Recall the definition of a group:

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{ called the inverse of x, with } x*y = y*x = e  \forall x \in G

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


In addition to the axioms, the following properties of groups are important to note:

  1. Uniqueness of the identity element
  2. Uniqueness of the inverse element
  3. Cancellation law
  4. Inverse property (extended)

Uniqueness of an element in mathematics means there exists only one such element with that property. We prove uniqueness by making an assumption that there are two elements in the set that satisfy the property, and show that if such a situation holds, then the two elements must be equal!

We use * to denote the binary operation between elements and “QED” to signal the end of the proof.

The remainder of the post aims to go through the proofs of these properties!


uniqueness of the identity element proof

We want to show that the identity element of a group is unique. Let’s assume the opposite.


Suppose for group G, there are two identity elements, e_1 and e_2. Then for any element, x \in G,  x*e_1 = e_1*x = x \text{ and } x*e_2 = e_2*x = x.

So e_1 = e_1*e_2 = e_2*e_1 = e_2 This shows that e_1 = e_2 (hence if there are two identity elements in a set, then those elements are the same thing just with different labelling)!


To break this proof, look at the following bullet points:

  • we can say  e_1 = e_1*e_2 because here e_2 acts as the identity element, and e_1 acts as a regular element of the set
  • e_1*e_2 = e_2*e_1 holds since  identity elements are commutative, by definition

(RECALL)  Identity: \exists e \in G,  \text{ called the identity element  such that } \forall x \in G, x*e = e*x = x. This shows the commutative nature of identity elements

  • finally we conclude  e_2*e_1 = e_2 since we use the fact that e_2 can also act as a regular element and e_1 can act as the identity element

uniqueness of inverse element proof

We want to show that the inverse element is unique. Suppose the opposite


Suppose there exists two inverse elements, h, k  \in G \text{ for } g \in G. Then h*g = g*h = e \text{ and } k*g = g*k = e where e is the identity element of G.

Then h = h*e = h*(g*k) = (h*g)*k = e*k = k. Hence, h=k.


Again, I’ll breakdown the proof below:

  • using the identity property, we can say h = h*e
  • since we defined the inverse h of g as: h*g = g*h = e, then we can substitute h*e for h*(g*k)
  • by associativity of G,  h*(g*k) = (h*g)*k
  • and by assumption k*g = g*k = e, we can substitute h*g for e
  • and finally, we use the identity property again to conclude h = h*e

cancellation laws proof

In a group G, if a*g = b*g then a=b for a, b, g \in G


Suppose a*g = b*g

Let h be the inverse of g, then h*g = g*h = e for identity element, e

Then a = a*e = a*(g*h) = (a*g)*h = (b*g)*h = b*(g*h) = b*e = b, as required


The breakdown is as follows:

  • a = a*e is possible using the identity property
  • a*e = a*(g*h) is possible since g*h = e*, hence used substitution
  • a*(g*h) = (a*g)*h by associativity
  • (a*g)*h = (b*g)*h since we assumed a*g = b*g
  • (b*g)*h = b*(g*h) by associativity
  • b*(g*h) = b*e by inverse property
  • b*e = b by identity property

inverse of g*h \text{ is } h^{-1} * g^{-1}


Let g, h \in G, then we know \exists g^{-1}, h^{-1} \in G satisfying g*g^{-1} = g^{-1}*g = e and h*h^{-1} = h^{-1}*h = e

We claim h^{-1} * g^{-1} is the inverse of g*h . To show this, we want to show that (g*h) * (h^{-1} * g^{-1}) = (h^{-1} * g^{-1})*(gh)

Since (g*h) * (h^{-1} * g^{-1}) = g*(h* h^{-1}) * g^{-1} = g*e* g^{-1} =  g*g^{-1} = e

And (h^{-1} * g^{-1})*(gh) = h^{-1} * (g^{-1}*g)*h =  h^{-1} *e*h = h^{-1}*h = e

Then (g*h) * (h^{-1} * g^{-1}) = (h^{-1} * g^{-1})*(gh)


The breakdown:

  • (g*h) * (h^{-1} * g^{-1}) = g*(h* h^{-1}) * g^{-1} by associativity
  • g*(h* h^{-1}) * g^{-1} = g*e* g^{-1} by inverse property
  • g*e* g^{-1} =  g*g^{-1} by identity property
  •  g*g^{-1} = e by identity property
  • (g*h) * (h^{-1} * g^{-1}) = (h^{-1} * g^{-1})*(gh) since both equal e


How clear is this post?