## Chapter 1: Elementary Group Theory.

### Exercise: 3.4

Complete Exercise of Elementary Group Theory - Exercise 3.3 : Class 12 Mathematics 2080 NEB.

Exercise 3.4 is about: Elementary Proerties of Group

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#### 1. If a and b are the elements of a group (G, *) such that

a) a * b = b, prove that a = e

**Solution:**

a * b = b

or, (a * b) * b-1 = b * b-1

or, a * (b * b-1) = b * b-1

or, a * e = e

or, a = e

b.a * b = e, prove that b = a^-1.

**Solution:**

a * b = e

or, a-1 * (a * b) = a-1 * e

or, (a-1 * a) * b = a-1 * e

or, e * b = a-1.

or, b = a-1.

#### 2. If the group ( G, ㅇ) is commutative show that (a ㅇ b)^-1 = a^-1 ㅇ b^-1, for all a, b ∊ G.

**Solution:**

(a o b) o (a-1 o b-1) = ((a o b) o a-1) o b-1 [by associative law]

= ((b o a) o a-1) o b-1 [by commutative law]

= (b o (a o a-1)) o b-1 [by associative law]

= (b o e) o b-1 [a o a-1 = e, identity element of G]

= b o b-1 [b o e = b]

= e.

Similarly, (a-1 o b-1) o (a o b) = e.

So, a-1 o b-1 is the inverse of a o b.

i.e. (a o b)-1 = a-1 o b-1.

#### 3. Prove that if every element of a group G is its own inverse, then G is abelian.

**Solution:**

Given G is a group such that a = a-1 for all a ԑ G. To prove G is abelian, let a,b ԑ G, then a * b ԑ G, where * is the binary operation of G.

Now, (a * b) = (a * b)-1.

= b-1 * a-1 = b * a.

Thus, a * b = b * a for all a,b ԑ G.

So group G is abelian.

#### 4. If ( G, ㅇ) is a group, then the group equation *x* ㅇ *x* = x has a unique solution *x* = *e*.

**Solution:**

x o x = x

Or, x o x = x o e, where e is the identity element of G.

By left cancellation law, we have,

x = e.

Since, identity element of a group is unique, so, x = e is a unique solution of given group equation.

#### 5. If G is a group such that (ab)^2 = a^2.b^2 for all a, b Є G, prove that G is an abelian group.

**Solution:**

To prove that G is an abelian group, we need to show that for any elements $a$ and $b$ in G, the operation $\ast $ (binary operation in G) is commutative, i.e., $a\ast b=b\ast \mathrm{a;}$.

Given: $(ab{)}^{2}={a}^{2}\cdot {b}^{2}$ for all $a,b\in G$

**Proof:**

Let $a,b\in G$ be arbitrary elements of the group.

From the given condition, we have: $(ab{)}^{2}={a}^{2}\cdot {b}^{2}$

Expanding the left-hand side of the equation: $(ab{)}^{2}=ab\cdot ab={a}^{2}\cdot {b}^{2}$

Since $(ab{)}^{2}={a}^{2}\cdot {b}^{2}$, we can cancel $a$ from the left and $b$ from the right side of the equation:

$ab=ba$

Thus, for any elements $a$ and $b$ in $G$, we have $a\ast b=b\ast a$, which means that $G$ is an abelian group.

### Elementary Group Theory - Exercise 3.4 : Class 12 Math PDF

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