I am just curios about about is that Relational Algebra under ($\times$) cartesian product count as a group ?
Since, let $A, B, C$ be three relations (tables) $A \times (B \times C) = (A \times B) \times C$
But the thing is $I$ don't know if there exist an Identity relation (table) $I$, such that $A \times I = A$ and there exist a relation (table) X (which X is inverse relation (table) of A) denote $A \times X = X \times A = I$
My guess is that I = $\phi$ which is the empty relation (table) but I am not sure if this holds.
So, is there really exists an $I$ that make the Relational Algebra under ($\times$) cartesian product a group.
I cannot see that looking at relations gives anything more than looking at sets in general, so let's just consider general sets.
As Arturo Magidin already has written, we do not have a group, since the element $\emptyset$ satisfying $X \times \emptyset = \emptyset = \emptyset \times X$ for any set $X$ is not invertible.
Also, since $|X \times Y| = |X| |Y|$ when $X$ and $Y$ are finite ($|X|$ denotes the number of elements in $X$), we cannot solve for example $A \times X = B$ if $|A| = 3$ and $|B| = 5$. There are no sets with a fractional number (e.g. $5/3$) of elements.
Some other properties are valid, though, at least under equivalence:
We can identify $(X \times Y) \times Z$ and $X \times (Y \times Z)$ by identifying $((x,y), z)$ with $(x,(y,z))$, thereby getting associativity.
We also have a multiplicative identity if we introduce some special element $\perp$ and identify $\{\perp\} \times X$ and $X \times \{\perp\}$ with $X$.
This gives use a monoid.