Discussion
Let \(G\) be the power set of \(\{1,2,3,4\}\). Let the binary operation \(\Delta\) on \(G\) be
\[ A \,\Delta\, B = (A \setminus B) \cup (B \setminus A). \]
Which ONE of the following statements is TRUE?
Answer: B
The TRUE statement is: \((G, \Delta)\) is an abelian group but not cyclic.
Symmetric difference is associative and commutative. The empty set is the identity, since \(A \,\Delta\, \emptyset = A\), and every set is its own inverse, since \(A \,\Delta\, A = \emptyset\). So \((G, \Delta)\) is an abelian group of order \(2^{4} = 16\) in which every non-identity element has order 2; it is isomorphic to \(\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2\).
A cyclic group of order 16 would need an element of order 16, so the group is not cyclic. For the same reason there is no element of order 4 or 8. \(\{1\}\) is not the identity: \(\{1\} \,\Delta\, \{1\} = \emptyset \neq \{1\}\), and \(\{1\} \,\Delta\, \{2\} = \{1,2\}\), not \(\{2\}\).
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