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Let $S$ be the set of all finite permutations of $\mathbb{N}$, i.e. they fix all but a finite set, and $A\subset S$ the set of all even permutations.

Theorem. The normal subgroups of $S_\infty$ are exactly the following four: $\{\mathrm{id}\}\subset A\subset S\subset S_\infty$ (see on Wikipedia).

Now, consider a model $M$ with domain $\mathbb{N}$ and let $\mathrm{Aut}(M)$ be the group of automorphisms of $M$. $\mathrm{Aut}(M)$ is a closed subgroup of $S_\infty$ and every closed subgroup of $S_\infty$ equals $\mathrm{Aut}(M)$, for some model $M$.

My question: Are there any (partial?) results generalizing the above theorem to $\mathrm{Aut}(M)$? I.e. what are the normal subgroups of $\mathrm{Aut}(M)$, for various structures $M$? Equivalently, if $A$ is a closed subgroup of $S_\infty$, what are the normal subgroups of $A$?

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  • $\begingroup$ I am curious to know where this Theorem is proved? (It's indeed well-known that $A$ is the only nontrivial normal subgroup in $S$, but then?) $\endgroup$ – Dima Pasechnik May 26 '16 at 22:52
  • $\begingroup$ @DimaPasechnik: I found it on Wikipedia: en.wikipedia.org/wiki/Symmetric_group#Normal_subgroups $\endgroup$ – Ioannis Souldatos May 26 '16 at 22:56
  • $\begingroup$ @DimaPasechnik: The reference on Wikipedia says "For more details see (Scott 1987, Ch. 11.3) or (Dixon & Mortimer 1996, Ch. 8.1)." $\endgroup$ – Ioannis Souldatos May 26 '16 at 22:57
  • $\begingroup$ @DimaPasechnik See the addition to my answer. $\endgroup$ – Igor Rivin May 27 '16 at 0:41
  • $\begingroup$ There was some work attempted on an overgroup $G$ of $S$ , the subgroup of all "bounded" permutations - bounded in the sense that for each $\pi\in G$ satisfies $|k-\pi(k)|\leq b(\pi)$ for all $k\in \mathbb{N}$. (everything what was published does not make much sense though :-)) $\endgroup$ – Dima Pasechnik May 27 '16 at 8:16
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Not an answer, but the closed subgroups of $S_\infty$ are classified in this paper by Bergman and Shelah:

George M. Bergman and Saharon Shelah, MR 2280223 Closed subgroups of the infinite symmetric group, Algebra Universalis 55 (2006), no. 2-3, 137--173.

To answer Dima Pasechnik's question in the comments, the Baer-Schreier-Ulam theorem follows from the result proved by Manfred Droste and R Gobel, 1979.

CLOSER TO AN ANSWER In their 2011 paper, MacPherson and Tent show (as conjectured in my comment) that these are often simple (the below is the beginning of the math review)

Dugald Macpherson and Katrin Tent, MR 2824528 Simplicity of some automorphism groups, J. Algebra 342 (2011), 40--52.:

Let $M$ be a countably infinite first-order relational structure and let $\mathrm{Aut}(M)$ be the automorphism group of $M$. The authors show that if $M$ is transitive, free, and homogeneous, and if $\mathrm{Aut}(M)$ is not the full symmetric group, then $\mathrm{Aut}(M)$ is simple. Transitive means that $\mathrm{Aut}(M)$ acts transitively on $M$; free means the class of finite substructures of $M$ has the free amalgamation property; and homogeneous means that isomorphisms between finite substructures extend to $\mathrm{Aut}(M)$.

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  • $\begingroup$ The paper that you cite characterizes the equivalence classes of the equivalence relation $G1 ≈ G2$ if there exists a finite set $U\subset S_\infty$, such that $⟨G1∪U⟩=⟨G2∪U⟩$, where $G_1,G_2$ are closed subgroups of $S_\infty$. I fail to see how it relates to the question at hand. $\endgroup$ – Ioannis Souldatos May 26 '16 at 22:50
  • $\begingroup$ There is certainly no classification of closed subgroups of $S_\infty$ in the usual sense (which suggest a full description), although there are important results about them (in the linked paper, they are split into 4 classes and some results are proved about these classes). $\endgroup$ – YCor May 26 '16 at 23:23
  • $\begingroup$ @YCor: Yes, but how this relates to finding the normal subgroups of a closed subgroup of $S_\infty$? I can not see any connection. I don't have access to the papaer, I just read the abstract. Does the classification give any information/hint about their normal subgroups? $\endgroup$ – Ioannis Souldatos May 26 '16 at 23:42
  • $\begingroup$ I was answering the reply, not your comment. I have no claim as you suggest (on the other hand, you can easily find an arxiv version of the mentioned paper) $\endgroup$ – YCor May 27 '16 at 0:15
  • $\begingroup$ @IoannisSouldatos I clearly state that my answer is not an answer, but a pointer to some information that you might be able to use to try to arrive at an answer. My guess is that the groups you are interested in are close to simple (just like $S_\infty$ itself). $\endgroup$ – Igor Rivin May 27 '16 at 0:18
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Answer was edited

