolvable(N) then // for N to contain a copy of A6 necessarily N is insolvable

NorSub:=NormalSubgroups(N);

AA:=[A:A in NorSub|IsIsomorphic(A‘subgroup,A6)];

if not IsEmpty(AA) then

i;

end if;

end if;

end for;

magma code $2$ output:

⬇

763 // SmallGroup(720,763) = S6

764 // SmallGroup(720,764) = PGL(2,9)

765 // SmallGroup(720,765) = M10

766 // SmallGroup(720,766) = A6 x C2

// The identifications of SmallGroup(24,i) here may be checked using

the StructureDescription() or IdGroup() command in gap.