*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 $p_1, p_2,\dots$.

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