identity and inverse

**demode** · July 20th 2010, 03:36 PM

Here is another part of the question:

For part (c), this is what I've done so far:

suppose both e and é are identities of $\mathcal{P}(M)$ and suppose $x \in \mathcal{P}(M)$ . Therefore

xe=x .....(1)
xé=x .....(2)

Since $e,é \in \mathcal{P}(M)$ , we can set x=é in equation (1) and set x=e in equation (2)

ée=é
eé=e

Thus e and é are equal and the identity is uniqe.

Is this correct? I think the problem here is that I've shown that the identity of $(\mathcal{P}(M), \star)$ is unique but I have not shown that it must have an identity. Can anyone help?

For part (d), I did this:

Let $a,b,c \in \mathcal{P}(M_n)$ .
Suppose b and c are both inverses of a. Then $ab=e$ also $ac=e$ . Hence $ab=bc \iff b=c$ .

Again, I think my approach just shows the uniqueness, and not the existence of an inverse... I appreciate any help with this.

**Ackbeet** · July 21st 2010, 02:22 AM

You need to exhibit the identity. What does the identity do, in this case? Well, given any set $x \in \mathcal{P}(M),$ you have that $x\star e=e\star x=x,$ right? Now, $x$ and $e$ must be the same kind of object; that is, members of $\mathcal{P}(M).$ Which member of $\mathcal{P}(M)$ will do what you need?

**demode** · July 22nd 2010, 03:08 AM

Originally Posted by Ackbeet

You need to exhibit the identity. What does the identity do, in this case? Well, given any set $x \in \mathcal{P}(M),$ you have that $x\star e=e\star x=x,$ right? Now, $x$ and $e$ must be the same kind of object; that is, members of $\mathcal{P}(M).$ Which member of $\mathcal{P}(M)$ will do what you need?

I'm still confused... I think $\mathcal{P}(M) = \{ \emptyset , \{1 \} \}$ , so

$1 \star \emptyset = (\emptyset \cup 1) \backslash (\emptyset \cap 1)$ . Then what?

**Plato** · July 22nd 2010, 03:50 AM

Originally Posted by demode

I'm still confused... I think $\mathcal{P}(M) = \{ \emptyset , \{1 \} \}$ , so

$1 \star \emptyset = (\emptyset \cup 1) \backslash (\emptyset \cap 1)$ . Then what?

$A \star \emptyset = \left( {A\backslash \emptyset } \right) \cup \left( {\emptyset \backslash A} \right) = A \cup \emptyset = A$

**Swlabr** · July 22nd 2010, 04:19 AM

So, Plato has told you that the identity is the empty set. You now need to prove this this identity is unique, and you need to find the inverse of an element.

To prove the identity is unique you do not need to use this example - any identity of a set is unique. This is because if you have two identities $e_1$ and $e_2$ then clearly $e_1e_2 = e_1$ but also $e_1e_2=e_2$ . Thus they are equal.

To find the inverse, you want to find a set $B$ such that $A \cup B = A \cap B$ . Note that $|A \cup B| \geq \max\{|A|, |B|\}$ and $|A \cap B| \leq \min\{|A|, |B|\}$ . What does this tell you? (Remember you want $A \cap B = A \cup B$ ).

**demode** · July 25th 2010, 02:10 AM

Originally Posted by Swlabr

To find the inverse, you want to find a set $B$ such that $A \cup B = A \cap B$ . Note that $|A \cup B| \geq \max\{|A|, |B|\}$ and $|A \cap B| \leq \min\{|A|, |B|\}$ . What does this tell you? (Remember you want $A \cap B = A \cup B$ ).

I think $A \cup B = A \cap B$ iff $A=B$ , which means every element is its own inverse.

To prove that the inverse for each element is unique I let $A, B, C \in \mathcal{P}(M_n)$ , where C and B are inverses of A. Then

$AB=e$ and $AC=e$ . Therefore $AB=AC \iff B=C$ .

Is it correct?

**Swlabr** · July 26th 2010, 12:29 AM

Originally Posted by demode

I think $A \cup B = A \cap B$ iff $A=B$ , which means every element is its own inverse.

To prove that the inverse for each element is unique I let $A, B, C \in \mathcal{P}(M_n)$ , where C and B are inverses of A. Then

$AB=e$ and $AC=e$ . Therefore $AB=AC \iff B=C$ .

Is it correct?

Yes, that is correct.

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