Elementary Group Theory – Exercise 3.2 : Class 12 Math

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:

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:

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

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:

1. The inverse element of 1 is 1 itself because 1 * 1 = 1 (mod 3).
2. 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 = 32

So, 32

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’ = 23

So, 23

is the inverse element of – 2.