Elementary Group Theory - Exercise 3.2 : Class 12 Math

$Elementary Group Theory - Exercise 3.2 : Class 12 Math$

Chapter 1: Elementary Group Theory.

Exercise: 3.2

Complete Exercise of Elementary Group Theory - Exercise 3.2 : Class 12 Mathematics 2080 NEB.

Exercise 3.2 is about: Algebraic Structure.

Read: NEB Class 12 Mathematics All Chapter Exercise

1. Given a Set Z = { 0, 1, 2, 3 } and a binary operation +4 is defined by the following Cayley's table.

+4	0	1	2	3
0	0	1	2	3
1	1	2	3	0
2	2	3	0	1
3	3	0	1	2

Solution:

Since, 0 + ₄0 = 0

1 + ₄0 = 0 + ₄1 = 1

2 + ₄0 = 0 + ₄2 = 3

3 + ₄0 = 0 + ₄3 = 3.

So, 0 is the identity element.

Since, 2 + ₄2 = 0

So, the inverse element of 2 is 2.

And 3 + ₄1 + 1 + ₄3 = 0

So, the inverse element of 3 is 1.

2. Let G = { 0, 1, 2}. Form a composition table for G under addition modulo 3 and multiplication modulo 3. Find the identity elements and the inverse elements of 1 and 2 in each case.

Solution:

To form the composition tables for addition and multiplication modulo 3, we need to compute the result of combining each pair of elements from the set G = {0, 1, 2} using the corresponding operation and then take the result modulo 3. Let's start with addition:

Addition Modulo 3 ( + ):

The composition table for addition modulo 3 is as follows:

+	0	1	2
0	0	1	2
1	1	2	0
2	2	0	1

Identity Element: The identity element for addition modulo 3 is 0 because for any element a in the set G, a + 0 = a (mod 3) and 0 + a = a (mod 3).

Inverse Elements:

The inverse element of 1 is 2 because 1 + 2 = 0 (mod 3) and 2 + 1 = 0 (mod 3).
The inverse element of 2 is 1 because 2 + 1 = 0 (mod 3) and 1 + 2 = 0 (mod 3).

Next, let's look at multiplication modulo 3:

Multiplication Modulo 3 ( * ):

The composition table for multiplication modulo 3 is as follows:

*	1	2
0	0	0
1	1	2
2	2	1

Identity Element: The identity element for multiplication modulo 3 is 1 because for any element a in the set G, a * 1 = a (mod 3) and 1 * a = a (mod 3).

Inverse Elements:

The inverse element of 1 is 1 itself because 1 * 1 = 1 (mod 3).
The inverse element of 2 is 2 itself because 2 * 2 = 1 (mod 3).

In both cases, addition and multiplication modulo 3, we find that the identity element is unique (0 for addition and 1 for multiplication). However, the inverse elements exist only for elements 1 and 2 in the case of addition modulo 3, while every element has an inverse in the case of multiplication modulo 3 (the set is closed under multiplication).

3. Let G = { ..., -3, -2, -1, 0, 1, 2, 3, ...} and the operation defined on G be of addition. Find the identity and the inverse of elements of G.

Solution:

Since, 1 + 0 = 0 + 1 = 1.

And 2 + 0 = 0 + 2 = 2.

So, 0 is the identity element of both 1 and 2.

Since, 1 + (-1) = (-1) + 1 = 0

Since, - 1 is the inverse element of 1.

And, 2 + (-2) = (-2) + 2 = 0

So, - 2 is the inverse element of 2.

4. Given the algebraic structure (G, ✖) with G = {-1, 1} where ✖ stands for the operation of multiplication, find the inverses of elements of G.

Solution:

Since, (-1).1 = 1. (-1) = - 1

And 1.1 = 1

So, 1 is the identity element of both 1 and -1.

Now, (-1) * (-1) = 1

And 1 * 1 = 1

So, the inverse elements of – 1 and 1 are -1 and 1 respectively.

5. Determine the identity element and inverse elements of 3 and -2 in each case given below.

a) Given an algebraic structure (Z, •) with binary operation • defined by m•n = m + n + 1 for all m, n ε Z.

Solution:

Let e be the identity element of 3 under binary operation defined by m*n = m + 1 + 1.

Then 3 * e = 3

Or, 3 + e + 3 [m * n = m + n + 1]

So, e = -1.

Again, let e’ be the identity element of – 2.

Then (-2) * e’ = - 2.

Or, (-2) + e’ + 1 = -2

Or, e’ + 1 = 0

So, e’ = - 1.

So, - 1is the required element of both 3 and – 2.

Again, le t ‘a’ be the inverse element of 3 under the given binary operation *

Then, 3 * a = e

Or, 3 + a + 1 = - 1 [m * n = m + n + 1, e = -1]

Or, a = - 1 – 4 = - 5.

So, - 5 is the inverse element of 3,

And let a’ be the inverse element of – 2 under ‘*’.

Then (-2) * a’ = e’

Or, (-2) + a’ + 1 = - 1.

Or, - 1 + a’ = - 1.

So, a ‘ = 0

So, 0 Is the inverse element of – 2.

b) Given an algebraic Structure (G, •) with G = R - {1}, the set of real numbers without the unit number and • stands for the binary operation defined by a•b = a+b-ab for a, b є G.

Solution:

Let e be the identity element of 3 under given binary operation *.

Then, 3 * e = 3.

Or, 3 + e – 3e = 3 [a * b = a + b – ab]

So, e = 0

So, 0 is the identity element of 3.

Again. Let e’ be the identity element of – 2 then,

(-2) * e’ – 3e’ = - 2.

So, e’ = 0

So, 0 is the identity element of – 2.

Let m be the inverse element of 3.

Then, 3 * m = e.

Or, 3 + m – 3m = 0 [a * b = a + b – ab, e = 0]

Or, - 2m = - 3

So, m = $\frac{3}{2}$

So, $\frac{3}{2}$

is the inverse of 3.

And let m’ be the inverse element of – 2 then,

(-2) * m’ = e’

(-2) + m’ – ( - 2)m’ = 0

Or, m’ + 2m’ = 2

Or, 3m’ = 2

So, m’ = $\frac{2}{3}$

So, $\frac{2}{3}$

is the inverse element of – 2.