# 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., $$0 + a = a = 0 + a{\text{ }}\forall a \in \mathbb{Z}$$

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.

__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.