# Normalizers in symmetric groups

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 3:47
• 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 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. 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 $U$? Apr 25 '16 at 13:27

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, $$|N(U)/U|. Indeed, they strengthen this bound if you add in a particular infinite family. The main result is this one:
Theorem: Let $$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.