a powerset define a group? Let $P(E)$ denote the power set of a set $E$: the set of subsets of $E$. Does the operation $A\cap B$ define a structure of group?
By denition, a group $G$ is a set with an operation $g.h$ (formally, a function $G\times G\rightarrow G$), with the following properties:
The property of the identity: for all $g\in G$, $e.g = g.e = g$.
Existence of inverses: for all $g\in G$ there is $h\in G$ (the inverse of $g$) such that $h.g = g.h = e$.
Associativity: for all $x,y,z\in G$, $x.(y.z) = (x.y).z$.
If the operation $g.h$ is commutative, that is, if $g.h = h.g$ for all $h,g\in G$ then the group is said to be abelian.
Τhanks in advanced!
 A: Here's a quick (hinted, rather than explicitly answered, since it makes a good lesson) walkthrough to answering the original question of whether the intersection operation defines a group:


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*If the operation has an identity — that is, an element $e\in P(E)$ (or, equivalently, $e\subseteq E$) such that for all sets $s\subseteq E$, $e\bigcap s=s$ — then what must that identity be?  (hint: try taking $s=E$; then your $e$ must satisfy $e\bigcap E=E$.  What is $s\bigcap E$ for any set $s$?)

*Given the identity $e$ that you found in step 1, can you show that for every $a$ there's a $b$ such that $a\bigcap b=e$?  Or can you find an $a$ for which $a\bigcap b$ can never be $e$ for any $b$?  (Hint: what happens if $a=\emptyset$?)
A: If you use symmetric difference $A\Delta B = (A\cup B) - (A\cap B)$, then yes. 
A: The operation is closed, associative and has identity the whole set $E$. But for $A$ a proper subset of $E$, we get $A\cap B\ne E$ for every subset $B$. So no, $(P(E),\cap)$ doesn't define a group structure: it's just a (commutative, non-cancellative) monoid with identity $E$ (which is also the only invertible element).
