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 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}
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 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.



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.



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.