I started doing algebra exercises before the course starts, and I'm befuddled with following problem, where it is asked to show if the operation is associative and commutative and whether it has an identity element and inverse elements.

Let set $X\neq\emptyset$ and $\mathcal{P}(X)$ denote all its subsets. Let $*$ be binary operation on $\mathcal{P}(X)$ which is defined with following equation:

$A*B = A\cup B$.

It is commutative: $A\cup B = B \cup A$

And also associative for $C \in \mathcal{P}(X)$: $A\cup (B\cup C ) = (A\cup B)\cup C$

Identity element $e = \emptyset$, for every set holds $\emptyset$. So, $A \cup e = \emptyset^{^*}$

Now I'm not sure about inverse element such that $A \cup B = e$

Is it possible in advanced algebra ?

I thought of something like $B:= e \setminus A$

*edit: Identity element equation should be $A \cup e = A$


Except for $e=\{\}$ no set has inverse.

If you take any nonempty set $A$ then union of $A$ with any set $B$ is nonempty. So $A$ can not have inverese. So $e$ is only one with inverse.

The power set $\mathcal{P}(X)$ with operation $*$ has the same algebraci structure as the set of natural numbers $\mathbb{N}_{0}$ with operation $+$, namely monoid.

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