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.


Let us test all the group axioms for Abelian group.

(G1) Closure Axiom. We know that the sum of any two integers is also an integer, i.e., for alla,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 and0 \in \mathbb{Z}, i.e., 0 + a = a = 0 +  a{\text{ }}\forall a \in \mathbb{Z}
Hence, additive identity exists.

(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 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 operation of multiplication is a group.


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, 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, associative axiom is satisfied.

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


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} andc + id \in \mathbb{C}, then by 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.