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Official PYQ · 2023

GATE CS 2023 Q41 — Monoids and Groups

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Let XX be a set and 2X2^{X} denote the powerset of XX.

Define a binary operation Δ\Delta on 2X2^{X} as follows:
AΔB=(A−B)∪(B−A). A \Delta B=(A-B) \cup(B-A) \text{. }
Let H=(2X,Δ)H=\left(2^{X}, \Delta\right). Which of the following statements about HH is/are correct?

Select all correct options
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Answer: A;D

To determine the correct statements about the algebraic structure H=(2X,Δ)H = (2^X, \Delta), we verify the group axioms for the symmetric difference operation Δ\Delta:

  1. Closure: For any A,B∈2XA, B \in 2^X, their symmetric difference AΔBA \Delta B is also a subset of XX, so AΔB∈2XA \Delta B \in 2^X.
  2. Associativity: The symmetric difference is associative, i.e., AΔ(BΔC)=(AΔB)ΔCA \Delta (B \Delta C) = (A \Delta B) \Delta C for all A,B,C∈2XA, B, C \in 2^X.
  3. Identity Element: The empty set ∅∈2X\emptyset \in 2^X acts as the identity element because:
    AΔ∅=(A−∅)∪(∅−A)=A∪∅=AA \Delta \emptyset = (A - \emptyset) \cup (\emptyset - A) = A \cup \emptyset = A
  4. Inverse Element: For any A∈2XA \in 2^X, its inverse A−1A^{-1} must satisfy AΔA−1=∅A \Delta A^{-1} = \emptyset. Testing AA as its own inverse:
    AΔA=(A−A)∪(A−A)=∅∪∅=∅A \Delta A = (A - A) \cup (A - A) = \emptyset \cup \emptyset = \emptysetThus, the inverse of AA is AA itself.

Since all group axioms are satisfied, H=(2X,Δ)H = (2^X, \Delta) is a group. This makes Option A correct and Option B incorrect.

Since the inverse of AA is AA itself, Option D is correct. Option C is incorrect because AΔAc=X≠∅A \Delta A^c = X \neq \emptyset (for any non-empty set XX).

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