# abelian finite groups - basic

G is the set of all subset of A. (For example - Say $A=\{1,2,3\}$ than $G=(\{1\},\{2\},\{3\},\{1,2\}...)$. ($A$ is at east two different elements).

the binary operation $*$ is intersection. I need to prove this is a group .

Identity element - will be the empty set.

Associative-Yes

Inverse element - Yes . I need that For each $g\in G$, there must be an element $g^{-1}\in G$ so that $g^{-1}*g=g*g^{-1}=e$. I think the empty set do that as well.

until here I hope I was right.

Now I want to show if this is finite or abelian.

Abelian - Yes- easy to show.

But now Im not sure if this is finite and how to show that...

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Don't you need any further assumptions on the set $A$? Otherwise $G$ will be infinite if $A$ is infinite. Furthermore you should rethink whether intersecting a set with the empty set gives you the same set again. – martin.koeberl Dec 27 '12 at 10:17
If $e$ is the identity element then $A*e=A$ but $A\cap\emptyset=\emptyset=e\neq A$ – Belgi Dec 27 '12 at 10:20
thats right. so what is the identity element? is it even a group??? – baaa12 Dec 27 '12 at 10:21
This is not a group when $*$ is the intersection. I would try the symmetric difference of two subsets as the operation. – Jyrki Lahtonen Dec 27 '12 at 10:22

If $e$ is the identity element then for all subsets $B$ of $A$: $B*e=B$ hence $B\cap e=B$ thus $B\subseteq e$.

But this is for all $B$ hence $e=A$ .

But for all subsets $B,C$ of $A$ we have $B\cap C\subseteq B$ hence if $B\neq A$ it have no inverse.

We conclude $G$ is not a group

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@JyrkiLahtonen - I corrected the second line, I think its ok now. thanks for the comment! – Belgi Dec 27 '12 at 10:30
sorry but what P(A) means? – baaa12 Dec 27 '12 at 10:31
@JyrkiLahtonen - I agree, I edited accordingly. thanks for commenting about it! – Belgi Dec 27 '12 at 10:33

Intersection in the power set of a set = the set of the set's subsets, does not make the power set into a group, as if the set has more than one element then there is no inverse for some element.

What you can do is to define the symmetric difference on the power set:

$$\forall\,H,K\in G=P(A)\,\,,\,\,H\Delta K:=(H\cup K)\setminus(H\cap K)=(H\setminus K)\cup(K\setminus H)$$

With the above operation the power set $\,G\,$ does become an (elementary) abelian group of order $\,2^n\,$.

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+1: This is the only (easy) way to turn the power set into a group that I know of. It may be that the real question is to show that the power set is not a group w.r.t. intersection. Let's wait for the OP to comment. – Jyrki Lahtonen Dec 27 '12 at 10:26