Skip to content

DA8. Abelian Group

Statement

For this assignment, we will explore the basic algebraic structure group. As we near the end of this course, we assume that you are well acquainted with the notations of different number systems such as ℤ, ℝ, ℕ, and ℚ.

Your task is to provide an example and a counter-example of the Abelian group by choosing any number system (mentioned above) and any algebraic operation such as addition (+), subtraction (-), multiplication (×), division (÷), addition modulo (+_n), or multiplication modulo (n X_n).

Choose operations that you find interesting and include reasoning for your choices.

Solution

Let’s start by defining the terms that included in the statement (Neso Academy, 2021):

  • Group: a group G denoted by \(\{G, \circ\}\) is a set G together with a binary operation \(\circ\) on G such that the following CAIN axioms hold:
    • Closure: \(\forall a, b \in G, a \circ b \in G\)
    • Associativity: \(\forall a, b, c \in G, (a \circ b) \circ c = a \circ (b \circ c)\)
    • Identity: \(\exists e \in G, \forall a \in G, a \circ e = e \circ a = a\)
    • Inverse: \(\forall a \in G, \exists a^{-1} \in G, a \circ a^{-1} = a^{-1} \circ a = e\)
  • When the CAIN properties are true for a set and operation on it; we call it a group.
  • Abelian Group: a group G is called an abelian group if the previous CAIN axioms hold, and the following Commutativity axiom holds:
    • Commutativity: \(\forall a, b \in G, a \circ b = b \circ a\)

So to prove that a set and operation on that set is an abelian group, we need to prove that the CAIN axioms hold and the Commutativity axiom holds; and then prove the opposite to provide a counter-example.

As an example, I will take the set \(\mathbb{Z}\) (integers) and the operation \(+\) (addition) on that set. Let’s prove that \(\{ \mathbb{Z}, + \}\) is an abelian group:

  • Closure: \(\forall a, b \in \mathbb{Z}, a + b \in \mathbb{Z}\), which is true because the sum of two integers is an integer.
  • Associativity: \(\forall a, b, c \in \mathbb{Z}, (a + b) + c = a + (b + c)\), which is true because the sum of three integers is the same regardless of the order of addition.
  • Identity: \(\exists e \in \mathbb{Z}, \forall a \in \mathbb{Z}, a + e = e + a = a\), which is true because the identity element of addition is 0 (as \(a + 0 = 0 + a = a\)) no matter what the value of \(a\) is.
  • Inverse: \(\forall a \in \mathbb{Z}, \exists a^{-1} \in \mathbb{Z}, a + a^{-1} = a^{-1} + a = e\), which is true because the inverse of an integer is its negative (as \(a + (-a) = (-a) + a = 0\)) no matter what the value of \(a\) is.
  • Commutativity: \(\forall a, b \in \mathbb{Z}, a + b = b + a\), which is true because the sum of two integers is the same regardless of the order of addition.

Thus, \(\{ \mathbb{Z}, + \}\) is an abelian group as the 5 axioms hold.

As a counter-example, I will take the set \(\mathbb{Z}^*\) (integers excluding zero) and the operation \(+\) (addition) on that set. Let’s prove that \(\{ \mathbb{Z}^*, + \}\) is not an abelian group:

  • Per the previous discussion, we confirm the Closure, Associativity and Commutativity axioms hold.
  • However, the Identity axiom does not hold because the identity element of addition is 0 (as \(a + 0 = 0 + a = a\)) and 0 is not in \(\mathbb{Z}^*\).

We have proven that \(\{ \mathbb{Z}^*, + \}\) violates one of the five axioms, thus it is not an abelian group (even it is not a group at all).

References