Normalizers in symmetric groups

Question

Question: Let $G$ be a finite group. Is it true that there is a subgroup $U$ inside some symmetric group $S_n$, such that $N(U)/U$ is isomorphic to $G$? Here $N(U)$ is the normalizer of $U$ in $S_n$.

Background: If true, this would for instance give a trivial proof of the Fried-Kollar Theorem that every finite group is the full automorphism group of a number field.

Results: If $U\le S_n$ acts regularly with respect to the natural action of $S_n$, then $N(U)/U\cong\text{Aut}(U)$. However, many finite groups are not the automorphism group of another finite group, like most cyclic groups. On the other hand, it is easy to get $N(U)/U\cong G$ for each abelian $G$ by choosing $U$ a direct product of semidirect products $C_{p_i}\rtimes C_{m_i}$ for suitable distinct primes $p_i$ and divisors $m_i$ of $p_i-1$, with the natural intransitive action of $U$ with orbit lengths $p_1, p_2,\dots$.

Added recently (answering Stefan Kohl's question from the comments): $Q_8$ is a normalizer quotient in $S_{81}$. Let $U=\mathbb F_3^4\rtimes H$ be the primitive group of degree $81$ where $H=C_5\rtimes C_8$ with $C_8$ inducing an automorphism group of order $2$ on $C_5$. Then $N_{S_{81}}(U)/U=Q_8$. This can be seen by hand, or using GAP:

gap> u:=PrimitiveGroup(81,27);;
gap> nu:=Normalizer(SymmetricGroup(81),u);;
gap> w:=nu/u;;
gap> Order(w);
8
gap> IsQuaternionGroup(w);
true

Remark 1: $N_{S_{81}}(U)$ is the semiaffine group ${A\Gamma L}_1(\mathbb F_{81})$. Remark 2: $U$ in Magma is PrimitiveGroup(81,26).

it would be a trivial proof of the FK theorem only if it's trivially true :) — YCor, Jul 20 '12 at 03:47
Hi Peter, welcome to MO! Do you happen to know, how to get the alternating groups for $n>6$? — j.p., Jul 24 '12 at 06:22
@jp: I haven't thought about such specific cases, because I was hoping for a general argument. However, it seem's to be more difficult than I originally expected. — Peter Mueller, Jul 24 '12 at 12:18
It seems, for $G$ with no nontrivial decomposition as a direct or permutational wreath product, that it is equivalent to the question whether there exists $U$ transitive satisfying your requirement. Most groups do not have such decompositions, so your question is very close to the same for $U$ transitive (and probably has the same answer). Anyway, a good start would be, as jp suggests, to test some particular cases. — YCor, Jun 12 '13 at 18:31
The smallest group for which I don't quickly see how to represent it in the form $N_{S_n}(U)/U$ for some subgroup $U$ of $S_n$ is the quaternion group $Q_8$ of order $8$. -- Do you know a suitable such $U$? — Stefan Kohl, Apr 25 '16 at 13:27

score 2 · Answer 1 · answered Feb 07 '20 at 15:02

I have come across a fairly recent result which pertains to this question. It is in this paper:

Guralnick, Robert M.; Maróti, Attila; Pyber, László, Normalizers of primitive permutation groups, Adv. Math. 310, 1017-1063 (2017). ZBL1414.20002.

An arXiv version is here. The paper examines the situation when $U$ is primitive. They show that, in all but a finite number of situations, $|N(U)/U|<n$. Indeed, they strengthen this bound if you add in a particular infinite family. The main result is this one:

Theorem: Let $U$ be a primitive subgroup of $S_n$, and let $N=N_{S_n}(U)$. Then $|N/U|< n$ unless $U$ is an affine primitive permutation group and the pair $(n, N/U)$ is one of: $$(3^4,O^−_4(2),(5^4,Sp_4(2)),(3^8,O^−_6(2)),(3^8,SO^−_6(2)),(3^8,O^+_6(2)),(3^8,SO^+_6(2)),(5^8,Sp_6(2)),(3^{16},O^−_8(2)),(3^{16},SO^−_8(2)),(3^{16},O^+_8(2)), \textrm{ or }(3^{16},SO^+_8(2)).$$ Moreover if $N/U$ is not a section of $\Gamma L_1(q)$ when $n=q$ is a prime power, then $|N/U|< n^{1/2}\log n$ for $n≥2^{14000}$.

@YCor's comment on the original question suggests that the primitivity assumption is not too onerous. It would be interesting (but probably very hard) to try and understand what might happen when $U$ is transitive and imprimitive.

Normalizers in symmetric groups

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