Show that, if \(T\) is the period of \(\mathbf{f}\), then \(g^{T+s}=g^{s} \cdot g^{T}\) and, in particular, \(g^{n T}=\left(g^{T}\right)^{n}\), so that the mappings \(g^{n T}\) ( \(n\) an integer) form a group.

Question: Show that the mappings \(g^{nT}\) form a group, where \(T\) is the period of a function \(\mathbf{f}\) and \(n\) is an integer. Answer: To prove that the mappings \(g^{nT}\) form a group, we need to show that they satisfy the group properties under the operation \(*\): 1. Closure: Given two elements in the group, \(g^{aT}\) and \(g^{bT}\), their product is also in the group: \((g^{aT} * g^{bT}) = (g^{aT} \cdot g^{bT}) = g^{(a+b)T}\). 2. Associativity: The mappings are associative under the operation \(*\): \((g^{aT} * g^{bT}) * g^{cT} = g^{aT} * (g^{bT} * g^{cT})\). 3. Identity element: The identity element is \(g^{0T}\), as it satisfies the property: \(g^{0T}(x) = x\) for all \(x\) in the domain. 4. Inverse: There exists an inverse \(g^{-aT}\) for each element \(g^{aT}\), such that \(g^{aT} * g^{-aT} = g^{0T}\), the identity element. As all group properties are satisfied, the mappings \(g^{nT}\) indeed form a group.