# Group Tables

Composition tables are useful in examining the following axioms in the manners explained below.

Closure Property: If all the elements of the table belong to the set $G$, then $G$ is closed under the composition a. If any of the elements of the table do not belong to the set, the set is not closed.

Existence of Identity: The element (in the vertical column) to the left of the row identical to the top row (border row) is called an identity element in $G$ with respect to operation “$*$”.

Existence of Inverse: If we mark the identity elements in the table then the element at the top of the column passing through the identity element is the inverse of the element in the extreme left of the row passing through the identity element and vice versa.

Commutative: If the table is such that the entries in every row coincide with the corresponding entries in the corresponding column, i.e. the composition table is symmetrical about the principal or main diagonal, the composition is said to have satisfied the commutative axiom, otherwise it is not commutative.

The process will be clearer with the help of following illustrative examples.

Example 1:

Prove that the set of cube roots of unity is an Abelian finite group with respect to multiplication.

Solution:
The set of cube roots of unity is $G = \left\{ {1,\omega ,{\omega ^2}} \right\}$. Let us form the composition table as given below.

 $\bullet$ $1$ $\omega$ ${\omega ^2}$ $1$ $1$ $\omega$ ${\omega ^2}$ $\omega$ $\omega$ ${\omega ^2}$ ${\omega ^3} = 1$ ${\omega ^2}$ ${\omega ^2}$ ${\omega ^3} = 1$ ${\omega ^4} = \omega$

(G1) Closure Axiom: Since each element obtained in the table is a unique element of the given set $G$, multiplication is a binary operation. Thus the closure axiom is satisfied.

(G2) Associative Axiom: The elements of $G$ arc all complex numbers and we know that the multiplication of a complex number is always associative. Hence the associative axiom is also satisfied.

(G3) Identity Axiom: Since row $1$ of the table is identical with the top border row of elements of the set, $1$ (the element to the extreme left of this row) is the identity element in $G$.

(G4) Inverse Axiom: The inverses of $1,\omega ,{\omega ^2}$  are $1,\omega$ and ${\omega ^2}$ respectively.

(G5) Commutative Axiom: Multiplication is commutative in $G$ because the elements equidistant with the main diagonal are equal to each other.
The number of elements in $G$ is 3. Hence $\left( {G, \times } \right)$ is a finite group of order 3.

Example 2:

Prove that $\left\{ {1, – 1,i, – i} \right\}$ is an Abelian multiplicative finite group of order 4.

Solution:

Let $G = \left\{ {1, – 1,i, – i} \right\}$. The following will be the composition table for $\left( {G, \times } \right)$.

 $\times$ $1$ $– 1$ $i$ $– i$ $1$ $1$ $– 1$ $i$ $– i$ $– 1$ $– 1$ $1$ $– i$ $i$ $i$ $i$ $– i$ $– 1$ $1$ $– i$ $– i$ $i$ $1$ $– 1$

(G1) Closure Axiom: Since all the entries in the composition table are elements of the set $G$, the set $G$ is closed under the operation multiplication. Hence the closure axiom is satisfied.

(G2) Associative Axiom: Multiplication for complex numbers is always associative.

(G3) Identity Axiom: Row 1 of the table is identical with that at the top border, hence the element $1$ in the extreme left column heading row $1$ is the identity clement.

(G4) Inverse Axiom: The inverse of $1$ is $1$. The inverse of $– 1$ is $– 1$. The inverse of $i$ is $– i$ and of $– i$ is $i$. Hence the inverse axiom is satisfied in $G$.

(G5) Commutative Axiom: Since in the table the 1st row is identical to the 1st column, the 2nd row is identical to the 2nd column, the 3rd row is identical to the 3rd column and the 4th row is identical to the 4th column. Hence the multiplication in $G$ is commutative.

The number of elements in $G$ is 4. Hence $G$ is an Abelian finite group of 4 with respect to multiplication.