# 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.

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How do you add sets? –  Tobias Kildetoft Mar 22 '13 at 16:15
math.stackexchange.com/questions/265834/… i read this post and i really gt confused .. –  art Mar 22 '13 at 16:15
thanks in advance for any comment ! –  art Mar 22 '13 at 16:15
If with $A+B$ you mean that union of sets, then the answer is no. The unity if exists can only by $\emptyset$, but then there does not have to exist an inverse element. Indeed, if $A\neq \emptyset$ then you cannot find $B$ such that $A+B = \emptyset$. –  Ilya Mar 22 '13 at 16:17
Please use $\LaTeX$ –  Avi Steiner Mar 22 '13 at 16:17

If you use symmetric difference $A\Delta B = (A\cup B) - (A\cap B)$, then yes.

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Since you get an abelian group (where $-$ will have a meaning), it might be a good idea to not use $-$ for set difference in the definition. –  Tobias Kildetoft Mar 22 '13 at 16:28
+1 Adding for whoever may be interested: the symmetric difference makes P(E) not only a group but in fact a $2$− elementary one: every non-trivial element is its own inverse or what's the same: $$AΔA=∅$$ –  DonAntonio Mar 22 '13 at 16:29
@DonAntonio: thanks, better now (+1)! –  Ilya Mar 22 '13 at 16:33
thanks for your help and especially the user for the better picture of my question!i know that we are in a group so we have to respect the rules of the group but i am not familiar with latex:(.Anyway,you are very helpfull all of you! –  art Mar 22 '13 at 17:25

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:

1. 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 $s$?)

2. 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 $a=\emptyset$?)

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