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.

Proof:

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)!

QED

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

Proof:

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, $h=k.$

QED

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 $a, b, g \in G$

Proof:

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

QED

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 $g*h \text{ is } h^{-1} * g^{-1}$

Proof:

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

QED

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?