permutation group, Lie group Let $S$ be any set, and denote by $G$ the collection of all subsets of $S$. For $A, B \in G$ let be $AB = (A - B) \cup (B - A)$. I know how to show that this set $G$, with this product operation is a group. How would I go about finding all sets $S$ for which this group or subgroup is the permutation group of some set? How would I go about finding all $S$ for which $G$ is the underlying group of some Lie group of dimension $> 0$?
 A: You're using the symmetric difference operation on the set of subsets of $\;S\;$ , namely $\;\mathcal P(X)\;$ . This is (an abelian, of course) $\;2$- elementary group, so your question is: when is a $\;2$- elementary group a permutation group on some set? 
The answer: only for $\;S_1\,,\,\,S_2\;$ , as in any other case $\;|S_n|\;$ is not a power of two.
A: Note that $AB$ is the set of all points lying in exactly one of $A$, $B$. So, $A(BC)$ is the set of all points lying in an odd number of $A, B, C$. Associativity follows immediately. The identity is the empty subset of $S$. The inverse of $A$ is $A$.
For $S$ the empty set, the corresponding group $G$ $($the identity group$)$ is isomorphic with a permutation group $($namely, that on a set with $1$ element$)$. For $S$ a set with exactly one element, the corresponding group $($with just two elements$)$ is isomorphic with a permutation group $($namely, that on a set with $2$ elements$)$. These are the only two. This follows from the fact that the groups $G$ constructed here are all abelian, while all the remaining permutation groups are non-abelian.
It is well known that every group is a subgroup of a permutation group. These included.
For no $S$ can the group $G$ constructed from $S$ be a Lie group of dimension $> 0$. Indeed, for every group $G$ here constructed, we have, for every $g \in G$, $g^2 = e$. Let, for contradiction, one of these, $G$ be a Lie group of dimension $>0$. Then $($since $\dim > 0$$)$ there exists a nonzero tangent vector at $e$, and so a nonzero left-invariant vector field $X$ on $G$. Let $\gamma$ be its integral curve, with initial point $e$. Choose $t$ such that $\gamma(t) \neq e$. But $\gamma(t) = \gamma(t/2)\gamma(t/2) = e$, where we used $g^2 = e$ in the second equality. $($Why is the first equality true?$)$ This is a contradiction.

A few remarks are in order. I could envision someone trying to use the following argument to show that only the first few $G$'s are isomorphic to permutation groups. For $S$ a set with $n$ elements, the number of elements of $G$ is $2^n$, while the number of elements of the permutation group on a set with $m$ elements is $m!$. But $2^n$ is the factorial of something only for $n=1$ and $2$. Unfortunately, this argument, as it stands, does not work for the case of $S$ an infinite set. $($But one might imagine that there is a cardinality argument that would replace it. Is there?$)$
Is there a Lie group of dimension $> 0$ based on $G$? Although technically this is "difficult" the idea is pretty simple. Lie groups all "look like $\mathbb{R}^n$" near the identity $e$, and $G$ sure does not look this way. Indeed, even "$G$ is discrete, while no Lie groups of $\dim > 0$ are" is not far from the answer. It is not complete, because "discrete" is a topological property, and $G$ does not come with a topology. But, on the other hand, we have the fact that no finite group $G$ underlies a Lie group of $\dim > 0$. To prove this fact, note that since charts are $1$-$1$ and map to an open subset of $\mathbb{R}$, the only candidate, for finite $G$, is $G$ the empty set. But the empty set is not a group.
