Question

Let \(X\) be a set and \(2^{X}\) denote the power set of \(X\). Define a binary operation \(\,\Delta\,\) on \(2^{X}\) by \(A \Delta B = (A-B)\cup(B-A)\) (symmetric difference). Let \(H=(2^{X},\Delta)\). Which of the following statements about \(H\) is/are correct?

• A. \(H\) is a group.
• B. Every element in \(H\) has an inverse, but \(H\) is NOT a group.
• C. For every \(A\in 2^{X}\), the inverse of \(A\) is the complement of \(A\).
• D. For every \(A\in 2^{X}\), the inverse of \(A\) is \(A\).

Hint:

Think of \((2^X,\Delta)\) as a vector space over \(\mathbb{F}_2\): sets correspond to 0–1 indicator vectors, and \(\Delta\) is bitwise XOR. The identity is \(\varnothing\), and every vector is its own inverse.

Answer 1

The Correct Option is A

To determine which statement about \( H = (2^X, \Delta) \) is correct, we need to understand the properties of symmetric difference and the structure of a group.

The symmetric difference \( A \Delta B \) is defined as \( (A - B) \cup (B - A) \) and it possesses several properties that are useful for our analysis:

1. It is commutative: \( A \Delta B = B \Delta A \).
2. It is associative: \( (A \Delta B) \Delta C = A \Delta (B \Delta C) \).
3. The identity element in this operation is the empty set \( \emptyset \), because \( A \Delta \emptyset = A \).
4. Every element \( A \in 2^X \) is its own inverse: \( A \Delta A = \emptyset \).

Let's analyze each statement given the properties above:

• \( H \) is a group: Since the symmetric difference operation \(\Delta\) is associative, commutative, has an identity element \( \emptyset \), and each element is its own inverse, \( H \) satisfies all the group axioms. Thus, this statement is correct.
• Every element in \( H \) has an inverse, but \( H \) is NOT a group: This statement is incorrect because \( H \), as described, satisfies the group axioms due to having associativity, an identity element, and every element having an inverse.
• For every \( A \in 2^{X} \), the inverse of \( A \) is the complement of \( A \): This statement is incorrect. The inverse of \( A \) under symmetric difference is actually \( A \) itself, not its complement.
• For every \( A \in 2^{X} \), the inverse of \( A \) is \( A \): This statement is correct because \( A \Delta A = \emptyset \), and thus \( A \) is its own inverse.

Therefore, the correct statement is that \( H \) is a group.