We are tasked with proving that $S$, the power set of a set $X$, forms a group under the operation of symmetric difference. Let's denote the symmetric difference operation as $\triangle$.

1. Closure

To show that $S$ is closed under symmetric difference, we need to prove that for any two subsets $A, B \in S$, their symmetric difference $A \triangle B$ is also a subset of $X$, i.e., $A \triangle B \in S$.

The symmetric difference of two sets $A$ and $B$, denoted $A \triangle B$, is defined as: $A \triangle B = (A \setminus B) \cup (B \setminus A)$ This is the set of elements that are in either $A$ or $B$, but not in both.

• Since $A, B \subseteq X$, both $A \setminus B$ and $B \setminus A$ are subsets of $X$.
• The union of subsets of $X$ is also a subset of $X$, so $A \triangle B \subseteq X$.

Thus, $A \triangle B \in S$, and closure holds.

2. Associativity

We need to show that the symmetric difference operation is associative, i.e., for any subsets $A, B, C \in S$, we have: $(A \triangle B) \triangle C = A \triangle (B \triangle C)$

Using the properties of symmetric difference, we can verify this:

• $A \triangle B$ can be thought of as the set of elements that belong to exactly one of $A$ or $B$, and similarly for $B \triangle C$, and $(A \triangle B) \triangle C$.

A property of the symmetric difference is that it is associative: $(A \triangle B) \triangle C = A \triangle (B \triangle C)$ This can be verified by checking the set memberships on both sides, but the key is that symmetric difference is well-known to satisfy associativity.

3. Identity Element

We need to find an identity element $E \in S$ such that for any subset $A \in S$, we have: $A \triangle E = A$ The symmetric difference with the empty set $\emptyset$ is the identity element: $A \triangle \emptyset = A \quad \text{for all subsets} \, A \subseteq X$ This follows from the fact that $A \setminus \emptyset = A$ and $\emptyset \setminus A = \emptyset$, so: $A \triangle \emptyset = A$ Therefore, the empty set $\emptyset$ serves as the identity element for the symmetric difference operation.

4. Inverse Element

We need to show that for each subset $A \in S$, there exists an inverse element $A^{-1} \in S$ such that: $A \triangle A^{-1} = \emptyset$ In fact, the inverse of a subset $A$ under symmetric difference is $A$ itself: $A \triangle A = \emptyset$ This is because the symmetric difference of a set with itself is the empty set (every element in $A$ is paired with itself, so no element remains in the difference).

Thus, every element $A \in S$ is its own inverse.

Conclusion

Since the symmetric difference satisfies the closure, associativity, identity, and inverse axioms, $(S, \triangle)$ is indeed a group.

Questions for further exploration:

1. How does the concept of symmetric difference extend to other algebraic structures like vector spaces or fields?
2. Can symmetric difference be defined on other sets (e.g., numbers, strings) and still form a group?
3. What is the significance of the empty set being the identity element in this group?
4. What are some real-world applications where symmetric difference is useful in group theory?
5. How would we prove that the group is abelian?

Tip:

Symmetric difference is a powerful operation in set theory, and it can often be used to simplify problems involving set relations. It's also particularly useful in areas like coding theory and cryptography!