*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).

triviallytrue :) – YCor Jul 20 '12 at 3:47transitivesatisfying 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