Structure of the full symmetric group on a countably infinite set

Question

Trying to get a handle on the full symmetric group $S$ of permutations on a countable set $X$. I had never really thought much about this group, but now I look at it for the first time it appears a more complex idea than I had naively assumed. I will confine my expression of puzzlement to one fairly straightforward question, but any further input on this group would be much appreciated.

For any element $g \in S$ we have a disjoint decomposition:

$$ X= \sigma_g \cup \tau_g $$ with $$ \sigma_g \cap \tau_g = \emptyset $$ where $\sigma_g$ is the set of points fixed by $g$

If $S_* = \{g \in S \mid \text{card}(\tau_g) \lt \aleph_0 \} $ then it seems that $S_*$ must be a normal subgroup of $S$.

(I have just noticed a comment by @ArturoMagidin which makes me more confident to make this claim. However the discussion in that location does not single out countably infinite groups in particular, and is heavily dependent on some literature references that are quite obscure. I would just like a simple, commonsense explanation, if that is feasible.)

Can anyone help me improve my rather hazy conception of the factor group ${S}/{S_*}$ and its relation to $S^* =\{g \in S \mid \text{card}(\sigma_g) \lt \aleph_0 \} $?

Answer 1

Check out this paper by Alperin/Covington/Macpherson and references there in (seems to be free from citeseer). They analyze the automorophism group of your mystery quotient, which should tell you a fair bit about the structure.

Answer 2

Just to clarify a few basic points.

The full symmetric group $S$ on a countably infinite set $X$ is indeed well-defined up to group isomorphism. (The same applies to sets $X$ with any given cardinality.)

$S_*$ is indeed a normal subgroup of $S$. It is easy to see that $g^{-1}hg \in S_*$ for all $g \in S$ and $h \in S_*$.

In fact it can be shown that $S$ has precisely four normal subgroups, $1$, $S$, $S_*$, and the the group $A_*$ of all even permutations in $S_*$. (Since the elements in $S_*$ move only finitely many points, they can be unambiguously defined as even or odd. But there is no meaningful way of attaching a parity to the elements of $S \setminus S_*$.) The group $A_*$ is simple.

I have no clear conception of the quotient group $S/S_*$ myself, so I cannot help you understand it intuitively! But it is a simple group.