Problem 4 - A Computational Introduction to Number Theory and Algebra

Question

Let \(G\) be an abelian group, and let \(g\) be an arbitrary, fixed element of \(G\). Assume that the group operation of \(G\) is written additively. We define a new binary operation \(\odot\) on \(G,\) as follows: for \(a, b \in G,\) let \(a \odot b:=a+b+g .\) Show that the set \(G\) under \(\odot\) forms an abelian group.

Step-by-Step Solution

Answer

Question: Prove that the binary operation \(\odot\) defined by \(a \odot b = a + b + g\) turns an abelian group \(G\) into an abelian group. Short Answer: To prove that \(G\) under the operation \(\odot\) is an abelian group, we need to show that it satisfies closure, associativity, identity element, inverse element, and commutativity properties: 1. Closure: \(a \odot b = a + b + g\) is in \(G\) since \(G\) is a group under the operation \(+\). 2. Associativity: \((a \odot b) \odot c = a \odot (b \odot c)\) as \(G\) is an abelian group under \(+\). 3. Identity element: \(e' = -g\) is the identity under the operation \(\odot\). 4. Inverse element: \(a^{-1'} = -a-g\) is the inverse element for each \(a \in G\) under the operation \(\odot\). 5. Commutativity: \(a \odot b = b \odot a\) as \(G\) is an abelian group under \(+\). Therefore, \(G\) with the operation \(\odot\) forms an abelian group.

Step 1: 1. Closure under G#

We need to show that for all \(a, b \in G\), the result of \(a \odot b\) is also in \(G\). Since \(G\) is a group under the operation \(+\), we know that \(a + b\) and \(a + b + g\) are in \(G\), because \(a, b, g \in G\). Therefore, \(a \odot b \in G\).

Step 2: 2. Associativity#

We need to show that \((a \odot b) \odot c = a \odot (b \odot c)\) for all \(a, b, c \in G\). Using the definition of \(\odot\), we have: \((a \odot b) \odot c = (a + b + g) \odot c = (a + b + g) + c + g = a + b + g + c + g\). Similarly, we have: \(a \odot (b \odot c) = a \odot (b + c + g) = a + (b + c + g) + g = a + b + c + g + g\). Since \(G\) is an abelian group and commutative, we know that \(a + b + g + c + g = a + b + c + g + g\). Therefore, \((a \odot b) \odot c = a \odot (b \odot c)\).

Step 3: 3. Identity element#

We need to find an element \(e' \in G\) such that \(a \odot e' = e' \odot a = a\) for all \(a \in G\). Let's consider the element \(e' = -g\): \(a \odot e' = a \odot (-g) = a + (-g) + g\). As \(G\) is an abelian group, we have \(a+(-g)+g=a+(g+(-g))=a+0=a\). Also, \(e' \odot a = (-g) \odot a = (-g) + a + g=a+(-g)+g=a\). Therefore, \(e' = -g\) is the identity element for the operation \(\odot\).

Step 4: 4. Inverse element#

We need to find an element \(a^{-1'} \in G\) for each \(a \in G\) such that \(a \odot a^{-1'} = a^{-1'} \odot a = e'\). Let's consider \(a^{-1'} = -a-g\): \(a \odot a^{-1'} = a \odot (-a-g) = a + (-a - g) + g = a + (-a)\). As \(G\) is an abelian group, we have \(a+(-a)=(-a)+a=0=-g=e'\). Also, \(a^{-1'} \odot a = (-a-g) \odot a = (-a - g) + a + g = -a +a=e'\). Therefore, \(a^{-1'} = -a-g\) is the inverse element for each \(a \in G\) under the operation \(\odot\).

Step 5: 5. Commutativity#

We need to show that \(a \odot b = b \odot a\) for all \(a, b \in G\). Since \(G\) is an abelian group under the operation \(+\), we have: \(a \odot b = a + b + g = b + a + g = b \odot a\). Hence, the operation \(\odot\) is commutative. All the properties required for \(G\) to be an abelian group under the operation \(\odot\) are verified, so the set \(G\) with the operation \(\odot\) forms an abelian group.