# Examples of Group

Example 1:

Show that the set of all integers …-4, -3, -2, -1, 0, 1, 2, 3, 4, ... is an infinite Abelian group with respect to the operation of addition of integers.

Solution:

Let us test all the group axioms for an Abelian group.

(G1) Closure Axiom: We know that the sum of any two integers is also an integer, i.e., for all $a,b \in \mathbb{Z}$, $a + b \in \mathbb{Z}$. Thus $\mathbb{Z}$ is closed with respect to addition.

(G2) Associative Axiom: Since the addition of integers is associative, the associative axiom is satisfied, i.e., for $a,b,c \in \mathbb{Z}$ such that $a + \left( {b + c} \right) = \left( {a + b} \right) + c$

(G3) Existence of Identity: We know that $0$ is the additive identity and $0 \in \mathbb{Z}$, i.e., $0 + a = a = 0 + a{\text{ }}\forall a \in \mathbb{Z}$

(G4) Existence of Inverse: If $a \in \mathbb{Z}$, then $- a \in \mathbb{Z}$. Also, $\left( { - a} \right) + a = 0 = a + \left( { - a} \right)$

Thus, every integer possesses additive inverse. Therefore $\mathbb{Z}$ is a group with respect to addition.

Since the addition of integers is a commutative operation, therefore $a + b = b + a{\text{ }}\forall a,b \in \mathbb{Z}$

Hence $\left( {\mathbb{Z}, + } \right)$ is an Abelian group. Also, $\mathbb{Z}$ contains an infinite number of elements.

Therefore $\left( {\mathbb{Z}, + } \right)$ is an Abelian group of infinite order.

Example 2:

Show that the set of all non-zero rational numbers with respect to the operation of multiplication is a group.

Solution:

Let the given set be denoted by ${Q_o}$. Then by group axioms, we have

(G1) We know that the product of two non-zero rational numbers is also a non-zero rational number. Therefore ${Q_o}$  is closed with respect to multiplication. Hence, the closure axiom is satisfied.

(G2) We know for rational numbers:
$\left( {a \cdot b} \right) \cdot c = a \cdot \left( {b \cdot c} \right)$ for all $a,b,c \in {Q_o}$
Hence, the associative axiom is satisfied.

(G3) Since $1$ the multiplicative identity is a rational number, hence the identity axiom is satisfied.

(G4) If $a \in {Q_o}$, then obviously, $\frac{1}{a} \in {Q_o}$. Also $\frac{1}{a} \cdot a = 1 = a \cdot \frac{1}{a}$
so that $\frac{1}{a}$ is the multiplicative inverse of $a$. Thus the inverse axiom is also satisfied. Hence ${Q_o}$ is a group with respect to multiplication.

Example 3:

Show that $\mathbb{C}$, the set of all non-zero complex numbers is a multiplicative group.

Solution:

Let $\mathbb{C} = \left\{ {z:z = x + iy,{\text{ }}x,y \in \mathbb{R}} \right\}$. Here $\mathbb{R}$ is the set of all real numbers and $i = \sqrt { - 1}$.

(G1) Closure Axiom: If $a + ib \in \mathbb{C}$ and $c + id \in \mathbb{C}$, then by the definition of multiplication of complex numbers

$\left( {a + ib} \right)\left( {c + id} \right) = \left( {ac - bd} \right) + i\left( {ad + bc} \right) \in \mathbb{C}$

Since $ac - bd,ad + bc \in \mathbb{R}$, for $a,b,c,d \in \mathbb{R}$. Therefore,$\mathbb{C}$ is closed under multiplication.

(G2) Associative Axiom:
$\left( {a + ib} \right)\left\{ {\left( {c + id} \right)\left( {e + if} \right)} \right\} = \left( {ace - adf - bcf - bde} \right) + i\left( {acf + ade + bce - bdf} \right)$
$= \left\{ {\left( {a + ib} \right)\left( {c + id} \right)} \right\}\left( {e + if} \right)$ for $a,b,c,d \in \mathbb{R}$ .

(G3) Identity Axiom: $e = 1\left( { = 1 + i0} \right)$ is the identity in $\mathbb{C}$.

(G4) Inverse Axiom: Let $\left( {a + ib} \right)\left( { \ne 0} \right) \in \mathbb{C}$, then

${\left( {a + ib} \right)^{ - 1}} = \frac{1}{{a + ib}} = \frac{{a - ib}}{{{a^2} + {b^2}}}$
${\left( {a + ib} \right)^{ - 1}} = \left( {\frac{a}{{{a^2} + {b^2}}}} \right) + i\left( {\frac{b}{{{a^2} + {b^2}}}} \right)$
${\left( {a + ib} \right)^{ - 1}} = m + in \in \mathbb{C}$

where $m = \left( {\frac{a}{{{a^2} + {b^2}}}} \right)$ and $n = \left( {\frac{b}{{{a^2} + {b^2}}}} \right)$

Hence $\mathbb{C}$ is a multiplicative group.