Binary operations are operations such as addition, subtraction, multiplication, division, modulus etc. that are applied to two quantities.

example 1: 2+5 is an example of an expression with addition as the binary operation

example 2: Let f and g be functions defined on sets A to B. Then the composition of the functions \text{ f(g(x)) } is a binary operation

We will use * to denote an arbitrary (general) binary operation.

A set G is “upgraded” into a group if it satisfied the following axioms under one binary operation (*) :

  1. Closure: \forall x, y \in  G, x*y \in G
  2. Associativity: \forall x, y, z \in G, (x*y)*z = x*(y*z)
  3.  Identity: \exists e \in G,  \text{ called the identity element  such that } \forall x \in G, x*e = e*x = x
  4. Inverse:  \exists y \in G, \text{is called an inverse element of } x \in G \text{ with } x*y = y*x = e

An Abelian group is a group that is follows the axioms 1 – 4 with the addition of one property: 

  1. Commutativity: \forall  x, y \in G,  x*y = y*x

For the remainder of this post, we will explore these axioms and look at some examples

Closure: \forall x, y \in  G, x*y \in G

This means we can take any elements in the set G and perform the operation defined by * and the result will also be an element in the group.

Consider this example: If G = \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 \} with regular addition, +, as the binary operation, then we want to know if we can take any two numbers in the set, add them, and get another number that is still in the set.

  • 1+3 = 4 (4 is in the set, hence condition is satisfied for this example)
  • -2 + 4 = 2 (2 is in the set, hence condition is satisfied for this example)

So far, it seems like G is closed under addition! If we can find at least one example of two numbers from G that, when added, result in a number that is not in the set, then we’ve shown that condition isn’t satisfied for all the elements of the set.

  • 2+9 = 11, which is not in the set. Hence, we can conclude that G is not closed under + since closure is not satisfied for ALL elements in the set!
  • If we were trying to prove that G is a group under +, then we can already stop and conclude that it isn’t closed under the operation.

Associativity: \forall x, y, z \in G, (x*y)*z = x*(y*z)

If a set G is associative, then we can take any three elements of a set G, perform the defined group operation, * , and the result will be the same irrespective of the order in which we applied the operation.

Consider the same example: G = \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 \} \text { under binary operation } +

  • Then (1+2)+3 = 3 + 3 = 6 \text { and } 1 + (2+3) = 1 + 5 = 6. So we can conclude that (1+2)+3 =  1 + (2+3) = 6. In other words, the placement of the brackets did not affect the result after the operation, +, was performed.
  • Similarly, (9+0) + 1 = 10 = 9 + (0+1)

If this holds for all elements in G, then we can conclude that G is associative!

  • If the binary operation was division, for the same set G, then the placement of the brackets would be important. Let * denote division, then (1*2)*4 = 8 and 1*(2*4) = 2. Hence the answer changed, so division is not associative for the set G

Identity: \exists e \in G, \text{ called the identity element such that } \forall x \in G, x*e = e*x = x

i.e. there exists an element, e   called the identity element, such that for any element x \in G, x*e = e*x = x

To understand this, let’s look at some cases for different groups:

  • G = \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 \} under binary operation, +,  has zero as the identity element. Since for any element in G, we can add zero and get that element: 1+0 = 1,  4+0 = 4,  \text{ etc }
  • Set T = {1, 3, 5, 7, ....} \text{ under } \times has 1 as the identity element since 1 \times 1 = 1, 1 \times 7 = 7, \text{ etc. }

Inverse: \exists y \in G, \text{ such that for an element, } x \in G, x*y = y*x = e

i.e. here exists an element, y \in G, \text{ such that } \text{ for an element } x \in G, x*y = y*x = e

  • Since G = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} \text{ under binary operation, +, } has zero as the identity element. Then the inverse of 4 \text{ is } -4 \text{ since } 4 + (-4) = 0
  • Similarly, the inverse of -3 \text{ is } 3 \text{ since } -3 + 3 = 0 and so on

Something worth noting for a group, (G, \times ) under multiplication: the inverse law takes into account that 0 cannot have an inverse

\exists y \in G, \text{ such that for an element, } x \in G, x \ne 0, x*y = y*x = e

So under some operation, an element and its inverse will give you the identity element!

  • If we consider the set G under multiplication, then the inverse property is not satisfied since 0 has no inverse element. In other words, there is \nexists x \in G \text{ such that } x \times 0 = 0 \times x = 1 (where 1 is the identity element of G under multiplication).

We denote a set G that forms a group under a binary operation as (G, *)

example 1: (G, +) is a set G with addition as the binary operation

example 1: (G, \times) is a set G with multiplication as the binary operation

To show that a set is a group, we want to show that all 4 axioms are satisfied for the defined operation * simultaneously. If we can find at least one counter-example, then we can immediately conclude that a set is not a group.

In our case, G was not a group under addition since it wasn’t closed. All other properties were satisfied.

commutativity: \forall  x, y \in G,  x*y = y*x

Commutativity tells us the order in which we apply the operation to our elements in some set doesn’t matter.

  • For our example, G = \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 \} under binary operation, +, we could add two elements, x,y \in G as x+y \text{ or } y+x and get the same result
  • For instance, -4 + 3 = 3 + (-4) = -1 or 5+2 = 2+5 = 7 . Note again, how 7 is not in the set, again reminding us that out set is not closed. In this case that our set was closed and commutative, G would be an Abelian group!

Now we’ll explore some sets that form groups and which don’t:

example 1: set of n x n matrices with real coefficients M_{n \times n } ( \mathbb{ R }) only form a group when every matrix is invertible. the identity matrix will be the identity element

example 2: the set of real numbers \mathbb{ R } without zero forms a group under multiplication with identity 1. If zero is included in the set, then 0 does not have an inverse, hence inverse property isn’t satisfied and set under multiplication is not a group

example 3: the set of rational numbers, \mathbb{ Q } under addition with 0 as the identity forms a group

 

How clear is this post?