Question

Let \( G \) be the power set of \( \{1, 2, 3, 4\} \). Let the binary operation \( \Delta \) on \( G \) be

• A. \( \{1\} \) is an identity element of \( (G, \Delta) \).
• B. \( (G, \Delta) \) is an abelian group but not cyclic.
• C. \( (G, \Delta) \) is a group and has an element of order 4.
• D. \( (G, \Delta) \) is a group and has an element of order 8.

Hint:

When working with sets and binary operations, the identity element and the order of elements are key properties that define the structure of the group.

Answer 1

The Correct Option is B

Step 1: Name the group.
Here $G$ is all subsets of $\{1,2,3,4\}$ and the operation is symmetric difference $\Delta$. Each element is a subset, and $G$ has $2^4 = 16$ members.

Step 2: Find the identity.
For any set $A$ we have $A \Delta \varnothing = A$. So the empty set is the identity, not $\{1\}$. That already makes the option naming $\{1\}$ as identity wrong.

Step 3: Check the order of elements.
For every subset $A$, $A \Delta A = \varnothing$. So each non identity element squares to the identity, meaning every element has order $1$ or $2$. There is no element of order $4$ or $8$.

Step 4: See the real structure.
The group is just four copies of $\mathbb{Z}_2$, that is $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. It is abelian. It is not cyclic because a cyclic group of order $16$ would need an element of order $16$, but the biggest order here is $2$.

Step 5: Conclusion.
So the only true statement is that $G$ is abelian but not cyclic, which is option B.
\[ \boxed{\text{Abelian but not cyclic (option B)}} \]