Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
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

2 Answers 2

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

share|improve this answer
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 $a=\emptyset$?)

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.