# Group or not?

1. Jul 31, 2010

### roam

1. The problem statement, all variables and given/known data

Let $$n \geq 1$$ be a positive integer and let $$M_n = \{ 1,...,n \}$$ be a set with n elements. Denote by $$\mathcal{P} (M_n)$$ the set of all subsets of Mn. For example $$\mathcal{P} (M_2) = \{ \{ \emptyset \}, \{ 1 \}, \{ 2 \}, \{ 1,2 \} \}$$.

Show that $$C=(\mathcal{P} (M_n) , \cap)$$ and $$U=(\mathcal{P} (M_n) , \cup)$$ each has an identity. Decide whether C and U are groups.

3. The attempt at a solution

For $$C=(\mathcal{P} (M_n) , \cap)$$, let $$a,b,c \in \mathcal{P} (M_n)$$

• Associativity: $$(a \cap b) \cap c = a \cup (b \cap c)$$ $$\checkmark$$
• Identity: is the empty set => $$a \cap \emptyset = a$$ $$\checkmark$$
• Inverse: I can't see what's the inverse of this group! for an element a we need an inverse b such that $$a \cap b = \emptyset$$. I think this is only true when a & b are completely distinct but I'm not sure...

Similarly $$U=(\mathcal{P} (M_n) , \cup)$$ satisfies the associativity and I think its identity is also $$\emptyset$$. But what is the inverse??

I need help finding the inverses, and please let me know if the rest of my working is correct.
Any help is really appreciated.

Last edited: Jul 31, 2010
2. Jul 31, 2010

### Hurkyl

Staff Emeritus
If you don't understand it for all n, try special cases. Maybe continue with your n=2 example. That looks like a very small finite set -- you should be able to answer any particular question at all with a few seconds effort.

3. Jul 31, 2010

### Dick

First try and think clearly. I don't think a intersect {} is a, if that's what you mean to say. But you are right that there is a problem with inverses. They aren't groups.

4. Aug 2, 2010

### roam

Oops, I made a mistake! So, for $$C=(\mathcal{P} (M_n) , \cap)$$

The identity for each element is itself, right? Because $$a \cap a = a$$. Therefore the inverse for every element is also itself. So it's a group?

And for $$U=(\mathcal{P} (M_n) , \cup)$$

$$A \cup \emptyset = A$$. Therefore $$\emptyset$$ is the identity.

But what is the inverse? I need "b" such that $$a \cup b = \emptyset$$. Even if there is no inverse, I guess I have to give some kind of explanation

5. Aug 2, 2010

### HallsofIvy

No, there is a unique identity for a group, not a different "identity" for each member. Is there a single set, x, such that $a\cap x= a$ for every a? (Look at $M_n$ itself.)

If set A had and inverse, B, say, then you would have to have $A\cup B= \emptyset$. How do the number of members of A and $A\cup B$ compare?

6. Aug 3, 2010

### roam

Yes, there is a single set, whenever n=1 then Mn has only one element. So the inverse of 1 is 1, it's also the identity. But this only true when n=1! So, $$C=(\mathcal{P} (M_n) , \cap)$$ not a group?

The only member of $$A\cup B$$ is the empty set whereas the number of members of A is greater than or equal to 1. But what does that prove?

7. Aug 3, 2010

### HallsofIvy

??? For any subset, A, of Mn, $A\cap M_n= A$.

It proves that the can't be equal! And since they must be the same set in order that B be "inverse" to A, there cannot be an inverse.