TITLE: Does the Symmetric difference operator define a group on the powerset of a set?
QUESTION [8 upvotes]: $G$ is the set of all subsets of a set $A$, under the operation of $\triangle\;$: Symmetric Difference of sets.
$A$ has at least two different elements.
I need to check if this is a group, and if it does to show if the group is abelian and/or finite.
Associativity - easy from Symmetric Difference.
Identity element - empty group.
Inverse element - each element is inverse to itself.
Am I right?
abelian?
finite?

REPLY [4 votes]: This is intended to be a comment to amWhy's answer. However, my comment is too long to fit in the comment box, plus I suspect my comment might be of sufficient interest to some people here to warrant a more public display.
Carolyn Bean (reference below) proved a number of interesting group-theoretic results related to the symmetric difference operation, a few of which I will state here. Let $X$ be a fixed set and let ${\cal P}(X)$ be the set of all subsets of $X.$ Bean proved that $\Delta$ can be characterized as the unique group operation $*$ on ${\cal P}(X)$ such that $A*B \subseteq A \cup B$ for all $A,$ $B \in {\cal P}(X).$ Define the co-symmetric difference operation $\nabla$ on ${\cal P}(X)$ by $A \nabla B = X - (A \Delta B)$ for all $A,$ $B \in {\cal P}(X).$ Then one can show that $\left(\,{\cal P}(X),\;\nabla\right)$ is also an abelian group. Bean proved that $\nabla$ can be characterized as the unique group operation $*$ on ${\cal P}(X)$ such that $A \cap B \subseteq A*B$ for all $A,$ $B \in {\cal P}(X).$ Bean additionally proved that $\left(\,{\cal P}(X),\;\Delta,\;\cap\right)$ and $\left(\,{\cal P}(X),\;\nabla,\;\cup\right)$ are isomorphic commutative rings with identity.
Carolyn Bean, Group operations on the power set, Journal of Undergraduate Mathematics 8 #1 (March 1976), 13-17.