Question

From the additive group Q to which one of the following groups does there exist a non-trivial group homomorphism ?

• A. \(\R^×\), the multiplicative group of non-zero real numbers
• B. \(\Z\), the additive group of integers
• C. \(\Z_2\), the additive group of integers modulo 2
• D. Q^×, the multiplicative group of non-zero rational numbers

Answer 1

The Correct Option is A

To determine which of the given groups has a non-trivial group homomorphism from the additive group of rational numbers \(\mathbb{Q}\), we need to explore the compatibility of each option with the properties of group homomorphisms.

Step 1: Understanding Group Homomorphisms

• A group homomorphism \(f: (G, \cdot) \to (H, *)\) is a function between two groups such that for all elements \(a, b \in G\), \(f(a \cdot b) = f(a) * f(b)\).
• A homomorphism is trivial if it maps every element to the identity element in the codomain.

Step 2: Analyzing the Options

• \(\Z\): The additive group of integers is torsion-free, whereas \(\mathbb{Q}\) has elements with arbitrary finite order (torsion elements). A non-trivial homomorphism cannot exist as the structure of \(\mathbb{Q}\) does not map interestingly into \(\Z\).
• \(\Z_2\): This group has only two elements, \(0\) and \(1\), and is finite. Since \(\mathbb{Q}\) is infinite and a homomorphism must preserve the order, a non-trivial homomorphism cannot exist from \(\mathbb{Q}\) to \(\Z_2\)
• Q^×: Similar reasoning as \(\R^×\) can apply, but since the nature of rational numbers and their arithmetic do not lead to a simple non-trivial mapping to non-zero rationals, no straightforward solution exists.
• \(\R^×\): The multiplicative group of non-zero real numbers supports exponential mappings. Consider \(\phi: \mathbb{Q} \to \R^×\) defined by \(\phi(q) = e^{\pi iq}\), which is non-trivial and a valid homomorphism.

Step 3: Conclusion

Among the given options, the multiplicative group of non-zero real numbers \(\R^×\) supports a non-trivial group homomorphism from \(\mathbb{Q}\). Hence, the correct answer is \(\R^×\).