Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be an infinite set and let $\operatorname{Sym}(X)$ be the symmetric group of $X.$ Let $N$ denote the set of all permutations $\pi\in\operatorname{Sym}(X),$ such that the complement of the set of fixed points of $\pi$ has cardinality strictly less than $\operatorname{card}(X).$

A theorem of Baer is cited in a paper I'm trying to read, which says or implies that $N\lhd \operatorname{Sym}(X)$ and is maximal in the family of normal subgroups of $\operatorname{Sym}(X).$

Baer's paper is in German so I cannot read it. I don't have much trouble with the fact that $N\lhd \operatorname{Sym}(X),$ but the maximality eludes me. Could you please help me with it?

If anyone's interested, the cited paper is

Baer, R., Die Kompositionsreihe der Gruppe aller eineindeutigen Abbildungen einer unendlichen Menge auf sich, Studia Math. 5 (1934), 15–17.

share|improve this question
1  
The normal subgroups of $Sym(X)$ are well-known, and can be found in Dixon and Mortimer's book on permutation groups, for example. Anyway, they are gotten by resticting the cardinality of the non-fixed points, and they are ordered (the same way the cardinality is), so it's not very hard (if you know this!) to see $N$ is maximal normal. –  user641 Feb 2 '12 at 23:49
2  
Scott's group theory also has proofs. –  Jack Schmidt Feb 3 '12 at 13:20

1 Answer 1

up vote 9 down vote accepted

This follows from the fact that, if a normal subgroup of ${\rm Sym}(X)$ contains a permutation $\pi$ which moves $\aleph_\nu$ points, then it contains all permutations moving $\aleph_\nu$ or fewer points. The Baer paper (available here) says very briefly that, in general, this can be proved in the same way as when $\nu=0$ and $X$ is countable; the countable case was proved earlier by Schreier and Ulam (Satz 1, Über die Permutationsgruppe der natürlichen Zahlenfolge, Studia Math. 4, 134-141 (1933), here.)

A proof of this fact can also be extracted from Dixon and Mortimer's book Permutation Groups (Springer, GTM #163, ISBN 0-387-94599-7, $\S 8.1$.) As with the original Schreier-Ulam proof, it works by starting with $\pi$ and repeatedly conjugating and multiplying, depending on the cycle structure of the permutations being worked with, until one is able to get a permutation with any desired cycle structure. A sketch of a proof (that is not quite the same as the one in Dixon and Mortimer) is:

  • Multiply $\pi$ by a conjugate of itself so as to get an element $\sigma$ with $\aleph_\nu$ 2-cycles, and possibly other cycles.

  • Then $\sigma$ is multiplied by a conjugate of itself to get rid of all cycles which are not of length 1 or 2, in such a way as to produce an element $\tau$ of order 2 with $\aleph_\nu$ 2-cycles and $\#X$ fixed points.

  • We can multiply $\tau$ by a conjugate of itself to cancel some of these 2-cycles, producing a permutation $\mu_\kappa$ with any number $\kappa\le \aleph_\nu$ of 2-cycles, and $\# X$ fixed points.

  • If $\aleph_\nu=\#X$, observe that we can multiply $\mu_{\aleph_\nu}$ by a conjugate of itself to get a permutation $\mu'_\lambda$ with $\#X$ 2-cycles and any number $\lambda\le \#X$ of fixed points.

  • The final step is to analyze the cycle structure of any permutation moving $\aleph_\nu$ points or fewer and show that it is a product of two permutations of order 1 or 2, each of which also moves $\aleph_\nu$ points or fewer. Up to conjugacy, each of those permutations is then either a $\mu_\kappa$ or a $\mu'_\lambda$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.