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Let $A$ be a non empty set. Let $P(A)$ denote the power set of $A$.

$P(A)$ can be given a group structure in multiple ways.

1. Using Disjoint union a group operation Source

may be worth noting in this context that the power set of $X$ is a group under the operation of disjoint union, with the empty set as identity. if the complement of $A \subset X$ is $A'$ then the disjoint union of $A$ and $B$ is: $$ A \circ B = (A \cap B') \cup (A' \cap B) $$ proving associativity is a nice exercise, though it follows much more simply if the disjoint union is viewed as ordinary addition (mod 2) in $\mathbb{F}_2^X$

2. Using Symmetric Difference as Group Operation Source

The power set $G$ of any set $A$ becomes an abelian group under the operation of symmetric difference:

  • Why abelian? Easy to justify, just use the definition in $(1)$ above: it's defined in a way that $g_1 \triangle g_2$ means exactly the same set as $g_2 \triangle g_1$, for any two $g \in G$.

  • As you note, the symmetric difference on $G$ is associative, which can be shown using the definition in $(1)$, by showing for any $f, g, > h \in G, (f\; \triangle\; g) \triangle \;h = f\;\triangle\; (g > \;\triangle\; h)$.

  • The empty set is the identity of the group (it would be good to justify this this, too), and

  • every element in this group is its own inverse. (Can you justify this, as well? Just show for any $g_i \in G, g_i\;\triangle \; g_i = > \varnothing$).

Question: Can we define a ring structure on power set of $A$.

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I believe the "disjoint union" alluded to by David Holden means "disjunctive union" which is a synonym for the symmetric difference. So, you aren't listing multiple ways, really.

Can we define a ring structure on power set of A.

As the wiki clearly states: The power set of any set becomes a Boolean ring with symmetric difference as the addition of the ring and intersection as the multiplication of the ring.

This is the standard ring structure on the powerset of a set. It may have other ring structures, but none would arise so naturally as this one, and would probably not have meaningful connection to the fact the underlying set is a powerset.

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  • $\begingroup$ This is offtopic but what do people(mathematician) mean when they say that this "structure arises naturally". Is it related to category theory or it is just an expression. $\endgroup$ – StammeringMathematician Oct 13 '18 at 15:53
  • $\begingroup$ @StammeringMathematician There is a precise meaning in category theory, but here I just mean it as an expression of natural language. I just mean that when looked at from a certain perspective, it's fairly obvious, and there aren't really any other competing structures, so it jumps out at us. $\endgroup$ – rschwieb Oct 13 '18 at 16:01

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