Question

Let $G = P(N)$, where the operation is
\[ A \Delta B = A \cup B - A \cap B \] Which of the following is true?

• A. $G$ is abelian but not cyclic
• B. $G$ has elements of order 4
• C. $G$ has elements of order 8
• D. $\emptyset$ is the identity element of $G$

Hint:

Power set under symmetric difference forms an abelian group. Every element has order 2 and identity is the empty set.

Answer 1

The Correct Option is A

Step 1: Understanding the Topic
This question asks about the properties of a group formed by the power set of natural numbers, $P(\mathbb{N})$, under the operation of symmetric difference ($\Delta$). We need to evaluate several statements regarding its properties like commutativity (abelian), cyclicity, the order of its elements, and its identity element.
Step 2: Key Approach - Verifying Group Properties
We will check the fundamental properties of the given algebraic structure $(G, \Delta)$ and then evaluate each option based on these properties.
Step 3: Detailed Explanation
A. Identity Element:
The identity element $E$ must satisfy $A \Delta E = A$ for any set $A \in G$. \[ A \Delta \emptyset = (A \cup \emptyset) - (A \cap \emptyset) = A - \emptyset = A \] Thus, the empty set $\emptyset$ is the identity element. Statement (D) is true.
B. Commutativity (Abelian Property):
The operation is abelian if $A \Delta B = B \Delta A$. \[ A \Delta B = (A \cup B) - (A \cap B) \] \[ B \Delta A = (B \cup A) - (B \cap A) \] Since set union and intersection are commutative, the symmetric difference is also commutative. Thus, the group is abelian.
C. Order of Elements:
The order of an element $A$ is the smallest positive integer $k$ such that $A^k = E$. Here, this means $A \Delta A \Delta \dots \Delta A$ ($k$ times) equals $\emptyset$. Let's check for $k=2$: \[ A \Delta A = (A \cup A) - (A \cap A) = A - A = \emptyset \] Since $A \Delta A = \emptyset$ for any non-empty set $A$, every non-identity element has an order of 2. This means there are no elements of order 4 or 8. Statements (B) and (C) are false.
D. Cyclic Property:
A group is cyclic if it can be generated by a single element. A cyclic group can have at most one element of order 2. Since our group $G$ is infinite and every non-identity element has order 2, it cannot be generated by any single element. Therefore, the group is not cyclic.
Step 4: Final Answer
Combining our findings, the group is abelian but not cyclic, and its identity element is the empty set. Therefore, statements (A) and (D) are the correct descriptions of the group.