TITLE: Alternating group on infinite sets
QUESTION [5 upvotes]: 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 $\text{Sym}\,X$. 
Questions

can we define an alternating group on $X?$
if so does it remain the only normal subgroup of $\text{Sym}\, X?$

REPLY [8 votes]: 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 $\lambda=\aleph_0$).