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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 $p_1, p_2,\dots$.

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it would be a trivial proof of the FK theorem only if it's trivially true :) –  YCor Jul 20 '12 at 3:47
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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 6:22
    
@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
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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

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