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?