# 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

1. Is it valid to have multiple non-unique identity elements in a group for each item in the collection?
2. 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.

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It is a direct consequence of the group axioms that the identity element and each inverse is unique. –  wildildildlife Mar 14 '11 at 20:58
are you sure that's what the exercise says? (i.e. find a set $X$ and $\mathcal{T}\subseteq\mathcal{P}(X)$ that is a group under union) I'm pretty sure this cant be done (except the trivial group). The identity would have to be a subset of every element of the group, therefore the inverse would have to make the elements smaller somehow... –  yoyo Mar 14 '11 at 21:24

If you have a group, then the identity element must be unique: if $e$ and $e'$ both satisfy the identity condition, $ex = xe = x$ for all $x$ and $e'x = xe' = x$ for all $x$, then you have $ee' = e$ (because $e'$ is an identity) and $ee' = e'$ (because $e$ is an identity), so $e=ee'=e'$, proving that in fact $e=e'$. So the identity element must be unique.

Likewise, if you have a group, then the inverse of an element must be unique: if both $y$ and $z$ are inverses of $x$, so $xy=yx = e$ and $xz=zx=e$, then $$z = ze = z(xy) = (zx)y = ey = y$$ so $z=y$.

Note, however, that while any $A'\subseteq A$ satisfies $A'\cup A = A$, this does not prove by itself that you have more than one identity element: you would need the same $A'$ to work for every subset $B$ of $X$ for $A'$ to be the identity; if you need to "change" the $A'$, then you are not constructing an identity for this possible group.

So the first step is to figure what subset(s) $E$, if any, satisfy that $E\cup A = A$ for all $A\subseteq\mathcal{X}$. If you can find more than one that works for all $A$, then you know it's not a group (because in a group, the identity element must be unique).

But if you can only find a single $E\subseteq\mathcal{X}$ that can work, then you know that one would have to be the identity, and now you can go ahead and see if you can find inverses for every element. If you can find an $A\subseteq \mathcal{X}$ that has at least two distinct inverses, then you would conclude that you cannot get a group (because inverses are unique in a group); or if you can find at least one $A\subseteq\mathcal{X}$ that has no inverse, then you would also know that this is not a group (because in a group, every element must have an inverse).

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Thanks! Illuminating as always. –  JasonMond Mar 15 '11 at 0:50
Any particular reason for the downvote? –  Arturo Magidin Mar 15 '11 at 13:40

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.

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