# Normal subgroups of infinite symmetric group

I recently took a course on group theory, which mentioned that the following proposition is equivalent to the continuum hypothesis: "The infinite symmetric group (i.e. the group of permutations on the set $\mathbb{N}$) has exactly 4 normal subgroups." Does anyone have any references or explanation for this?

• Well, one of the normal subgroups is the set of permutations of finite support. – Arturo Magidin Jul 4 '12 at 0:58
• The support are the elements that are not fixed by the permutation. I.e., I'm refering to the subgroup of permutations that are the identity on a cofinite subset of $\mathbb{N}$. – Arturo Magidin Jul 4 '12 at 1:01
• L. Onofri (1929) Teoria delle sostituzioni che operano su una infinità numerabile di elementi, Memoria III. Annali di Matematica Pura ed Applicata. Vol. 7(1), 103-130, 1929. Independently and later: J. Schreier, S. Ulam (1933) Über die Permutationsgruppe der natürlichen Zahlenfolge. Studia Mathematica (1933) Vol. 4(1), p.134-141, 1933. – YCor Jul 15 '19 at 11:17

You are inquiring about the Schreier-Ulam theorem. This old MO post contains an answer of mine with the statement of the result; here is a link to the original paper (thanks to t.b.). I would be happy to supplement this and/or that answer with a link to a free, electronically available English language proof, if anyone knows one.

• How does this relate to CH? Does this theorem assume it? As I see it, it seems that the theorem, if true with no assumptions, would imply that the statement in question is false... – tomasz Jul 4 '12 at 1:18
• @tomasz: CH does not enter into it. Schreier-Ulam states that the only nontrivial normal subgroups of $S_{\Omega}$ are $\cup A_n$ and $\cup S_n$. The other two (to make up the count of four) are the trivial and total subgroup. – Arturo Magidin Jul 4 '12 at 1:27
• The original paper by Schreier and Ulam is available here for those who read German. The homepage of the Polish Virtual Library is certainly worth bookmarking. – t.b. Jul 4 '12 at 1:28
• @mlbaker: Seems like it. On the other hand, I'd be surprised if counting the normal subgroups of permutations of $\mathbb{R}$ was not contingent on CH. One would expect (at least) a normal subgroup of the form "all permutations fixing $\aleph_\alpha$ elements" for each $\aleph_\alpha < \mathfrak{c}$. – Jason DeVito Jul 4 '12 at 2:14
• @mlbaker: I don't see how CH would come into it in any case. The arguments are all finitistic, since we are dealing with groups. – Arturo Magidin Jul 4 '12 at 4:07

For a general infinite set $X$, the normal subgroups of Sym$(X)$ are:

1. Sym$(X)$;

2. the trivial subgroup;

3. the even permutations of $X$ with finite support;

4. for each cardinality $c$ with $\aleph_0 \le c \le |X|$, the group of all permutations of $X$ with support less than $c$.

There is a straightforward proof in Chapter 8 of the book "Permutation Groups" by J.D. Dixon and B.M. Mortimer, where the result is attributed to Baer.

I don't think the proof uses CH or GCH although the result itself is affected by CH.

The widely known reference for this result is Schreier-Ulam (1933). But this was previously proved by Luigi Onofri (1929), in the third of a series of 3 articles, of 1927, 1928, and 1929, not mentioned in Schreier-Ulam's article, in which he also introduces (§4 of the first opus) its natural topology on the group of permutations of a countable set.

L. Onofri. Teoria delle sostituzioni che operano su una infinità numerabile di elementi. Memoria 3a. Annali di Matematica Pura ed Applicata December 1929, Volume 7, Issue 1, pp 103–130. (restricted access: Springerlink).

This is in §141, p124:

I gruppi $G_1$ e $G_2$ sono gli unici sottogruppi invarianti del totale $G$. (The groups $G_1$ and $G_2$ are the unique invariant subgroups of the whole group $G$.)

Here $G$ is the group of permutations of an infinite countable set $I$, $G_1$ its subgroup of finitely supported permutations ("substitutions operating on finitely many elements" in his language) and $G_2$ the subgroup of even elements in $G_1$, and "invariant" means invariant under conjugation.

Onofri's proof has 3 steps (I try to use italics when I translate in English without translating in modern math language):

(a) he defines (§100, p104) a subset $H$ of $G$ to be infinitely transitive of infinite grade if it is transitive by pre-composition of injective self-maps whose image with infinite complement, or, to be closer to his language, if for any two infinite systems of (distinct) elements $[x_1,x_2,\dots]$, $[y_1,y_2,\dots]$ in $I$ such that the complements (called residual systems) of both $\{x_i:i\ge 1\}$ and $\{y_i:i\ge 1\}$ are infinite, there exists $s\in H$ ("a substitution $s$ of $H$") such that $s(x_i)=y_i$, $(i=1,2,\dots)$. Here "infinite grade" refers to infiniteness of the complement.

