# Proof: Order of Inverse Element in a Group Let \( G \) be a group and \( a \in G \). To prove: \[ o(a^{-1}) = o(a) \] where \( o(a) \) denotes the order of element \( a \). --- ## 1. Complete Solution with Calculations ### Step 1: Definition of Order By definition, the order of an element \( a \in G \), denoted \( o(a) \), is the smallest positive integer \( n \) such that \[ a^n = e \] where \( e \) is the identity element of \( G \). ### Step 2: Expressing the Order of \( a^{-1} \) Let \( m = o(a^{-1}) \). By definition: \[ (a^{-1})^m = e \] ### Step 3: Relating Orders Using Inverses Note that: \[ (a^{-1})^m = (a^m)^{-1} \] So, \[ (a^m)^{-1} = e \implies a^m = e \] Thus, any positive integer \( m \) such that \( (a^{-1})^m = e \) must satisfy \( a^m = e \). ### Step 4: Comparing Minimality Since \( o(a) = n \) is the minimal positive integer with \( a^n = e \), it follows that \( m \) (the order of \( a^{-1} \)) must be a multiple of \( n \): \[ m \geq n \] ### Step 5: Reverse Argument Similarly, since \[ a^n = e \implies (a^{-1})^n = (a^n)^{-1} = e^{-1} = e \] Thus, \( (a^{-1})^n = e \), so the minimal positive integer \( m \) with \( (a^{-1})^m = e \) satisfies \( n \geq m \). ### Step 6: Equality of Orders From Steps 4 and 5: \[ m \geq n \text{ and } n \geq m \implies m = n \] Therefore, \[ o(a^{-1}) = o(a) \] --- ## 2. Explanation of Each Part - **Step 1:** Defined the order of an element as the smallest positive integer with the property that the element raised to that power equals the identity. - **Step 2:** Introduced \( m \) as the order of the inverse element, defined analogously. - **Step 3:** Used exponent rules in groups to show that raising the inverse to a power is the same as inverting the element raised to that power. - **Step 4:** Concluded that if the inverse raised to \( m \) is the identity, then so is the element raised to \( m \), so \( m \) must be at least as large as the order of the element. - **Step 5:** Showed the reverse: if the element raised to its order is the identity, then so is the inverse raised to the same power, so the order of the element is at least as large as the order of its inverse. - **Step 6:** Combined the inequalities to conclude that the two orders are equal. --- ## 3. Final Conclusion For any group \( G \) and any element \( a \in G \), the order of \( a \) and its inverse \( a^{-1} \) are equal: \[ o(a^{-1}) = o(a) \]