Here is a characterization I came up with:

Definition

  • Let $G=Aut(M)$ (equiv. $G$ is a closed subgroup of $S_\infty$) and $N$ is a subgroup of $G$. Define $x\sim_N y$ iff there exists an automorphism $g\in N$ such that $g.x=y$.
  • $G$ respects $\sim_N$ if for all $x,y$, $$x\sim_N y\text{ iff } g.x\sim_N g.y$$

Lemma Let $M$ be a countable model, $G=Aut(M)$ and $N$ be a closed subgroup of $G$. The following are equivalent:

  1. $N\vartriangleleft G$
  2. $G$ respects $\sim_N$

Proof. Unfold the definitions \begin{align*} N & \vartriangleleft G \text{ iff }\\ % \forall g\in G\; \forall n\in N\; \exists n'\in N,\; & g^{-1}ng=n' \text{ iff }\\ \forall x\in M \forall g\in G\; \forall n\in N\; \exists n'\in N,\; & g^{-1}ng.x=n'.x \text{ iff }\\ \forall x,y \in M \forall g\in G,\; & (\exists n\in N,\; y=g^{-1}ng.x) \leftrightarrow (\exists n'\in N,\; y=n'.x) \text{ iff }\\ \forall x,y \in M \forall g\in G,\; & (\exists n\in N,\; g.y=ng.x) \leftrightarrow (\exists n'\in N,\;y=n'.x) \text{ iff }\\ \forall x,y \in M \forall g\in G,\; & (g.x\sim_N g.y) \leftrightarrow (x\sim_N y) \end{align*}

End of Proof.

Stabilizers

As a special case we prove that if $X$ is a $G$-invariant set, then point-wise stabilizer of $X$ is a normal subgroup of $G$.

Definition If $G$ is a group acting on $\mathbb{N}$ and $n\in\mathbb{N}$, then $G_n$ denotes the stabilizer of $n$ in $G$. If $X$ is a subset of $\mathbb{N}$, $G_{X}$ denotes the pointwise-stabilizer of $X$.

Lemma 1 For all $n,g$, $G_{g(n)}= g G_n g^{-1}$.

Proof. $f\in G_{g(n)}$ iff $f(g(n))=g(n)$ iff $g^{-1}(f(g(n)))=n$ iff $g^{-1}(f(g))\in G_n$ iff $f\in g G_n g^{-1}$. $\square$

Corollary 2 For $X\subset\mathbb{N}$, $G_{g(X)}=g G_X g^{-1}$.

Lemma 3 Let $G=Aut(M)$, for some $M$, and let $X$ be a $G$-invariant set. Then the pointwise-stabilizer of $X$ is a normal subgroup in $G$.

Proof. Let $g\in G$. Since $X$ is G-invariant, $g(X)=X$. By Corollary 2, $G_X=G_{g(X)}=g G_X g^{-1}$, which proves the result. $\square$.

Some Comments

  • By Lemma 3, the number of normal subgroups of $G$ correlates with the number of $G$-orbits. We can construct an example of $G$ with $\aleph_0$-many normal subgroups, even with $2^{\aleph_0}$-many normal subgroups.
  • In view of Lemma 3, $Aut(M)$ is a simple group, only if $Aut(M)$ acts trasitively on $M$, i.e. has only one orbit. So, it is necessary in the result mentioned by Igor Rivin that the $Aut(M)$ acts transitively on $M$.
  • By a theorem of Dana Scott, every $Aut(M)$-orbit is defined by an $L_{\omega_1,\omega}$-sentence (in the same vocabulary as $M$). If $N$ is a normal subgroup of $Aut(M)$, what can we say about the action of $N$ on $M$?
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