Question

Consider the group \( (\mathbb{Q}, +) \) and its subgroup \( (\mathbb{Z}, +) \).
For the quotient group \( \mathbb{Q}/\mathbb{Z} \), which one of the following is FALSE?

• A. \( \mathbb{Q}/\mathbb{Z} \) contains a subgroup isomorphic to \( (\mathbb{Z}, +) \).
• B. There is exactly one group homomorphism from \( \mathbb{Q}/\mathbb{Z} \) to \( (\mathbb{Q}, +) \).
• C. For all \( n \in \mathbb{N} \), there exists \( g \in \mathbb{Q}/\mathbb{Z} \) such that the order of \( g \) is \( n \).
• D. \( \mathbb{Q}/\mathbb{Z} \) is not a cyclic group.

Answer 1

The Correct Option is A

The problem involves analyzing properties of the quotient group \( \mathbb{Q}/\mathbb{Z} \). To determine which statement is false, let's evaluate each option:

1. \( \mathbb{Q}/\mathbb{Z} \) contains a subgroup isomorphic to \( (\mathbb{Z}, +) \).
   • A subgroup isomorphic to \( (\mathbb{Z}, +) \) would mean \( \mathbb{Q}/\mathbb{Z} \) has an infinite cyclic subgroup. However, all elements \( g \) in \( \mathbb{Q}/\mathbb{Z} \) have finite order, specifically the order of \( g \) is the smallest positive \( n \) such that \( ng = 0 \) in \( \mathbb{Q}/\mathbb{Z} \). Thus, no element generates an infinite cyclic subgroup.
   • This statement is false because \( \mathbb{Q}/\mathbb{Z} \) does not have a subgroup isomorphic to \( \mathbb{Z} \).
2. There is exactly one group homomorphism from \( \mathbb{Q}/\mathbb{Z} \) to \( (\mathbb{Q}, +) \).
   • Consider any homomorphism \( f: \mathbb{Q}/\mathbb{Z} \to \mathbb{Q} \). Since \( \mathbb{Q} \) is torsion-free and all elements of \( \mathbb{Q}/\mathbb{Z} \) have finite order, the image of any element in \( \mathbb{Q}/\mathbb{Z} \) must be zero for the homomorphism to be well-defined, which means only the trivial homomorphism exists.
   • This statement is true.
3. For all \( n \in \mathbb{N} \), there exists \( g \in \mathbb{Q}/\mathbb{Z} \) such that the order of \( g \) is \( n \).
   • Consider the element \( \frac{1}{n} + \mathbb{Z} \in \mathbb{Q}/\mathbb{Z} \). It is clear that \( n \times (\frac{1}{n} + \mathbb{Z}) = n \cdot \frac{1}{n} + n\mathbb{Z} = 1 + \mathbb{Z} = 0 \) in \( \mathbb{Q}/\mathbb{Z} \), so this element has order \( n \).
   • This statement is true.
4. \( \mathbb{Q}/\mathbb{Z} \) is not a cyclic group.
   • A group is cyclic if there exists a single element whose powers generate the whole group. However, for \( \mathbb{Q}/\mathbb{Z} \), each element having finite order implies it cannot generate all elements of arbitrary order.
   • This statement is true, as \( \mathbb{Q}/\mathbb{Z} \) cannot be generated by a single element.

After evaluating all the options, the statement that is false is: \(\mathbb{Q}/\mathbb{Z}\) contains a subgroup isomorphic to \((\mathbb{Z}, +)\).