It is well known that the only normal subgroup of $S_n$ is $A_n$ when $n\geqslant 5$, and that $A_n$ is also simple. Furthermore, $A_{\infty}$, the even permutations on $\mathbb{N}$, is also simple. This lead me to wonder about the following:

Take a general set $X$ with cardinality $\kappa>\aleph_0$ from which we can generate the group $\text{Sym}\,X$.


  • can we define an alternating group on $X?$

  • if so does it remain the only normal subgroup of $\text{Sym}\, X?$


If $\operatorname{Sym} X$ denotes the group of finite-support permutations, you can define $\operatorname{Alt} X$ as the group of even finite-support permutations. If $N\subseteq \operatorname{Sym} X$ is a normal subgroup, then $N\cap \operatorname{Sym} F$ is normal in $\operatorname{Sym} F$ for any finite $F\subset X$. It follows that if $N$ contains any nontrivial even permutation $\sigma$, it must contain all of $\operatorname{Alt} X$ (since for any finite set $F$ with at least $5$ elements containing the support of $\sigma$, it must contain all of $\operatorname{Alt} F$, and every element of $\operatorname{Alt} X$ is in some such $\operatorname{Alt} F$). Similarly, if $N$ contains any odd permutation, it must be all of $\operatorname{Sym} X$. So the only nontrivial proper normal subgroup of $\operatorname{Sym} X$ is $\operatorname{Alt} X$.

If you want to consider the group of all permutations of $X$ (with arbitrary support), then there are more normal subgroups. For instance, for any infinite cardinal $\lambda\leq|X|$, the subgroup of permutations with support of cardinality $<\lambda$ is a normal subgroup. In fact, these together with the finite-support alternating group are all the nontrivial proper normal subgroups of the full permutation group (I don't know the proof of this off the top of my head; this is known as the "Baer-Schreier-Ulam theorem"). In particular, this indicates that there is no reasonable notion of "sign" for permutations with infinite support (there is no "$<\lambda$-support alternating subgroup" among the normal subgroups unless $\lambda=\aleph_0$).

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  • $\begingroup$ Many thanks - could you answer the first question, if possible? $\endgroup$ – K. 622 Feb 2 '16 at 4:25
  • $\begingroup$ @K.622: I'm not sure exactly what the first question means. The finite support even permutations certainly still form a normal subgroup. However, there is no notion of the "sign" of an infinite-support permutation--see my latest edit. $\endgroup$ – Eric Wofsey Feb 2 '16 at 4:33
  • $\begingroup$ @EricWofsey Out of curiosity, do you know if Baer-Schreier-Ulam is provable in ZF alone? $\endgroup$ – Noah Schweber Feb 2 '16 at 4:53
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    $\begingroup$ @NoahSchweber: As I said, I don't know the proof off the top of my head, but the approach to proving it I can imagine would heavily use choice. Certainly the exact statement of it that I gave can't be correct in ZF (for instance, the set of permutations with support of size $<\lambda$ need not be closed under composition, since $\lambda$ could be a sum of two smaller cardinals!). $\endgroup$ – Eric Wofsey Feb 2 '16 at 4:58
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    $\begingroup$ @K.622 Just as a remark, if you take $\operatorname{Alt} X$ to be the commutator subgroup group in the case of permutations without any support restrictions, then $\operatorname{Alt} X=\operatorname{Sym} X$, and in fact every element of $\operatorname{Sym} X$ is a commutator. That and some other interesting things about $\operatorname{Sym} X$ can be found in this paper. (In particular every cayley graph is bounded! It is open if there is a countable infinite group with that property) $\endgroup$ – user29123 Feb 2 '16 at 14:42

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