Defining sets and union as a group I came across an exercise in this book where the question was to define a collection of sets and the union operator as a group. The two parts of the question were to (1) find the identity element and (2) find the inverse element.
Assuming some collection of sets that's closed over set union $\mathcal{X}$ and some element $A \in \mathcal{X}$, my answer to the first question was that any set $A' \subseteq A$ and $\emptyset$ could be the identity element. But I run into trouble if I try to build an inverse out of this. For example, there's no way to obtain $\emptyset$ from a non-empty $A$ using union. And it's obvious that if uniqueness of identity and inverse is a requirement for defining groups, then for any $A$, $A$ is also its own identity and inverse.
My question is


*

*Is it valid to have multiple non-unique identity elements in a group for each item in the collection?

*Is it valid to have a unique inverse element for each item in the collection even if the identity elements are not unique?


Judging by what I read on wikipedia, uniqueness doesn't seem to be a criterion.

Let $(\mathcal{X}, \cup, E, -)$ be a group. Take arbitrary $A\in \mathcal{X}$. $A\cup E = E$, then $A\supseteq E$. $A\cup (-A) = E$, then $A\subseteq E$. Then $A=E$. This group is trivial.
$\cup$ is idempotent. Any idempotent group is trivial. Proof. Let $(G, +, 0, -)$ be a group. Take arbitrary $g\in G$. Then $g+g=g, -g+g+g=-g+g, 0+g=0, g=0$. Qed.
