Power Set Operations and Group Theory

The question asks us to examine whether the power set \(P(X)\) of a non-empty set \(X\) forms a group under two defined operations: intersection (⋆) and symmetric difference (Δ).

Recall that a set \(G\) with an operation \( * \) forms a group if it satisfies four properties:

1. Closure: For all \(a, b \in G\), \(a * b \in G\).
2. Associativity: For all \(a, b, c \in G\), \((a * b) * c = a * (b * c)\).
3. Identity Element: There exists an element \(e \in G\) such that for all \(a \in G\), \(a * e = e * a = a\).
4. Inverse Element: For all \(a \in G\), there exists an element \(a^{-1} \in G\) such that \(a * a^{-1} = a^{-1} * a = e\).

Operation ⋆ (Intersection) Analysis

The operation is defined as \(A \sstarf B = A \cap B\) for \(A, B \in P(X)\).

• Closure: For any subsets \(A, B \subseteq X\), their intersection \(A \cap B\) is also a subset of \(X\). So, \(A \cap B \in P(X)\). Closure holds.
• Associativity: The intersection of sets is associative: \((A \cap B) \cap C = A \cap (B \cap C)\). Associativity holds.
• Identity Element: We need an element \(E \in P(X)\) such that \(A \cap E = E \cap A = A\) for all \(A \in P(X)\). For \(A \cap E = A\) to be true for all subsets \(A\), \(A\) must be a subset of \(E\). The only subset of \(X\) that contains all other subsets of \(X\) is \(X\) itself. So, the identity element would be \(X\). Since \(X\) is non-empty, \(X \in P(X)\). Thus, \(X\) is the identity element for the intersection operation on \(P(X)\). Identity holds.
• Inverse Element: For any \(A \in P(X)\), we need \(A^{-1} \in P(X)\) such that \(A \cap A^{-1} = X\) (the identity element). If \(A\) is a proper non-empty subset of \(X\) (i.e., \(A \ne \emptyset\) and \(A \ne X\)), then \(A \cap A^{-1}\) will always be a subset of \(A\). For \(A \cap A^{-1}\) to equal \(X\), it must be that \(A = X\) and \(A^{-1} = X\). Therefore, only \(X\) has an inverse (itself) under this operation (as \(X \cap X = X\)). For any other element \(A \ne X\) in \(P(X)\), there is no subset \(A^{-1}\) such that \(A \cap A^{-1} = X\). For example, if \(X = \{1, 2\}\), \(P(X) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}\). The identity is \(\{1, 2\}\). Let \(A = \{1\}\). We need \(\{1\} \cap A^{-1} = \{1, 2\}\). No subset \(A^{-1}\) satisfies this.

Since not every element in \(P(X)\) has an inverse under the intersection operation, \(P(X)\) is not a group under ⋆.

Operation Δ (Symmetric Difference) Analysis

The operation is defined as \(A \Delta B = (A \cup B) \setminus (A \cap B)\) for \(A, B \in P(X)\). This is also known as \(A \oplus B\) in the context of vector spaces over \(Z_2\).

• Closure: For any subsets \(A, B \subseteq X\), their union \(A \cup B\) and intersection \(A \cap B\) are subsets of \(X\). Their difference \((A \cup B) \setminus (A \cap B)\) is also a subset of \(X\). So, \(A \Delta B \in P(X)\). Closure holds.
• Associativity: The symmetric difference is associative: \((A \Delta B) \Delta C = A \Delta (B \Delta C)\). This can be verified using element-wise arguments (an element is in the symmetric difference of three sets if and only if it is in exactly one or exactly three of the sets) or set identities. Associativity holds.
• Identity Element: We need an element \(E \in P(X)\) such that \(A \Delta E = E \Delta A = A\) for all \(A \in P(X)\). Let's test the empty set \(\emptyset\): \(A \Delta \emptyset = (A \cup \emptyset) \setminus (A \cap \emptyset) = A \setminus \emptyset = A\). Since \(A \Delta \emptyset = A\) for all \(A \in P(X)\), the empty set \(\emptyset\) is the identity element for the symmetric difference operation. \(\emptyset \in P(X)\). Identity holds.
• Inverse Element: For any \(A \in P(X)\), we need \(A^{-1} \in P(X)\) such that \(A \Delta A^{-1} = \emptyset\) (the identity element). \(A \Delta A^{-1} = (A \cup A^{-1}) \setminus (A \cap A^{-1}) = \emptyset\). This equation holds if and only if \(A \cup A^{-1} = A \cap A^{-1}\). This equality is only possible if \(A \cup A^{-1}\) and \(A \cap A^{-1}\) are the same set, which implies that \(A\) and \(A^{-1}\) must be equal, i.e., \(A = A^{-1}\). Let's check if \(A\) is its own inverse: \(A \Delta A = (A \cup A) \setminus (A \cap A) = A \setminus A = \emptyset\). Thus, every element \(A \in P(X)\) is its own inverse under the symmetric difference operation. Inverse holds.

Since \(P(X)\) satisfies closure, associativity, identity, and inverse properties under the symmetric difference operation, \(P(X)\) is a group under Δ.

Conclusion

Based on the analysis:

Therefore, \(P(X)\) is a group under Δ, but not under ⋆.