He proves (§110, p108) that the only subgroup of the group of permutations of $I$ that is infinitely transitive of infinite grade, is the whole group of permutations.

(b) He proves (§136, p121) that the only normal subgroup not contained in the subgroup of finitely supported permutations, is the whole group, and his strategy consists in proving that such a subgroup is infinitely transitive of infinite grade.

His statement is actually Un complesso C contenente una sostituzione h su infiniti elementi e le sue trasformate mediante G, coincide con il totale.

A literal translation is: A complex containing a substitution h on infinitely many elements and its conjugate transforms G coincides with the total.

An adapted translation, from my understanding, is: "A subsemigroup (of $G$, the group of permutations of $I$) containing an infinitely supported permutation as well as its conjugates coincides with the whole group of permutations."

Here I should explain why I'm translating complesso (literally "complex") into "subsemigroup": "complesso" is evoked, not defined in the first memoir (1927, §25, p89): indeed he defines a somewhat inefficient terminology for subsemigroups of the group of permutations "(sub)groups" for subgroups, "pseudogroups" those subsemigroups that are not subsemigroups, and even the splits pseudogroups as "simple" and "composite", where "simple" means that no element is invertible. Then he seems to use "complex" means something which either a group, a "simple" pseudogroup, or a "composite pseudogroup". I'm still a bit confused since he occasionally uses "complesso" for cosets, but anyway the proof is convincing if one interprets as "subsemigroup".

(c) The hardest part is done, since the result now reduces to classify the normal subgroups contained in the subgroup of finitely supported permutations; this short concluding step is §141, p124. Actually, he then (§142) observes that the alternating subgroup is the "only invariant subgroup" of the group of finitely supported permutations while the alternating subgroup itself has "no normal subgroup except the identity". It's a bit sloppy on whether he considers the trivial/whole subgroups as normal ("invariant") subgroups, but the proof is totally correct as far as I could check.

He, by the way, observes (§139) after step (b) the corollary that the group of permutations of $I$ has no proper subgroup of finite index, and in particular (§140) that there is no way to coherently extend parity of finite permutations to arbitrary permutations (since this would produce a subgroup of index 2).

First, if you want an English reference for the proof, Group Theory, by W.R.Scott has a proof.

The first theorem of this kind was due to Schreier and Ulam, which was for the symmetric group on a countably infinite set. Then Baer extended this result to arbitrary infinite cardinals.

For a symmetric group on a set of cardinality $$\kappa$$, the nontrivial normal subgroups are the alternating and symmetric 'finite support' subgroups (i.e. all finite permutations of even order and finite support, and all permutations of finite support), and the symmetric 'bounded support' subgroups, namely those of the form { $$x$$ is a permutation on $$\kappa$$ with support of cardinality at most $$\alpha$$ }, for each infinite cardinal $$\alpha<\kappa$$.

As for the continuum hypothesis, if you consider $$Sym(\mathbb{R})$$, the nontrivial normal subgroups are precisely the two of finite support, and the one of permutations of countable support if, and only if, the continuum hypothesis is true in your model of set theory.

The proof works as follows:

1. Every permutation is a product of two involutions (permutations $$g$$ such that $$g^2=1$$)

2. Given an involution of support cardinality $$\alpha$$ in a normal subgroup of $$\mathrm{Sym}(X)$$, by conjugating and multiplying in $$\mathrm{Sym}(X)$$, we can show that involutions of support cardinality $$\alpha+\alpha$$, and hence involutions of support cardinality $$\alpha\aleph_0$$ also exist in $$\mathrm{Sym}(X)$$, but we cannot produce involutions any larger than $$\alpha\aleph_0$$ this way, since the support of $$gh$$ is at most of cardinality $$|\mathrm{supp}(g)|+|\mathrm{supp}(h)|$$.

3. Once we have an involution of support cardinality $$\alpha$$ in a normal subgroup, that normal subgroup must contain all involutions of support cardinality $$\alpha$$.

4. With finite cardinals, this means that given any nontrivial finite permutation in a normal subgroup, the subgroup of alternating permutations of finite support must be contained in the normal subgroup; and if there is an odd permutation then the subgroup of all permutations of finite support must be contained in the normal subgroup.

5. With infinite cardinals, this means that if we have an infinite involution of support cardinality $$\alpha$$ in a normal subgroup, then we have all involutions of support $$\le\alpha$$ in this normal subgroup, and hence all permutations of support $$\le\alpha$$ in this normal subgroup

6. Finally, given a $$n$$-cycle for finite $$n$$, we can multiply this by a conjugate of it to get an involution of "comparable" support, and similarly for a countable cycle.