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Let $S=\{1,2,3,4\}$ and the operation $*$ so defined in $\mathcal{P}(S)\times\mathcal{P}(S)$: $$(A,B)*(C,D) = (A \Delta C, B \cap D) \quad\forall (A,B)(C,D)\in\mathcal{P}(S)\times\mathcal{P}(S)$$

  1. What type of structure is this?
  2. Determine the set $U(\mathcal{P}(S)\times\mathcal{P}(S))$ of invertible elements;
  3. What of the following is a closed part in $\mathcal{P}(S)\times\mathcal{P}(S)$ $$L = \{\{1,3\}\} \times\mathcal{P}(S))$$ $$M = \mathcal{P}(S)) \times\{\{1,3\}\}$$ $$N = \{\{1,3\}, \emptyset\} \times \mathcal{P}(S))$$

This is associative, commutative and it has an identity, cause: $$\exists (E, E') | \forall (A,B),\quad(A,B)*(E, E') = (A,B)$$ $$(A \Delta E, B \cap E')=(A,B)$$ $$A \Delta E=A\Leftrightarrow E=\emptyset$$ $$B \cap E'=B\Leftrightarrow E'=B$$ I don't need to test it twice, cause the operation is commutative. So $(E, E') = (\emptyset, B)$, is correct? If correct I can search for invertible elements, and so: $$(I, I')*(A, B)=(\emptyset, B)$$ $$(I \Delta A, I' \cap B)=(\emptyset,B)$$ $$I \Delta A=\emptyset\Leftrightarrow I=A$$ $$I' \cap B=B\Leftrightarrow I'=B$$ Is correct? If correct, this means that every element is invertible and so the structure is commutative group. Am I right? What about last question?

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    You haven't found an identity element - it can't depend on $A$ or $B$ because it needs to work for all elements. There is an identity though. It's also not true that $I'\cap B=B\implies I'=B$, although the other implication is true.2012-03-26
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    So identity is $(\emptyset, \{1,2,3,4\})$?2012-03-26
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    Yes, although I'd write $(\varnothing,S)$ to save space! This now means that not every element is invertible.2012-03-26
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    It is not a group, inverses in general do not exist.2012-03-26
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    Do you mean, there aren't invertible elements at all?2012-03-26
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    @Mariano: No, it means not every element is invertible. For example, the identity is certainly invertible.2012-03-26
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    No, @Andre means that there are some elements which are not invertible. For example, for any $A\subseteq S$, the element $(A,\emptyset)$ is not invertible, since it is impossible to find another subset $D\subseteq S$ such that $D\cap \emptyset = S$.2012-03-26
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    And so what's the invertible elements apart the identity?2012-03-29

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