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

Step 1: Identify the operation.
The given operation is the symmetric difference: \[ A\Delta B=(A\cup B)-(A\cap B) \] It satisfies \(A\Delta A=\varnothing\) and \(A\Delta\varnothing=A\). Step 2: Verify group axioms on \(2^X\).
Closure: \(A\Delta B\subseteq X\Rightarrow A\Delta B\in 2^X\).
Associativity: Symmetric difference is associative: \[ A\Delta(B\Delta C)=(A\Delta B)\Delta C \] (this follows from characteristic functions mod 2, or standard set identities).
Identity: \(\varnothing\) since \(A\Delta\varnothing=A\) for all \(A\).
Inverse: For every \(A\), \(A\Delta A=\varnothing\Rightarrow A^{-1}=A\).
Also \(\Delta\) is commutative, so \(H\) is an abelian group.
\(\Rightarrow\) (A) True. (D) True (each set is its own inverse).
Step 3: Check the remaining statements.
(B) claims every element has an inverse but \(H\) is not a group—contradicted by Step 2. \(\Rightarrow\) False.
(C) says the inverse of \(A\) is its complement \(A^{c}\). But \(A\Delta A^{c}=X\neq\varnothing\) (unless \(X=\varnothing\)). Hence the complement is not the identity result, so not the inverse. \(\Rightarrow\) False. \[ \boxed{\text{Correct Options: (A) and (D)}} \]