91Ó°ÊÓ

Firstly we need to define the operation \((a, b) \longmapsto a + b\) on \(R_n\). Since \(\pi: \mathbb{Z} \longrightarrow R_n\) maps the elements of \(\mathbb{Z}\) to their non-negative remainders when divided by \(n\), we can define the operation as follows: For \(a, b \in R_n\), let \(a + b = \pi(c)\), where \(c \in \mathbb{Z}\) and \(c = a +_{\mathbb{Z}} b\), the sum in \(\mathbb{Z}\).

Now we need to show that this definition is well-defined, that is, the result of \(a + b\) is uniquely determined for any \(a, b \in R_n\). Let \(a, b \in R_n\), and let \(c, d \in \mathbb{Z}\) be such that \(\pi(c) = a\) and \(\pi(d) = b\). Then \(\pi(c +_\mathbb{Z} d) = a+b\) according to our definition. Since the remainders when divided by \(n\) are unique for any two integers, this operation is well-defined.

We are asked to show that the operation satisfies the property: for all \(x, y \in \mathbb{Z}\), \(\pi(x + y) = \pi(x) + \pi(y)\). Let \(x, y \in \mathbb{Z}\) and \(a, b \in R_n\) be such that \(\pi(x)=a\) and \(\pi(y)=b\). Then according to our definition, \(\pi(x+y) = \pi(a +_{\mathbb{Z}} b) = a + b\). Since \(a = \pi(x)\) and \(b = \pi(y)\), we have \(\pi(x + y) = \pi(x) + \pi(y)\), which proves the required property.

Now we need to check the group properties of \(R_n\) under the defined operation: 1. Closure: Since the operation is defined to be the sum of two elements in \(R_n\), it follows that \(R_n\) is closed under the operation. 2. Associativity: For any \(a, b, c \in R_n\), we have \((a + b) + c = \pi((a +_{\mathbb{Z}} b) +_{\mathbb{Z}} c) = \pi(a +_{\mathbb{Z}} (b +_{\mathbb{Z}} c)) = a + (b + c)\) by the operation property. 3. Identity: Since \(\pi(0)\) maps to the remainder 0, 0 is the identity element under the operation. 4. Inverse: For any \(a \in R_n\), let \(c = n - a\) so that \(a + c = \pi(a +_{\mathbb{Z}} c) = \pi(n) = 0\). So, each element has an inverse in \(R_n\). Furthermore, the operation is commutative since for any \(a, b \in R_n\), we have \(a + b = \pi(a +_{\mathbb{Z}} b) = \pi(b +_{\mathbb{Z}} a) = b + a\), which implies that \(R_n\) is an abelian group. In conclusion, we have shown that there exists a unique operation on \(R_n\) satisfying the required property, and that with this operation, \(R_n\) is an abelian group.