Is the following structure a group?

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. –  Matt Pressland Mar 26 '12 at 12:10
So identity is $(\emptyset, \{1,2,3,4\})$? –  Mariano Mar 26 '12 at 12:36
Yes, although I'd write $(\varnothing,S)$ to save space! This now means that not every element is invertible. –  Matt Pressland Mar 26 '12 at 12:59
It is not a group, inverses in general do not exist. –  André Nicolas Mar 26 '12 at 12:59
Do you mean, there aren't invertible elements at all? –  Mariano Mar 26 '12 at 13:42

Perhaps a better way to understand this is that it is a product of two magmas on $\mathcal{P}(S)$: the two coordinates do not interact with one another. On one of the coordinates the operation is the symmetric difference, and on the other the operation is the intersection. They are both associative and commutative operations, so we have the product of two commutative semigroups.
That means that the product, $(\mathcal{P}(S),\triangle)\times(\mathcal{P}(S),\cap) = (\mathcal{P}(S)\times\mathcal{P}(S), \triangle\times\cap)$ is a group if and only if it is a group in "each coordinate".
$(\mathcal{P}(S),\triangle)$ is in fact a group: the empty set is the identity element, as $A\triangle \varnothing = A$ for all $A\in\mathcal{P}(S)$, and every element is its own inverse. (In fact, this is the "additive" structure of the boolean ring on $\mathcal{P}(S)$).
$(\mathcal{P}(S),\cap)$ is a commutative monoid (a commutative semigroup with identity): $S$ itself is the identity, since $B\cap S = B$ for all $B\in \mathcal{P}(S)$.
However, it is not a group; note that the identity element is the "largest" element of $\mathcal{P}(S)$ (ordered by inclusion), but the operation is decreasing: for all $A$ and $B$, $A\cap B\subseteq A$; that means that if $A\in\mathcal{P}(S)\subsetneq S$, then $A\cap B\subsetneq S$ for all $B\in\mathcal{P}(S)$; thus, elements other than $S$ cannot have $\cap$-inverses.