Lorian Wilder

To prove that any cyclic group is abelian, follow these steps: **Step 1:** Let \( G \) be a cyclic group. By definition, there exists an element \( g \in G \) (called a generator) such that every element of \( G \) can be written as an integer power of \( g \). That is, for all \( a \in G \), there exists an integer \( m \) such that \( a = g^m \). **Step 2:** Take any two elements \( a, b \in G \). Since \( G \) is cyclic, there exist integers \( m \) and \( n \) such that: \[ a = g^m \quad \text{and} \quad b = g^n. \] **Step 3:** Compute the product \( ab \) and \( ba \): \[ ab = g^m \cdot g^n = g^{m + n}, \] \[ ba = g^n \cdot g^m = g^{n + m}. \] **Step 4:** Since integer addition is commutative (\( m + n = n + m \)), we have: \[ g^{m + n} = g^{n + m}. \] **Step 5:** Therefore, \( ab = ba \). This holds for all \( a, b \in G \), so \( G \) is abelian. **Conclusion:** Every cyclic group is abelian because the commutativity of integer addition ensures that all elements (expressed as powers of a generator) commute.