**Group theory** is the study of a set of elements present in a group, in Maths. A group is a collection of elements or objects that are consolidated together to perform some operation on them. If any two of its elements are combined through an operation to produce a third element belonging to the same set and meets the four hypotheses namely closure, associativity, invertibility and identity, they are called **group axioms**.

**For instance: **A** **group of integers which are performed under multiplication operation. A geometrical group theory in the branch of Mathematics, is basically the study of groups that are finitely produced with the use of the research of the relationships between the algebraic properties of these groups and also topological and geometric properties of the spaces.

## Group Axioms Properties

Suppose Dot(.) is an operation and G is the group, then the axioms of group theory are defined as;

**Closure:**If â€˜xâ€™ and â€˜yâ€™ are two elements in a group, G, then x.y will also come into G.**Associativity:**If â€˜xâ€™, â€˜yâ€™ and â€˜zâ€™ are in group G, then x . (y . z) = (x . y) . z.**Invertibility:**For every â€˜xâ€™ in G, there exists some â€˜yâ€™ in G, such that; x . y = y . x.**Identity:**For any element â€˜xâ€™ in G, there exists an element â€˜Iâ€™ in G, such that: x. I = I . x, where â€˜Iâ€™ is called the identity element of G.

The most common example, which satisfies these axioms, is the addition of two integers, which results in an integer itself. Hence, the closure property is satisfied. Also, the addition of integers satisfies the associative property. There exists an identity element name as zero in the group, which when added with any number, gives the original number. Also, for every integer, there exists an inverse, in such a way, when they are added gives zero as result. So, all the group axioms are satisfied in case of addition operation of two integers.

### Proof of Group Axiom

**1:** If G is a group which has a and b as its elements, such that a,bâˆˆG, then (a Ã— b)^{-1} = a^{-1} Ã— b^{-1}

**Proof:**

To prove: (a Ã— b) Ã— b^{-1} Ã— a^{-1}= I, where I is the identity element of G.

Consider the L.H.S of the above equation, we have,

L.H.S = (a Ã— b) Ã— b^{-1} Ã— b^{-1}

=> a Ã— (b Ã— b^{-1}) Ã— b^{-1}

=> a Ã— I Ã— a^{-1} (by associative axiom)

=> (a Ã— I) Ã— a^{-1} (by identity axiom)

= a Ã— a^{-1} (by identity axiom)

= I (by identity axiom)

= R.H.S

Hence, proved.

**2:** If in a group G, â€˜xâ€™, â€˜yâ€™ and â€˜zâ€™ are three elements such that x Ã— y = z Ã— y, then x = z.

**Proof:** Let us assume that x Ã— y = z Ã— y. (i)

Since â€˜yâ€™ is an element of group G, this implies there exist some â€˜aâ€™ in G with identity element I, such that;

y Ã— a = I (ii)

On multiplying both sides of (i) by â€˜aâ€™ we get,

x Ã— y Ã— a = z Ã— y Ã— a

x Ã— (y Ã— a) = z Ã— (y Ã— a) (by associativity)

From eq.(ii);

a Ã— I = c Ã— I [using (ii)]

a = c (by identity axiom)

This is also known as cancellation law.

Hence, proved.