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).

Answer 1

Alexei Entin and Cindy Tsang prove a positive answer to this question in their preprint Normalizer Quotients of Symmetric Groups and Inner Holomorphs. One of the main tools is Yves de Cornulier's answer to the MO question Is every finite group the outer automorphism group of a finite group? and its subsequent variant by Benjamin Sambale.

Answer 2

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.